Contents

(Course Documents)

(Math 2673 Summer 2004)

(Exam 1)

Exam 1 Math 2673 Summer 2004

Name:

E-mail (optional):

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $\vu = \left\langle2, 1, \tm2\right\rangle,$ $\vv = \left\langle1, 2, \tm1\right\rangle,$ and $\vw = \left\langle\tm3, 0, 1\right\rangle$. Also, let $\theta$ be the angle between $\vv$ and $\vw$.

[2]()

[3] Find $\cos\theta.$

[4] Find $\sin\theta.$

[2]()

[4] Find the area of the parallelogram determined by $\vv$ and $\vw$.

[4] Find the volume of the parallelepiped determined by $\vu,$ $\vv,$ and $\vw$.

[2]()

[3] How many vectors in the $xy$-plane are orthogonal to $\vv?$

[6] Find two vectors $\va$ and $\vb$ with the following three properties:

$\va + \vb = \vw;$

$\va$ and $\vv$ are parallel;

$\vb$ and $\vv$ are orthogonal.

Let $\vf(t) = \left\langle\t3 ,\sin t\right\rangle.$

[4] Sketch the graph of $\vf$ in the direction of increasing $t.$

[4] Find a function $g:\R\to\R$ such that the graph of $g$ coincides with the graph of $\vf.$

Let $\vf(t) = \left\langle\t2 ,\e t;,\sin t\right\rangle.$

[3] Find $\vf'(t).$

[4] Find the equation of the line that is tangent to the graph of $\vf$ where $t = \fric<\pi,3>.$

[3] Calculate $\dintl0,1;\vf(t)dt.$

A projectile is launched due north at an angle of $\fric<\pi,4>$ to the horizontal at an initial speed of $100$ ft/sec. The wind is blowing northeast at a rate of 50 ft/sec.

[2]()

[3] Find the initial velocity vector.

[6] Find the vector functions that describe velocity and motion.

[2]()

[3] Find the maximum height.

[3] Find the horizontal range.

[3] Find the speed of impact.

[4] Give the equation of the plane that contains the points $(1,\tm1,1),$ $(3,0,1),$ and $(\tm2,1,0).$

[4] Do the planes with the equations $2x + y - z = 1$ and $\tm x + 3y + 4z = 3$ intersect? If so, where?

Consider the rectangular equation $\x2 + \y2 + \z2 + 2x - 4y = 4.$

[3] Describe the object defined by the equation.

[3] Write the equation using cylindrical coordinates.

[3] Write the equation using spherical coordinates.

[2]( Calculate the following limits.)

[4] $\limv{(x,y)},{(1,1)};\dfric<\x2 + 2xy + 4y,\x2 y + 1>$

[4] $\limv{(x,y)},{(0,0)};\dfric<xy,\x2 + \y2 >$

[6] Use the $\varepsilon-\delta$ definition of limit to prove that $\limp{(x,y)},{(1,1)};2x + y p = 3.$

Let $f(x,y) = \x2 y - 2x - y.$

[6] Find all local extrema and saddle points.

[6] Find the absolute extrema of $f$ on the region bounded by the triangle with vertices $(0,0),$ $(3,0),$ and $(0,\tm3).$

Exam 1 Math 2673 Summer 2004

Let $\vu = \left\langle2, 1, \tm2\right\rangle,$ $\vv = \left\langle1, 2, \tm1\right\rangle,$ and $\vw = \left\langle\tm3, 0, 1\right\rangle$. Also, let $\theta$ be the angle between $\vv$ and $\vw$.

[2]()

[3] Find $\cos\theta.$

$\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<\tm4,\sqrt{60}>$ $=$ $\fric<\tm2,\sqrt{15}>$

[4] Find $\sin\theta.$

$\sin\theta$ $=$ $\fric<\Vert\vv\times\vw\Vert,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<{\left\Vert\left\langle2,2,6\right\rangle\right\Vert},\sqrt{60}>$ $=$ $\fric<\sqrt{44},\sqrt{60}>$ $=$ $\fric<\sqrt{11},\sqrt{15}>$

[4] Find the area of the parallelogram determined by $\vv$ and $\vw$.

$\Vert\vv\times\vw\Vert$ $=$ $\Vert\left\langle2,2,6\right\rangle\Vert$ $=$ $\sqrt{44}$

[4] Find the volume of the parallelepiped determined by $\vu,$ $\vv,$ and $\vw$.

$2$ $1$ $\tm2$
   
$1$ $2$ $\tm1$
   
$\tm3$ 0 $1$
$=$ $\vert 2(2 - 0) - (1 - 3) - 2(0 + 6)\vert$ $=$ $6$

[3] How many vectors are parallel to the $xy$-plane and orthogonal to $\vv?$

A vector $\vx$ $=$ $\left\langle x,y,0\right\rangle$ which is parallel to the $xy$-plane is orthogonal to $\vv$ if and only if $\vv\cdot\vx$ $=$ $x + 2y$ $=$ 0. Therefore, there are infinitely many vectors that are both parallel to the $xy$-plane and orthogonal to $\vv.$

[6] Find two vectors $\va$ and $\vb$ with the following three properties:

$\va + \vb = \vw;$

$\va$ and $\vv$ are parallel;

$\vb$ and $\vv$ are orthogonal.

Let $\va$ $=$ proj$_{{}_{\vw}}\vv$ $=$ $\left\langle\tm\fric<2,3>,\tm\fric<4,3>,\fric<2,3>\right\rangle$ and $\vb$ $=$ $\vw - \va$ $=$ $\vw -$   proj$_{{}_{\vw}}\vv$ $=$ $\left\langle\tm\fric<7,3>,\fric<4,3>,\fric<1,3>\right\rangle.$

Let $\vf(t) = \left\langle\t3 ,\sin t\right\rangle.$

[4] Sketch the graph of $\vf$ in the direction of increasing $t.$

[4] Find a function $g:\R\to\R$ such that the graph of $g$ coincides with the graph of $\vf.$

$g(x) = \sin\sqrt[3]{x}$

Let $\vf(t) = \left\langle\t2 ,\e t;,\sin t\right\rangle.$

[3] Find $\vf'(t).$

$\vf'(t) = \left\langle2t,\e t;,\cos t\right\rangle.$

[2]( [4] Find the equation of the line that is tangent to the graph of $\vf$ where $t = \fric<\pi,3>.$)

$\vf\left(\fric<\pi,3>\right)$ $=$ $\left\langle\fric<\pi^{2},9>,\e{}^{\fric<\pi,3>};,
\fric<\sqrt{3},2>\right\rangle$

$\vf'\left(\fric<\pi,3>\right)$ $=$ $\left\langle\fric<2\pi,3>,\e{}^{\fric<\pi,3>};,\fric<1,2>\right\rangle$

$x = \fric<2\pi,3>t + \fric<\pi^{2},9>$

$y = \e{}^{\fric<\pi,3>};t + \e{}^{\fric<\pi,3>};$

$z = \fric<1,2>t + \fric<\sqrt{3},2>$

[3] Calculate $\dintl0,1;\vf(t)dt.$

$\dintl0,1;\vf(t)dt$ $=$ $\left\langle\dintl0,1;\t2 dt,\dintl0,1;\e t;dt,
\dintl0,1;\sin t dt\right\rangle$ $=$ $\left\langle\biggml0,1;\fric<1,3>\t3 p,\biggml0,1;\e t;p,
\biggml0,1;\tm\cos t p\right\rangle$ $=$ $\left\langle\fric<1,3>,e - 1, 1 - \cos 1\right\rangle$

A projectile is launched due north at an angle of $\fric<\pi,4>$ to the horizontal at an initial speed of $10$ ft/sec. The wind is blowing northeast at a rate of 5 ft/sec.

The 10 and 5 above were originally 100 and 50. The solution, however, is written for 5 and 10. Rather than change the solution, I changed the problem. Please see me if you have any questions.

[2]()

[3] Find the initial velocity vector.

$\vv(0) = \left\langle0,\tm5\sqrt{2},5\sqrt{2}\right\rangle$

[6] Find the vector functions that describe velocity and motion.

$\va(0) = \left\langle5,\tm5,\tm32\right\rangle$

$\vv(t)$ $=$ $\left\langle5t,\tm5t,\tm32t\right\rangle + \vv(0)$

$=$ $\left\langle5t,\tm5t - 5\sqrt{2},
\tm32t + 5\sqrt{2}\right\rangle$

$\vs(t)$ $=$ $\left\langle\fric<5,2>\t2 ,\tm\fric<5,2>\t2 - 5\sqrt{2}t,
\tm16\t2 + 5\sqrt{2}t\right\rangle + \vs(0)$

$\vs(t)$ $=$ $\left\langle\fric<5,2>\t2 ,\tm\fric<5,2>\t2 - 5\sqrt{2}t,
\tm16\t2 + 5\sqrt{2}t\right\rangle$

[3] Find the maximum height.

$\tm32t + 5\sqrt{2} = 0$

$t = \fric<5\sqrt{2},32>$

$\tm16\fricp<5\sqrt{2},32>2 + 5\sqrt{2}\fricp<5\sqrt{2},32> $

$=$ $\tm\fric<25,32> + \fric<50,32> $

$=$ $\fric<25,32>$

[3] Find the position of impact.

$\tm16\t2 + 5\sqrt{2}t = 0$

$\fric<5\sqrt{2},16>$

$\vs\left(\fric<5\sqrt{2},16>\right)$ $=$ $\left\langle\fric<5,2>\fricp<5\sqrt{2},16>2 ,
\tm\fric<5,2>\fricp<5\sqrt{2},16>...
...\fricp<5\sqrt{2},16>2
+ 5\sqrt{2}\left(\fric<5\sqrt{2},16>\right)\right\rangle$ $=$ $\left\langle\fric<125,256>,\tm\fric<925,256>,0\right\rangle$

[3] Find the speed of impact.

$\left\Vert\vv\left(\fric<5\sqrt{2},16>\right)\right\Vert$ $=$ $\fric<95,8>$

[4] Give the equation of the plane that contains the points $(1,\tm1,1),$ $(3,0,1),$ and $(\tm2,1,0).$

$\vv$ $=$ $\left\langle3 - 1, 0 + 1, 1 - 1\right\rangle$ $=$ $\left\langle2,1,0\right\rangle$

$\vw$ $=$ $\left\langle3 + 2, 0 - 1, 1 - 0\right\rangle$ $=$ $\left\langle5,\tm1,1\right\rangle$

$\vn$ $=$ $\vv\times\vw$ $=$ $\left\langle1,\tm2,\tm7\right\rangle$

$1(x - 1) - 2(y + 1) - 7(z - 1) = 0$

[2]( [4] Do the planes with the equations $2x + y - z = 1$ and $\tm x + 3y + 4z = 3$ intersect? If so, find the line of intersection?)

$2x$ $+$ $y$ = 1
$\tm x$ $+$ $3y$ = 3

$2x$ $+$ $y$ = 1
$\tm2x$ $+$ $6y$ = 6

$7y = 7$

$y = 1$

$x = 0$

$z = 0$

$(0, 1, 0)$

$2x$ $-$ $z$ = 1
$\tm x$ $+$ $4z$ = 3

$2x$ $-$ $z$ = 1
$\tm2x$ $+$ $8z$ = 6

$7z = 7$

$z = 1$

$x = 1$

$y = 0$

$(1, 0, 1)$

$\vv = \left\langle1,\tm1,1\right\rangle$

The line of intersection is given by

$x = t$

$y = \tm t + 1$

$z = t.$

Consider the rectangular equation $\x2 + \y2 + \z2 + 2x - 4y = 4.$

[3] Describe the object defined by the equation.

$\x2 + \y2 + \z2 + 2x - 4y = 4$

$(x + 1)^{2} + (y - 2)^{2} + \z2 = 9$

This is the equation of the sphere with center $(\tm1,2,0)$ and radius 3.

[3] Write the equation using cylindrical coordinates.

$\x2 + \y2 + \z2 + 2x - 4y = 4$

$\r2 \cos^{2}\theta + \r2 \sin^{2}\theta + \z2
+ 2r\cos\theta - 4r\sin\theta = 4$

$\r2 + \z2 + 2r\cos\theta - 4r\sin\theta = 4$

[3] Write the equation using spherical coordinates.

$\x2 + \y2 + \z2 + 2x - 4y = 4$

$\rho^{2} + 2\rho\sin\phi\cos\theta - 4\rho\sin\phi\sin\theta = 4$

[2]( Calculate the following limits.)

[4] $\limv{(x,y)},{(1,1)};\dfric<\x2 + 2xy + 4y,\x2 y + 1>$ $=$ $\fric<7,2>$

[4] $\limv{(x,y)},{(0,0)};\dfric<xy,\x2 + \y2 >$

$\limv{(x,0)},{(0,0)};\dfric<xy,\x2 + \y2 >$ $=$ 0

$\limv{(x,x)},{(0,0)};\dfric<xy,\x2 + \y2 >$ $=$ $\limv{(x,x)},{(0,0)};\dfric<\x2 ,2\x2 >$ $=$ $\fric<1,2>$

$\limv{(x,y)},{(0,0)};\dfric<xy,\x2 + \y2 >$ DNE

[6] Use the $\varepsilon-\delta$ definition of limit to prove that

$\limp{(x,y)},{(1,1)};2x + y p = 3.$

Let $\varepsilon > 0$ and choose $\delta > 0$ such that $\delta < \fric<\varepsilon,3>$. Suppose that $d((x,y),(1,1)) < \delta$. Then

[2]()

$\sqrt{(x - 1)^{2} + (y - 1)^{2}} < \delta$

which implies

$\sqrt{(x - 1)^{2}}$ $=$ $\vert x - 1\vert$ $<$ $\delta$

and

$\sqrt{(y - 1)^{2}}$ $=$ $\vert y - 1\vert$ $<$ $\delta.$

So

$2\vert x - 1\vert + \vert y - 1\vert < 3\delta$

$2\vert x - 1\vert + \vert y - 1\vert < \varepsilon$

$\vert 2(x - 1) + y - 1\vert < \varepsilon$

$\vert 2x - 2 + y - 1\vert < \varepsilon$

$\vert 2x + y - 3\vert < \varepsilon$

as desired.

Let $f(x,y) = \x2 y - 2x - y.$

[2]( [6] Find all local extrema and saddle points.)

$f_{{}_{x}}(x,y) = 2xy - 2$

$f_{{}_{y}}(x,y) = \x2 - 1$

$f_{{}_{xy}}(x,y) = 2x$

$f_{{}_{xx}}(x,y) = 2y$

$f_{{}_{yy}}(x,y) = 0$

CP: $(1,1),$ $(\tm1,\tm1)$

$D(1,1) = \tm4$

$D(\tm1,\tm1) = \tm4$

SP: $(1,1,\tm2),$ $(\tm1,\tm1,2)$

[6] Find the absolute extrema of $f$ on the region bounded by the triangle with vertices $(0,0),$ $(3,0),$ and $(0,\tm3).$

Note that neither critical point is in or on the triangle so we analyze the function on the boundary.

[2]()

$0\leq x\leq3$ and $y = 0$

$f(x,0) = \tm2x$

Max: 0 at $(0,0)$

Min: $\tm6$ at $(3,0)$

$x = 0$ and $\tm3\leq y\leq 0$

$f(0,y) = \tm y$

Max: $3$ at $(0,\tm3)$

Min: 0 at $(0,0)$

$0\leq x\leq3$ and $y = x - 3$

$f(x,x - 3)$

$=$ $\x2 (x - 3) - 2x - (x - 3)$

$=$ $\x3 - 3\x2 - 3x + 3$

$g(x) = \x3 - 3\x2 - 3x + 3$

$g'(x) = 3\x2 - 6x - 3$

$g'(x) = 3(\x2 - 2x - 1)$

$3(\x2 - 2x - 1)$ $=$ 0

$x = \fric<2\pmo\sqrt{8},2>$

$x = 1\pmo\sqrt{2}$

CP: $1 + \sqrt{2}$

$g(0) = 3$

$g(3) = \tm6$

$g\left(1 + \sqrt{2}\right) = \tm2 - 4\sqrt{2}$

Max: $3$ at $(0,\tm3)$

Min: $\tm2 - 4\sqrt{2}$ at $\left(1 + \sqrt{2},\tm2 + \sqrt{2}\right)$

(Exam 2)

Exam 2 Math 2673 Spring 2004

Name:

E-mail (optional):

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

[3]( Define $\vf:\R\to\vVss3 ,$ $\vg:\R\to\vVss3 ,$ and $h:\R\to\R$ as follows:)

$\vf(t) = \left\langle\t2 ,\e t;,t\right\rangle$

$\vg(t) = \left\langle\sqrt{t},\ln t;,t\right\rangle$

$h(t) = \t3 $

[2]( For each of the following, give the rule for the given function and find the domain (if possible).)

[2] $(h\vf)(t)$

[2] $\vg[h(t)]$

[2]()

[2] $(\vf + \vg)(t)$

[2] $\vg[\vf(t)]$

[2]( Differentiate.)

[2] $\vf(t)$

[2] $\vg(t)$

[2]( Integrate.)

[2] $\dintl0,1;\vf(t)dt$

Suppose that $\mathcal{C}$ is a plane curve defined by a vector function $\vr$. Then the binormal vector at the point defined by $t$ is $\vB(t)$ $=$ $\vT(t) \times \vN(t).$

Let $\mathcal{C}$ be the curve defined by ${\vr(t)} = \left<2\sin t, \sqrt{5}t, 2\cos t\right>.$

[3] Find the unit tangent vector $\vT(t).$

[3] Find the unit normal vector $\vN(t).$

[3] Find the binormal vector $\vB(t).$

[3] Find the curvature $\kappa(t).$

A projectile is launched with an initial speed of 100 feet per second at an angle of $\fric<\pi,6>$ to the horizontal. Assume that the only force acting on the object is gravity.

[2] Find the initial velocity vector.

[2] Find the vector function that describes velocity.

[2] Find the vector function that describes motion.

[2] Find the maximum height.

[2] Find the horizontal range.

[2] Find the speed of impact.

[4] Let $\mathcal{C}$ be the curve defined by the vector function $\vr(t) = \left\langle\cos t,\sin t,\sqrt{3}t\right\rangle$. Paramaterize the curve with respect to arc length.

[2]( Define $f:\Rss3 \to\R$ by ${f(x,y,z)} = \sqrt{\x2 + \y2 - z}$.)

[2] Give the domain of $f.$

[2] Calculate $f(2,1,1).$

Define $f:\Rss2 \to\R$ by $f(x,y) = 4\x2 + 9\y2 $.

[2]( Sketch the $c$-level curve for each of the following values of $c.$)

[2] $c = 1$

[2] $c = 36$

[3] Sketch the graph of $f.$

Exam 2 Math 2673 Spring 2004

[3]( Define $\vf:\R\to\vVss3 ,$ $\vg:\R\to\vVss3 ,$ and $h:\R\to\R$ as follows:)

$\vf(t) = \left\langle\t2 ,\e t;,t\right\rangle$

$\vg(t) = \left\langle\sqrt{t},\ln t,t\right\rangle$

$h(t) = \t3 $

[2]( For each of the following, give the rule for the given function and find the domain (if possible).)

[2] $(h\vf)(t)$ $=$ $\left\langle\t5 ,\t3 \e t;,\t4 \right\rangle$

Domain: $\R$

[2] $\vg[h(t)]$ $=$ $\left\langle\sqrt{\t3 },\ln\t3 ,\t3 \right\rangle$

Domain: $(0,\infty)$

[2] $(\vf + \vg)(t)$ $=$ $\left\langle\t2 + \sqrt{t},\e t; + \ln t, 2t\right\rangle$

Domain: $(0,\infty)$

[2] $\vg[\vf(t)]$

This is undefined since the range of $\vf$ consists of vectors and the domain of $\vg$ consists of real numbers.

[2]( Differentiate.)

[2] $\vf(t) = \left\langle\t2 ,\e t;,t\right\rangle$

$\vf'(t) = \left\langle 2t ,\e t;,1\right\rangle$

[2] $\vg(t) = \left\langle\sqrt{t},\ln t,t\right\rangle$

$\vg'(t) = \left\langle\fric<1,2\sqrt{t}>,\fric<1,t>,1\right\rangle$

[2]( Integrate.)

[2] $\dintl0,1;\vf(t)dt$

$\dintl0,1;\left\langle\t2 ,\e t;,t\right\rangle;dt$

$=$ $\left\langle\dintl 0,1;\t2 dt,
\dintl 0,1;\e t;dt,
\dintl 0,1; tdt\right\rangle$

$=$ $\left\langle\biggml 0,1;\fric<1,3>\t3 p,
\biggml 0,1;\e t;p,
\biggml 0,1;\fric<1,2>\t2 p\right\rangle$

$=$ $\left\langle\fric<1,3>, e - 1,\fric<1,2>\right\rangle$

[4] Let $\mathcal{C}$ be the curve defined by the vector function $\vf(t) = \left\langle\cos t,\sin t,\sqrt{3}t\right\rangle$. Paramaterize the curve with respect to arc length.

$\vf'(t) = \left\langle\tm\sin t,\cos t,\sqrt{3}\right\rangle$.

$s$ $=$ $\dintl0,t;\sqrt{(\tm\sin\tau)^{2} + (\cos\tau)^{2} + 3}d\tau$ $=$ $\dintl0,t;2d\tau$ $=$ $\biggml0,t;2\tau p$ $=$ $2t$

$t$ $=$ $\fric<1,2>s$

$\vg(s) = \left\langle\tm\sin\left(\fric<1,2>s\right),
\cos\left(\fric<1,2>s\right),\fric<\sqrt{3},2>s\right\rangle$.

[2]( A projectile is launched with an initial speed of 100 feet per second at an angle of $\fric<\pi,6>$ to the horizontal. Assume that the only force acting on the object is gravity.)

[2] Find the initial velocity vector.


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$\vv_{{}_{0}}$ $=$ $50\sqrt{3}\vi + 50\vj$

[2] Find the vector function that describes velocity.

$\va(t) = \tm32\vj$

$\vv(t) = \tm32t\vj + \vv(0)$

$\vv(t) = \tm32t\vj + 50\sqrt{3}\vi + 50\vj$

$\vv(t) = 50\sqrt{3}\vi + (\tm32t + 50)\vj$

[2] Find the vector function that describes motion.

$\vs(t) = 50\sqrt{3}t\vi + (\tm16\t2 + 50t)\vj + \vs(0)$

$\vs(t) = 50\sqrt{3}t\vi + (\tm16\t2 + 50t)\vj$

[2] Find the maximum height.

$\vv(t) = 50\sqrt{3}\vi + (\tm32t + 50)\vj$

$\tm32t + 50 = 0$

$t = \fric<25,16>$

$\vs(t) = 50\sqrt{3}t\vi + (\tm16\t2 + 50t)\vj$

$\tm16\left(\fric<25,16>\right)^{2}
+ 50\left(\fric<25,16>\right)$ $=$ $\fric<625,16>$

$\fric<625,16>$ ft

[2] Find the horizontal range.

$\vs(t) = 50\sqrt{3}t\vi + (\tm16\t2 + 50t)\vj$

$\tm16\t2 + 50t$ $=$ 0

$t(\tm16t + 50$ $=$ 0

$t$ $=$ $\fric<25,8>$

$50\sqrt{3}\cdot\fric<25,8>$ $=$ $\fric<625\sqrt{3},4>$

[2] Find the speed of impact.

$\vv(t) = 50\sqrt{3}\vi + (\tm32t + 50)\vj$

$\vv\left(\fric<25,8>\right)$ $=$ $50\sqrt{3}\vi + (\tm32\left(\fric<25,8>\right) + 50)\vj$

$\left\Vert\vv\left(\fric<25,8>\right)\right\Vert$ $=$ $100$

100 ft/sec.

[2]( Define $f:\Rss3 \to\R$ by ${f(x,y,z)} = \sqrt{\x2 + \y2 - z}$.)

[2] Give the domain of $f.$

$\{(x,y,z)\in\Rss3 : \x2 + \y2 > z\}$

[2] Calculate $f(2,1,1).$

$f(2,1,1)$ $=$ $2$

Let $\mathcal{C}$ be the curve defined by ${\vr(t)} = \left<2\sin t, \sqrt{5}t, 2\cos t\right>.$

[2]()

[3] Find the unit tangent vector $\vT(t).$

$\vr'(t) = \left<2\cos t, \sqrt{5}, \tm2\sin t\right>.$

$\Vert\vr'(t)\Vert$ $=$ $\sqrt{4\cos^{2}t,5,4\sin^{2}t}$ $=$ $3$

$\vT(t)$

$=$ $\dfric<\vr'(t),\Vert\vr'(t)\Vert>$

$=$ $\fric<1,3>\left<2\cos t,\sqrt{5},\tm2\sin t\right>$

[3] Find the unit normal vector $\vN(t).$

$\vT'(t)$ $=$ $\fric<1,3>\left<\tm2\sin t, 0, \tm2\cos t\right>$

$\Vert\vT'(t)\Vert$ $=$ $\fric<2,3>$

$\vN(t)$ $=$ $\fric<3,2>\vT'(t)$ $=$ $\left<\tm\sin t, 0, \tm\cos t\right>$

[3] Find the binormal vector $\vB(t).$

$\vB(t)$

$=$ $\vT(t) \times \vN(t)$

$=$ $\fric<1,3>\left<\tm\sqrt{5}\cos t, 2,\sqrt{5}\sin t \right>$

[3] Find the curvature $\kappa(t).$

$\kappa(t)$ $=$ $\fric<\Vert\vT'(t)\Vert,\Vert\vr'(t)\Vert>$ $=$ $\dfric<{\fric<2,3>},3>$ $=$ $\fric<2,9>$

Define $f:\Rss2 \to\R$ by $f(x,y) = 4\x2 + 9\y2 $.

[2]( Sketch the $c$-level curve for each of the following values of $c.$)

[2] $c = 1$

$4\x2 + 9\y2 $ $=$ $1$

$\dfrac{\x2 }{\fricp<1,2>2 } + \dfrac{\y2 }{\fricp<1,3>2 }$ $=$ $1$


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[2] $c = 36$

$4\x2 + 9\y2 $ $=$ $36$

$\fric<\x2 ,9> + \fric<\y2 ,4>$ $=$ $1$


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\begin{picture}(0,0)
\axesfivebyfive
\p...
...h=\figurewidth,linecolor=figurecolor]%
(0,0)(60,40)
\end{picture}}
\end{picture}

[3] Sketch the graph of $f.$

Total Points:

(Exam 3)

Exam 3 Math 2673 Spring 2004

Name:

E-mail (optional):

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

For each of the following, give the coordinates of the point in the the other two coordinate systems.

[2] Rectangular coordinates: $\left(\fric<\sqrt{2},2>,\fric<\sqrt{2},2>,\sqrt{3}\right)$

[2] Cylindrical coordinates: $\left(\fric<\sqrt{3},2>,\fric<\pi,6>,\fric<1,2>\right)$

[2] Spherical coordinates: $\left(3,\fric<\pi,3>,\fric<\pi,2>\right)$

For each of the following, give the equation of the described object in rectangular, cylindrical, and spherical coordinates.

[6] The sphere centered at ${(0,0,0)}$ with radius $1.$

[6] The cylinder whose intersection with the $xy$-plane is the circle centered at ${(0,0)}$ with radius 3.

[2]( For each of the following, calculate the limit or show that it does not exist.)

[4] $\limv{(x,y)},(0,0);\dfric<\x2 y + 3,x + y + 1>$

[4] $\limv{(x,y)},(0,0);\dfric<\x2 y,\x4 + \y2 >$

Let $f(x,y) = \x2 y - y.$

[2]( [5] Find all local extrema and saddle points.)

[2]( [5] Find absolute extrema of $f$ on the region enclosed by the triangle with vertices $(\tm1,0),$ $(0,0),$ and $(0,1).$)

[6] Use the definition of limit to show that

$\limp{(x,y)},(1,\tm2);2x + y + 1p = 1.$

[8] Prove that $f(x,y) = 2x + \y2 + 1$ is differentiable for all $(a,b)\in\Rss2 $ and find $f'_{{}_{(a,b)}}(x,y).$

Exam 3 Math 2673 Spring 2004

[2]( For each of the following, give the coordinates of the point in the the other two coordinate systems.)

[2]

Rectangular coordinates: $\left(\fric<\sqrt{2},2>,\fric<\sqrt{2},2>,\sqrt{3}\right)$

$\r2 $ $=$ $\x2 + \y2 $ $=$ $\fric<1,2> + \fric<1,2>$ $=$ $1$

$r = 1$

$\tan\theta$ $=$ $\fric<x,y>$ $=$ $1$

$\theta$ $=$ $\fric<\pi,4>$

$\rho^{2}$ $=$ $\r2 + \z2 $ $=$ $1 + 3$ $=$ $4$

$\rho = 2$

$r = \rho\sin\phi$

$1 = 2\sin\phi$

$\sin\phi = \fric<1,2>$

$\phi = \fric<\pi,6>$

Cylindrical: $\left(1,\fric<\pi,4>,\sqrt{3}\right)$

Spherical: $\left(2,\fric<\pi,4>,\fric<\pi,6>\right)$

[2]

Cylindrical coordinates: $\left(\fric<\sqrt{3},2>,\fric<\pi,6>,\fric<1,2>\right)$

$x$ $=$ $r\cos\theta$ $=$ $\fric<\sqrt{3},2>\cos\fric<\pi,6>$ $=$ $\fric<3,4>$

$y$ $=$ $r\sin\theta$ $=$ $\fric<\sqrt{3},2>\sin\fric<\pi,6>$ $=$ $\fric<\sqrt{3},4>$

$\rho^{2}$ $=$ $\r2 + \z2 $ $=$ $\fric<3,4> + \fric<1,4>$ $=$ $1$

$\rho$ $=$ $1$

$r$ $=$ $\rho\sin\phi$

$\fric<\sqrt{3},2>$ $=$ $\sin\phi$

$\phi$ $=$ $\fric<\pi,3>$

Rectangular: $\left(\fric<3,4>,\fric<\sqrt{3},4>,\fric<1,2>\right)$

Spherical: $\left(1,\fric<\pi,6>,\fric<\pi,3>\right)$

[2] Spherical coordinates:

$\left(3,\fric<\pi,3>,\fric<\pi,2>\right)$

$x$ $=$ $\rho\sin\phi\cos\theta$ $=$ $3\sin\fric<\pi,2>\cos\fric<\pi,3>$ $=$ $\fric<3,2>$

$y$ $=$ $\rho\sin\phi\sin\theta$ $=$ $3\sin\fric<\pi,2>\sin\fric<\pi,3>$ $=$ $\fric<3\sqrt{3},2>$

$z$ $=$ $\rho\cos\phi$ $=$ $3\cos\fric<\pi,2>$ $=$ 0

$r$ $=$ $\rho\sin\phi$ $=$ $3\sin\fric<\pi,2>$ $=$ $3$

Rectangular: $\left(\fric<3,2>,\fric<3\sqrt{3},2>,0\right)$

Cylindrical: $\left(3,\fric<\pi,3>,0\right)$

For each of the following, give the equation of the described object in rectangular, cylindrical, and spherical coordinates.

[3]( [6] The sphere centered at ${(0,0,0)}$ with radius $1.$)

\(\text{%
Rectangular:
\(\x2 + \y2 + \z2 = 1\)}\)

Cylindrical: $\r2 + \z2 = 1$

Spherical: $\rho = 1$

[3]( [6] The cylinder whose intersection with the $xy$-plane is the circle centered at ${(0,0)}$ with radius 3.)

\(\text{%
Rectangular:
\(\x2 + \y2 = 9\)}\)

Cylindrical: $r = 3$

Spherical: $\rho\sin\phi = 3$

[2]( For each of the following, calculate the limit or show that it does not exist.)

[4] $\limv{(x,y)},(0,0);\dfric<\x2 y + 3,x + y + 1>$ $=$ $3$

[4] $\limv{(x,y)},(0,0);\dfric<\x2 y,\x4 + \y2 >$

$\limv{(x,0)},(0,0);\dfric<\x2 y,\x4 + \y2 >$ $=$ $\limv{(x,0)},(0,0);\dfric<0,\x4 >$ $=$ 0

$\limv{(\sqrt{y},y)},(0,0);\dfric<\x2 y,\x4 + \y2 >$ $=$ $\limv{(\sqrt{y},y)},(0,0);\dfric<\y2 ,2\y2 >$ $=$ $\fric<1,2>$

$\limv{(x,y)},(0,0);\dfric<\x2 y,\x4 + \y2 >$ DNE

[6] Use the definition of limit to show that

$\limp{(x,y)},(1,\tm2);2x + y + 1p = 1.$

Let $\varepsilon > 0$ and choose $\delta$ such that $0 < \delta < \fric<\varepsilon,3>$. Suppose that $d((x,y),(1,\tm2)) < \delta$. Then

[2]()

$\sqrt{(x - 1)^{2} + (y + 2)^{2}} < \delta$

and so

$\sqrt{(x - 1)^{2}}$ $=$ $\vert x - 1\vert$ $<$ $\delta$

and

$\sqrt{(y + 2)^{2}}$ $=$ $\vert y + 2\vert$ $<$ $\delta.$

Hence,

$2\vert x - 1\vert + \vert y + 2\vert$ $<$ $3\delta$

$\vert 2(x - 1) + (y + 2)\vert$ $<$ $3\delta$

$\vert 2x - 2 + y + 2\vert$ $<$ $\varepsilon$

$\vert 2x + y\vert$ $<$ $\varepsilon$

$\vert(2x + y + 1) - 1\vert$ $<$ $\varepsilon$

as desired.

Let $f(x,y) = \x2 y - y.$

[2]( [5] Find all local extrema and saddle points.)

$f_{{}_{x}}(x,y) = 2xy$

$f_{{}_{y}}(x,y) = \x2 - 1$

$f_{{}_{xx}}(x,y) = 2y$

$f_{{}_{yy}}(x,y) = 0$

$f_{{}_{xy}}(x,y) = 2x$

Critical Points: $(\tm1,0),$ $(1,0)$

$D(\tm1,0)$ $=$ $\tm4$

$D(1,0)$ $=$ $\tm4$

Saddle Points: $(\tm1,0,0),$ $(1,0,0)$

[5] Find absolute extrema of $f$ on the region enclosed by the triangle with vertices $(\tm1,0),$ $(0,0),$ and $(0,1).$

Note that since one critical point of $f$ is on the boundary of the triangle and the other is not in or on the triangle, we need only analyze $f$ on the boundary of the triangle. Find the maximum and minimum values of

$g(x)$ $=$ $f(x,0)$ on $[\tm1,0];$

$g(y)$ $=$ $f(0,y)$ on $[0,1];$

$g(x)$ $=$ $f(x,x + 1)$ on $[\tm1,0].$

Note that $g(x)$ $=$ $f(x,0)$ $=$ 0 for all $x\in[\tm1,0].$

Note that $g(y)$ $=$ $f(0,y)$ $=$ $\tm y$ for all $y\in[0,1].$

$g(x)$ $=$ $f(x,x + 1)$ $=$ $\x2 (x + 1) - (x + 1)$ $=$ $\x3 + \x2 - x - 1$

$g'(x)$ $=$ $3\x2 + 2x - 1$ $=$ $(3x - 1)(x + 1)$

Critical Numbers: $\fric<1,3>\notin[\tm1,0],$ $\tm1\in[\tm1,0]$

$g(\tm1)$ $=$ 0

Absolute Max: 0 at $(x,0)$ for all $x\in[\tm1,0]$

Absolute Min: $\tm1$ at $(0,1)$

[8] Prove that $f(x,y) = 2x + \y2 + 1$ is differentiable for all $(a,b)\in\Rss2 $ and find $f'_{{}_{(a,b)}}(x,y).$

Note that

$f_{{}_{x}}(x,y) = 2$

and

$f_{{}_{y}}(x,y) = 2y.$

Consider

$\limv{(h,k)},(0,0);
\dfric<{f(a + h,b + k) - f(a,b)
- \left(f_{{}_{x}}(a,b)h + f_{{}_{y}}(a,b)k\right)},\Vert(h,k)\Vert>$

$=$ $\limv{(h,k)},(0,0);
\dfric<2(a + h) + (b + k)^{2} + 1 - (2a + \b2 + 1)
- (2h + 2bk),\sqrt{\h2 + \k2 }>$

$=$ $\limv{(h,k)},(0,0);
\dfric<2a + 2h + \b2 + 2bk + \k2 + 1 - 2a - \b2 - 1 - 2h - 2bk,
\sqrt{\h2 + \k2 }>$

$=$ $\limv{(h,k)},(0,0);\dfric<\k2 ,\sqrt{\h2 + \k2 }>$

$\leq$ $\limv{(h,k)},(0,0);\dfric<\k2 ,\sqrt{\k2 }>$

$\leq$ $\limv{(h,k)},(0,0);\dfric<\k2 ,\vert k\vert>$

$=$ $\limv{(h,k)},(0,0);\vert k\vert$

$=$ 0

as desired.

$f'_{{}_{(a,b)}}(x,y)$ $=$ $2x + 2by$

(Exam 34)

Exam 1 Math 2673 Spring 2004

Name:

E-mail (optional):

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

[2]( For each of the following, describe the region in space being defined.)

[4] $\tm6 \leq \x2 - 2x + \y2 + 6y + \z2 \leq \tm1$

[4] $\y2 + \z2 = 9$

[2]( Let $\vv = \left\langle1,\tm1,4\right\rangle$ and $\vw = \left\langle2,2,0\right\rangle$. Find the following.)

[2] $2\vv - \vw$ $=$

[2] $\vv\cdot\vw$ $=$

[2]()

[3] $\Vert\vv\Vert$ $=$

[3] Find the angle between $\vv$ and $\vw.$

[2]( Let $\vv = \left\langle1,1,1\right\rangle$ and $\vw = \left\langle1,1,0\right\rangle$.)

[3] comp$_{{}_{\vv}}\vw$ $=$

[3] proj$_{{}_{\vv}}\vw$ $=$

[3] $\vv\times\vw$ $=$

[4] Find the area of the parallelogram determined by $\left\langle1,2,3\right\rangle$ and $\left\langle\tm2,4,1\right\rangle.$

[4] Find the volume of the parallelpiped determined by $\left\langle\tm3,0,5\right\rangle,$ $\left\langle1,2,3\right\rangle,$ and $\left\langle\tm2,4,1\right\rangle.$

[4] Find the equation of the plane that contains the points $(1,\tm1,\tm1),$ $(2,0,1),$ and $(4,\tm2,3).$

[3]( [3] Find equations of the line containing the point ${(1,\tm1,0)}$ in the direction of $\left\langle\tm3,1,1\right\rangle$.)

[4] Does the line with parametric equations $x = 4t + 1,$ $y = 2t - 3,$ and $z = t + 2$ intersect the plane with equation $2(x - 5) + y + 1 + 6(z - 3) = 0$? If so, where?

[5] Find an equation of the line where the following planes intersect.

$\phantom{\tm}2x - y + 3z = 1$

$\tm3x + y + 5z = 7$

Exam 1 Math 2673 Spring 2004

For each of the following, describe the region in space being defined.

[4] $\tm6 \leq \x2 - 2x + \y2 + 6y + \z2 \leq \tm1$

$\tm6 \leq (x - 1)^{2} - 1 + (y + 3)^{2} - 9 + \z2 \leq \tm1$

$4 \leq (x - 1)^{2} + (y + 3)^{2} + \z2 \leq 9$

This inequality describes the set of all points on or in the sphere centered at $(1,\tm3,0)$ with radius 3 that are not in the sphere centered at $(1,\tm3,0)$ with radius 2.

[4] $\y2 + \z2 = 9$

This is the circular cylinder with radius 3 and having the $x$-axis as its central axis.

Let $\vv = \left\langle1,\tm1,4\right\rangle$ and $\vw = \left\langle2,2,0\right\rangle$. Find the following.

[2] $2\vv - \vw$ $=$ $\left\langle2 - 2,\tm2 - 2,8 - 0\right\rangle$ $=$ $\left\langle0,\tm4,8\right\rangle$

[2] $\vv\cdot\vw$ $=$ $2 - 2 + 0$ $=$ 0

[3] $\Vert\vv\Vert$ $=$ $\sqrt{1 + 1 + 16}$ $=$ $\sqrt{18}$ $=$ $3\sqrt{2}$

[3] Find the angle between $\vv$ and $\vw.$

Let $\theta$ be the angle between and $\vv$ and $\vw$. Since $\vv\cdot\vw$ $=$ $0,$ $\vv$ and $\vw$ are orthogonal.

Let $\vv = \left\langle1,1,1\right\rangle$ and $\vw = \left\langle1,1,0\right\rangle$.

[3] comp$_{{}_{\vv}}\vw$ $=$ $\dfric<\vv\cdot\vw,\Vert\vv\Vert>$ $=$ $\dfric<2,\sqrt{3}>$

[3] proj$_{{}_{\vv}}\vw$ $=$ $\dfricp<\vv\cdot\vw,\Vert\vv\Vert> \dfric<\vv,\Vert\vv\Vert>$ $=$ $\left\langle\fric<2,3>,\fric<2,3>,\fric<2,3>\right\rangle$

[3] $\vv\times\vw$ $=$ $\left\langle\tm1,1,0\right\rangle$

[4] Find the area of the parallelogram determined by $\left\langle1,2,3\right\rangle$ and $\left\langle\tm2,4,1\right\rangle.$

$\left\Vert\left\langle1,2,3\right\rangle
\times
\left\langle\tm2,4,1\right\rangle\right\Vert$ $=$ $\left\Vert\left\langle\tm10,\tm7,8\right\rangle\right\Vert$ $=$ $\sqrt{213}$

[4] Find the volume of the parallelpiped determined by $\left\langle\tm3,0,5\right\rangle,$ $\left\langle1,2,3\right\rangle,$ and $\left\langle\tm2,4,1\right\rangle.$

$\left\vert
\left\langle\tm3,0,5\right\rangle
\cdot
\left(\left\langle1,2,3\right\rangle
\times
\left\langle\tm2,4,1\right\rangle\right)\right\vert$ $=$ $\left\vert
\left\langle\tm3,0,5\right\rangle
\cdot
\left\langle\tm10,\tm7,8\right\rangle\right\vert$ $=$ $70$

[4] Find the equation of the plane that contains the points $(1,\tm1,\tm1),$ $(2,0,1),$ and $(4,\tm2,3).$

Let $\vv = \left\langle1, 1, 2\right\rangle,$ $\vw = \left\langle3, \tm1, 4\right\rangle,$ and $\vn$ $=$ $\vv\times\vw$ $=$ $\left\langle6,2,\tm4\right\rangle$. The equation of the plane with normal vector $\left\langle6,2,\tm4\right\rangle$ and containing the point $(1,\tm1,\tm1)$ is $6(x - 1) + 2(y + 1) - 4(z + 1) = 0.$

[3]( [3] Find equations of the line containing the point ${(1,\tm1,0)}$ in the direction of $\left\langle\tm3,1,1\right\rangle$.)

$x = \tm3t + 1$

$y = t - 1$

$z = t$

[4] Does the line with parametric equations $x = 4t + 1,$ $y = 2t - 3,$ and $z = t + 2$ intersect the plane with equation $2(x - 5) + y + 1 + 6(z - 3) = 0$? If so, where?

$2[(4t + 1) - 5] + 2t - 3 + 1 + 6[(t + 2) - 3] = 0$

$16t - 16 = 0$

$t = 1$

The plane and the line intersect at the point $(5, \tm1, 3)$.

[2]( [5] Find an equation of the line where the following planes intersect.)

$\phantom{\tm}2x - y + 3z = 1$

$\tm3x + y + 5z = 7$

Let $z = 0$.

$2x$ $-$ $y$ = 1
$\tm3x$ $+$ $y$ = 7

$\tm x = 8$

$x = \tm8$

$y = \tm17$

So the point $(\tm8,\tm17,0)$ is on the line of intersection of the two planes.

Normal vectors of the two planes are $\vn_{{}_{1}} = \langle2,\tm1,3\rangle$ and $\vn_{{}_{2}} = \langle\tm3,1,5\rangle$. So a vector parallel to the line of intersection is $\vn_{{}_{1}} \times \vn_{{}_{2}}
= \langle \tm8,\tm19,\tm1\rangle$. Equations of the line passing through the point $(\tm8,\tm17,0)$ with parallel vector $\langle \tm8,\tm19,\tm1\rangle$ are

$x = \tm8 - 8t$

$y = \tm17 - 19t$

$z = \tm t.$

(Final)

Final Exam Math 2673 Summer 2004

Name:

E-mail (optional):

Directions: Show all of your work in the space provided and justify all of your answers. You may use written materials but you are not permitted to ask another person for help (except for the instructor).

[2] Find a vector-valued function whose graph is the line connecting the points

$(1,\tm1,2)$ and $(5,1,6).$ $(1,0,2)$ and $(1,2,6).$ $(2,\tm3,1)$ and $(0,3,5).$ $(5,4,\tm2)$ and $(1,\tm2,3).$ $(2,3,2)$ and $(\tm1,2,5).$ $(2,0,8)$ and $(0,2,7).$ $(3,\tm5,3)$ and $(2,1,6).$ $(2,\tm3,1)$ and $(\tm1,\tm2,\tm5).$ $(\tm1,1,\tm2)$ and $(3,3,9).$ $(2,2,6)$ and $(5,1,4).$ $(0,3,9)$ and $(0,\tm6,2).$ $(8,8,7)$ and $(7,5,4).$ $(3,3,3)$ and $(\tm1,0,6).$

[2] Paramaterize the curve by arc length.

[2]( [2] For each of the following, determine whether or not the lines are coplanar. If so, give the equation of the plane. If not, explain why they are not.)

$x$ $=$ $t$ $+$ 1
$y$ $=$ $t$ $-$ 2
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $3s$ $+$ 1
$z$ $=$ $s$ $+$ 2

$x$ $=$ $2t$ $-$ 3
$y$ $=$ $\tm t$ $-$ 2
$z$ $=$ $2t$ $+$ 3

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $s$ $-$ 3
$z$ $=$ $s$ $+$ 2

$x$ $=$ $\tm2t$ $+$ 1
$y$ $=$ $3t$    
$z$ $=$ $t$ $-$ 4

$x$ $=$ $3s$ $+$ 1
$y$ $=$ $\tm s$ $-$ 3
$z$ $=$ $2s$ $-$ 1

$x$ $=$ $\tm t$ $-$ 2
$y$ $=$ $2t$ $+$ 6
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $-$ 2
$y$ $=$ $\tm s$    
$z$ $=$ $\tm s$ $+$ 3

$x$ $=$ $t$ $+$ 1
$y$ $=$ $t$ $-$ 2
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $3s$ $+$ 1
$z$ $=$ $s$ $+$ 2

$x$ $=$ $2t$ $-$ 3
$y$ $=$ $\tm t$ $-$ 2
$z$ $=$ $2t$ $+$ 3

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $s$ $-$ 3
$z$ $=$ $s$ $+$ 2

$x$ $=$ $\tm2t$ $+$ 1
$y$ $=$ $3t$    
$z$ $=$ $t$ $-$ 4

$x$ $=$ $3s$ $+$ 1
$y$ $=$ $\tm s$ $-$ 3
$z$ $=$ $2s$ $-$ 1

$x$ $=$ $\tm t$ $-$ 2
$y$ $=$ $2t$ $+$ 6
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $-$ 2
$y$ $=$ $\tm s$    
$z$ $=$ $\tm s$ $+$ 3

$x$ $=$ $t$ $+$ 1
$y$ $=$ $t$ $-$ 2
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $3s$ $+$ 1
$z$ $=$ $s$ $+$ 2

$x$ $=$ $2t$ $-$ 3
$y$ $=$ $\tm t$ $-$ 2
$z$ $=$ $2t$ $+$ 3

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $s$ $-$ 3
$z$ $=$ $s$ $+$ 2

$x$ $=$ $\tm2t$ $+$ 1
$y$ $=$ $3t$    
$z$ $=$ $t$ $-$ 4

$x$ $=$ $3s$ $+$ 1
$y$ $=$ $\tm s$ $-$ 3
$z$ $=$ $2s$ $-$ 1

$x$ $=$ $\tm t$ $-$ 2
$y$ $=$ $2t$ $+$ 6
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $-$ 2
$y$ $=$ $\tm s$    
$z$ $=$ $\tm s$ $+$ 3

$x$ $=$ $t$ $+$ 1
$y$ $=$ $t$ $-$ 2
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $3s$ $+$ 1
$z$ $=$ $s$ $+$ 2

[2] Do the following planes intersect? If so, find the line of intersection.

$2x$ $+$ $3y$ $-$ $z$ $=$ $7$
$3x$ $-$ $8y$ $+$ $2z$ $=$ $11$

$2x$ $+$ $2y$ $-$ $z$ $=$ $7$
$2x$ $-$ $8y$ $+$ $2z$ $=$ $10$

$x$ $+$ $y$ $-$ $z$ $=$ $6$
$2x$ $+$ $3y$ $+$ $z$ $=$ $8$

$2x$ $-$ $y$ $+$ $z$ $=$ $\tm7$
$3x$ $+$ $3y$ $-$ $z$ $=$ $6$

$x$ $+$ $y$ $+$ $3z$ $=$ $\tm1$
$x$ $+$ $2y$ $-$ $2z$ $=$ $8$

$3x$ $-$ $y$ $-$ $2z$ $=$ $\tm3$
$x$ $+$ $y$ $+$ $z$ $=$ $7$

$2x$ $+$ $3y$ $-$ $z$ $=$ $7$
$3x$ $-$ $8y$ $+$ $2z$ $=$ $11$

$2x$ $+$ $3y$ $-$ $z$ $=$ $7$
$3x$ $-$ $8y$ $+$ $2z$ $=$ $11$

$2x$ $+$ $2y$ $-$ $z$ $=$ $7$
$2x$ $-$ $8y$ $+$ $2z$ $=$ $10$

$x$ $+$ $y$ $-$ $z$ $=$ $6$
$2x$ $+$ $3y$ $+$ $z$ $=$ $8$

$2x$ $-$ $y$ $+$ $z$ $=$ $\tm7$
$3x$ $+$ $3y$ $-$ $z$ $=$ $6$

$x$ $+$ $y$ $+$ $3z$ $=$ $\tm1$
$x$ $+$ $2y$ $-$ $2z$ $=$ $8$

$3x$ $-$ $y$ $-$ $2z$ $=$ $\tm3$
$x$ $+$ $y$ $+$ $z$ $=$ $7$

[2]( For each of the following, sketch the graph of the function and use arrows to indicate the direction of increasing $t.$)

[2]

$\vr(t) = \left\langle\e t;,t\right\rangle$

$\vr(t) = \left\langle t,\e t;\right\rangle$

$\vr(t) = \left\langle t,\sin t\right\rangle$

$\vr(t) = \left\langle \sin t,t\right\rangle$

$\vr(t) = \left\langle \cos t,t\right\rangle$

$\vr(t) = \left\langle t,\cos t\right\rangle$

$\vr(t) = \left\langle t,\ln t\right\rangle$

$\vr(t) = \left\langle \ln t,t\right\rangle$

$\vr(t) = \left\langle t,\tan t\right\rangle$

$\vr(t) = \left\langle \tan t,t\right\rangle$

$\vr(t) = \left\langle \sin t,t\right\rangle$

$\vr(t) = \left\langle \ln t,t\right\rangle$

$\vr(t) = \left\langle t,\sin t\right\rangle$

[2]

$\vr(t) = \left\langle 3\cos t,2\sin t,t\right\rangle$

$\vr(t) = \left\langle 2\cos t,3\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm3\cos t,2\sin t,t\right\rangle$

$\vr(t) = \left\langle 3\cos t,\tm2\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm3\cos t,\tm2\sin t,t\right\rangle$

$\vr(t) = \left\langle =\tm2\cos t,3\sin t,t\right\rangle$

$\vr(t) = \left\langle 2\cos t,\tm3\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm2\cos t,\tm3\sin t,t\right\rangle$

$\vr(t) = \left\langle 3\cos t,2\sin t,t\right\rangle$

$\vr(t) = \left\langle 2\cos t,3\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm2\cos t,3\sin t,t\right\rangle$

$\vr(t) = \left\langle 2\cos t,\tm3\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm2\cos t,\tm3\sin t,t\right\rangle$

[4] Let $\vf(t) = \left\langle\e t;\cos t,\sqrt{\t2 + 1}\right\rangle.$

$\vf(t) = \left\langle\e t;\sin t,\sqrt{\t2 + 1}\right\rangle.$

$\vf(t) = \left\langle\e t;\cos t,\sqrt{\t2 - 1}\right\rangle.$

$\vf(t) = \left\langle\e t;\sin t,\sqrt{\t2 - 1}\right\rangle.$

$\vf(t) = \left\langle\e t;\cos t,\sqrt{1 - \t2 }\right\rangle.$

$\vf(t) = \left\langle\e t;\sin t,\sqrt{1 - \t2 }\right\rangle.$

$\vf(t) = \left\langle\ln t,\sqrt{\t2 + 1}\right\rangle.$

$\vf(t) = \left\langle\ln t,\sqrt{\t2 - 1}\right\rangle.$

$\vf(t) = \left\langle\ln t,\sqrt{1 - \t2 }\right\rangle.$

$\vf(t) = \left\langle\sec^{3}t,\sqrt{\t2 + 1}\right\rangle.$

$\vf(t) = \left\langle\sec^{3}t,\sqrt{\t2 - 1}\right\rangle.$

$\vf(t) = \left\langle\sec^{3}t,\sqrt{1 - \t2 }\right\rangle.$

$\vf(t) = \left\langle\e t;\cos t,\sqrt{\t2 + 1}\right\rangle.$

$\dintl1,2;\vf(t)dt$ $=$

[2] Sketch the graph of $f(x,y) = \e x;.$

$f(x,y) = \e y;.$

$f(x,y) = \sin x.$

$f(x,y) = \sin y.$

$f(x,y) = \cos x.$

$f(x,y) = \cos y.$

$f(x,y) = \ln x.$

$f(x,y) = \ln y.$

$f(x,y) = \x2 .$

$f(x,y) = \y2 .$

$f(x,y) = x + y.$

$f(x,y) = x - y.$

$f(x,y) = y - x.$

[2] Find the following limit or show that it does not exist.

$\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$

[2] Use the $\varepsilon-\delta$ definition of limit to prove $\limp{(x,y)},{(1,1)};3x + 2y p = 5.$ $\limp{(x,y)},{(2,1)};3x + 2y p = 8.$ $\limp{(x,y)},{(\tm1,1)};3x + 2y p = \tm1.$ $\limp{(x,y)},{(1,\tm1)};3x + 2y p = 1.$ $\limp{(x,y)},{(2,2)};3x + 2y p = 10.$ $\limp{(x,y)},{(0,1)};3x + 2y p = 2.$ $\limp{(x,y)},{(1,0)};3x + 2y p = 3.$ $\limp{(x,y)},{(2,1)};2x + 3y p = 7.$ $\limp{(x,y)},{(2,1)};x + 3y p = 5.$ $\limp{(x,y)},{(1,1)};5x - 2y p = 3.$ $\limp{(x,y)},{(\tm2,1)};3x + 2y p = \tm4.$ $\limp{(x,y)},{(0,1)};2x + 3y p = 3.$ $\limp{(x,y)},{(2,\tm1)};3x + 2y p = 4.$

[3] Use the $\varepsilon-\delta$ definition of limit to prove $\limv{(x,y)},{(1,1)};xy = 1.$ $\limv{(x,y)},{(2,1)};xy = 2.$ $\limv{(x,y)},{(1,2)};xy = 2.$ $\limv{(x,y)},{(2,2)};xy = 4.$ $\limv{(x,y)},{(3,1)};xy = 3.$ $\limv{(x,y)},{(1,3)};xy = 3.$ $\limv{(x,y)},{(2,3)};xy = 6.$ $\limv{(x,y)},{(3,2)};xy = 6.$ $\limv{(x,y)},{(4,1)};xy = 4.$ $\limv{(x,y)},{(1,4)};xy = 4.$ $\limv{(x,y)},{(2,4)};xy = 8.$ $\limv{(x,y)},{(4,2)};xy = 8.$ $\limv{(x,y)},{(4,3)};xy = 12.$

[3] Let $f(x,y,z) = \x2 y + 4y + 3\z2 $. Show that $f$ is differentiable at $(a,b,c)$ for all $(a,b,c)\in\Rss3 $ and find $f'_{{}_{(a,b,c)}}(x,y,z).$

Let $f(x,y) = 3\x2 y + 3xy + \y2 .$ $f(x,y) = \x2 y + xy + \y2 .$ $f(x,y) = 2\x2 y + 2xy + \y2 .$ $f(x,y) = 4\x2 y + 4xy + \y2 .$ $f(x,y) = 5\x2 y + 5xy + \y2 .$ $f(x,y) = 3\x2 y + 3xy + \y2 .$ $f(x,y) = \x2 y + xy + \y2 .$ $f(x,y) = 2\x2 y + 2xy + \y2 .$ $f(x,y) = 4\x2 y + 4xy + \y2 .$ $f(x,y) = 5\x2 y + 5xy + \y2 .$ $f(x,y) = 3\x2 y + 3xy + \y2 .$ $f(x,y) = \x2 y + xy + \y2 .$ $f(x,y) = 2\x2 y + 2xy + \y2 .$

[3] Find all local extrema and saddle points of $f.$

[2] Find all extrema of $f$ on the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1).$

[2] Let $R = [1,2]\times[0,1]$ $R = [1,2]\times[1,1]$ $R = [\tm1,1]\times[1,2]$ $R = [0,1]\times[1,2]$ $R = [0,1]\times[1,1]$ $R = [\tm2,0]\times[0,1]$ $R = [\tm2,1]\times[0,1]$ $R = [\tm2,1]\times[1,3]$ $R = [0,1]\times[2,3]$ $R = [\tm1,0]\times[1,2]$ $R = [1,2]\times[0,1]$ $R = [1,2]\times[1,1]$ $R = [\tm1,1]\times[1,2]$ and calculate $\diintlp R;\x2 y + \y3 dA.$

[2] Let $C$ be the circle centered at $(0,0)$ with radius $1$ $\sqrt{2}$ $2$ $\sqrt{3}$ $3$ $4$ $5$ $\sqrt{5}$ $6$ $7$ $\sqrt{7}$ $8$ $9$ and calculate $\diintlp C;x\e\x2 + \y2 ;dA.$

[2] Let $T$ be the triangle with vertices $(0,0),$ $(4,0),$ and $(4,1)$ and calculate $\diintlp T;\x2 y + \y3 dA.$

Let $f(x,y,z) = \x2 y + 3xy + y\z2 $. Let $\R$ be the rectangular prism determined by $\vi,$ $\vj,$ and $\vk$. Let $T$ be the tetrahedron with vertices $(0,0,0),$ $(2,0,0),$ $(0, 1, 0)$, and $(0,0,3)$. Let $C$ be the right circular cylinder whose base is the unit circle and whose height is 5. Finally, let $S$ be the unit sphere.

[2]()

[2] $\diiintl R;f(x,y,z)dV$

[3] $\diiintl T;f(x,y,z)dV$

[2]()

[3] $\diiintl C;f(x,y,z)dV$

[3] $\diiintl S;f(x,y,z)dV$

(Final1)

Final Exam Math 2673 Summer 2004

Name:

E-mail (optional):

Directions: Show all of your work in the space provided and justify all of your answers. You may use written materials but you are not permitted to ask another person for help (except for the instructor).

[2] Find a vector-valued function whose graph is the line connecting the points $(1,\tm1,2)$ and $(5,1,6).$ $(1,0,2)$ and $(1,2,6).$ $(2,\tm3,1)$ and $(0,3,5).$ $(5,4,\tm2)$ and $(1,\tm2,3).$ $(2,3,2)$ and $(\tm1,2,5).$ $(2,0,8)$ and $(0,2,7).$ $(3,\tm5,3)$ and $(2,1,6).$ $(2,\tm3,1)$ and $(\tm1,\tm2,\tm5).$ $(\tm1,1,\tm2)$ and $(3,3,9).$ $(2,2,6)$ and $(5,1,4).$ $(0,3,9)$ and $(0,\tm6,2).$ $(8,8,7)$ and $(7,5,4).$ $(3,3,3)$ and $(\tm1,0,6).$

[2] Paramaterize the curve by arc length.

[2]( [3] For each of the following, determine whether or not the lines are coplanar. If so, give the equation of the plane. If not, explain why they are not.)

$x$ $=$ $t$ $+$ 1
$y$ $=$ $t$ $-$ 2
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $3s$ $+$ 1
$z$ $=$ $s$ $+$ 2

$x$ $=$ $2t$ $-$ 3
$y$ $=$ $\tm t$ $-$ 2
$z$ $=$ $2t$ $+$ 3

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $s$ $-$ 3
$z$ $=$ $s$ $+$ 2

$x$ $=$ $\tm2t$ $+$ 1
$y$ $=$ $3t$    
$z$ $=$ $t$ $-$ 4

$x$ $=$ $3s$ $+$ 1
$y$ $=$ $\tm s$ $-$ 3
$z$ $=$ $2s$ $-$ 1

$x$ $=$ $\tm t$ $-$ 2
$y$ $=$ $2t$ $+$ 6
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $-$ 2
$y$ $=$ $\tm s$    
$z$ $=$ $\tm s$ $+$ 3

$x$ $=$ $t$ $+$ 1
$y$ $=$ $t$ $-$ 2
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $3s$ $+$ 1
$z$ $=$ $s$ $+$ 2

$x$ $=$ $2t$ $-$ 3
$y$ $=$ $\tm t$ $-$ 2
$z$ $=$ $2t$ $+$ 3

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $s$ $-$ 3
$z$ $=$ $s$ $+$ 2

$x$ $=$ $\tm2t$ $+$ 1
$y$ $=$ $3t$    
$z$ $=$ $t$ $-$ 4

$x$ $=$ $3s$ $+$ 1
$y$ $=$ $\tm s$ $-$ 3
$z$ $=$ $2s$ $-$ 1

$x$ $=$ $\tm t$ $-$ 2
$y$ $=$ $2t$ $+$ 6
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $-$ 2
$y$ $=$ $\tm s$    
$z$ $=$ $\tm s$ $+$ 3

$x$ $=$ $t$ $+$ 1
$y$ $=$ $t$ $-$ 2
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $3s$ $+$ 1
$z$ $=$ $s$ $+$ 2

$x$ $=$ $2t$ $-$ 3
$y$ $=$ $\tm t$ $-$ 2
$z$ $=$ $2t$ $+$ 3

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $s$ $-$ 3
$z$ $=$ $s$ $+$ 2

$x$ $=$ $\tm2t$ $+$ 1
$y$ $=$ $3t$    
$z$ $=$ $t$ $-$ 4

$x$ $=$ $3s$ $+$ 1
$y$ $=$ $\tm s$ $-$ 3
$z$ $=$ $2s$ $-$ 1

$x$ $=$ $\tm t$ $-$ 2
$y$ $=$ $2t$ $+$ 6
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $-$ 2
$y$ $=$ $\tm s$    
$z$ $=$ $\tm s$ $+$ 3

$x$ $=$ $t$ $+$ 1
$y$ $=$ $t$ $-$ 2
$z$ $=$ $3t$ $+$ 1

$x$ $=$ $2s$ $+$ 2
$y$ $=$ $3s$ $+$ 1
$z$ $=$ $s$ $+$ 2

[3] Do the following planes intersect? If so, find the line of intersection.

$2x$ $+$ $3y$ $-$ $z$ $=$ $7$
$3x$ $-$ $8y$ $+$ $2z$ $=$ $11$

$2x$ $+$ $2y$ $-$ $z$ $=$ $7$
$2x$ $-$ $8y$ $+$ $2z$ $=$ $10$

$x$ $+$ $y$ $-$ $z$ $=$ $6$
$2x$ $+$ $3y$ $+$ $z$ $=$ $8$

$2x$ $-$ $y$ $+$ $z$ $=$ $\tm7$
$3x$ $+$ $3y$ $-$ $z$ $=$ $6$

$x$ $+$ $y$ $+$ $3z$ $=$ $\tm1$
$x$ $+$ $2y$ $-$ $2z$ $=$ $8$

$3x$ $-$ $y$ $-$ $2z$ $=$ $\tm3$
$x$ $+$ $y$ $+$ $z$ $=$ $7$

$2x$ $+$ $3y$ $-$ $z$ $=$ $7$
$3x$ $-$ $8y$ $+$ $2z$ $=$ $11$

$2x$ $+$ $3y$ $-$ $z$ $=$ $7$
$3x$ $-$ $8y$ $+$ $2z$ $=$ $11$

$2x$ $+$ $2y$ $-$ $z$ $=$ $7$
$2x$ $-$ $8y$ $+$ $2z$ $=$ $10$

$x$ $+$ $y$ $-$ $z$ $=$ $6$
$2x$ $+$ $3y$ $+$ $z$ $=$ $8$

$2x$ $-$ $y$ $+$ $z$ $=$ $\tm7$
$3x$ $+$ $3y$ $-$ $z$ $=$ $6$

$x$ $+$ $y$ $+$ $3z$ $=$ $\tm1$
$x$ $+$ $2y$ $-$ $2z$ $=$ $8$

$3x$ $-$ $y$ $-$ $2z$ $=$ $\tm3$
$x$ $+$ $y$ $+$ $z$ $=$ $7$

[2]( For each of the following, sketch the graph of the function and use arrows to indicate the direction of increasing $t.$)

[3] $\vr(t) = \left\langle\e t;,t\right\rangle$

$\vr(t) = \left\langle t,\e t;\right\rangle$

$\vr(t) = \left\langle t,\sin t\right\rangle$

$\vr(t) = \left\langle \sin t,t\right\rangle$

$\vr(t) = \left\langle \cos t,t\right\rangle$

$\vr(t) = \left\langle t,\cos t\right\rangle$

$\vr(t) = \left\langle t,\ln t\right\rangle$

$\vr(t) = \left\langle \ln t,t\right\rangle$

$\vr(t) = \left\langle t,\tan t\right\rangle$

$\vr(t) = \left\langle \tan t,t\right\rangle$

$\vr(t) = \left\langle \sin t,t\right\rangle$

$\vr(t) = \left\langle \ln t,t\right\rangle$

$\vr(t) = \left\langle t,\sin t\right\rangle$

[3]

$\vr(t) = \left\langle 3\cos t,2\sin t,t\right\rangle$

$\vr(t) = \left\langle 2\cos t,3\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm3\cos t,2\sin t,t\right\rangle$

$\vr(t) = \left\langle 3\cos t,\tm2\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm3\cos t,\tm2\sin t,t\right\rangle$

$\vr(t) = \left\langle =\tm2\cos t,3\sin t,t\right\rangle$

$\vr(t) = \left\langle 2\cos t,\tm3\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm2\cos t,\tm3\sin t,t\right\rangle$

$\vr(t) = \left\langle 3\cos t,2\sin t,t\right\rangle$

$\vr(t) = \left\langle 2\cos t,3\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm2\cos t,3\sin t,t\right\rangle$

$\vr(t) = \left\langle 2\cos t,\tm3\sin t,t\right\rangle$

$\vr(t) = \left\langle \tm2\cos t,\tm3\sin t,t\right\rangle$

[3] Let $\vf(t) = \left\langle\e t;\cos t,\sqrt{\t2 + 1}\right\rangle.$ $\vf(t) = \left\langle\e t;\sin t,\sqrt{\t2 + 1}\right\rangle.$ $\vf(t) = \left\langle\e t;\cos t,\sqrt{\t2 - 1}\right\rangle.$ $\vf(t) = \left\langle\e t;\sin t,\sqrt{\t2 - 1}\right\rangle.$ $\vf(t) = \left\langle\e t;\cos t,\sqrt{1 - \t2 }\right\rangle.$ $\vf(t) = \left\langle\e t;\sin t,\sqrt{1 - \t2 }\right\rangle.$ $\vf(t) = \left\langle\ln t,\sqrt{\t2 + 1}\right\rangle.$ $\vf(t) = \left\langle\ln t,\sqrt{\t2 - 1}\right\rangle.$ $\vf(t) = \left\langle\ln t,\sqrt{1 - \t2 }\right\rangle.$ $\vf(t) = \left\langle\sec^{3}t,\sqrt{\t2 + 1}\right\rangle.$ $\vf(t) = \left\langle\sec^{3}t,\sqrt{\t2 - 1}\right\rangle.$ $\vf(t) = \left\langle\sec^{3}t,\sqrt{1 - \t2 }\right\rangle.$ $\vf(t) = \left\langle\e t;\cos t,\sqrt{\t2 + 1}\right\rangle.$

$\dintl1,2;\vf(t)dt$ $=$

A projectile is launched with initial velocity vector $\vv = \left\langle30,40,50\right\rangle$. The wind is blowing southeast at a rate of 50 ft/sec.

[4] Find the vector functions that describe velocity and motion.

[2] Find the maximum height.

[2] Find the position of impact.

[2] Find the speed of impact.

[2] Sketch the graph of $f(x,y) = \e x;.$ $f(x,y) = \e y;.$ $f(x,y) = \sin x.$ $f(x,y) = \sin y.$ $f(x,y) = \cos x.$ $f(x,y) = \cos y.$ $f(x,y) = \ln x.$ $f(x,y) = \ln y.$ $f(x,y) = \x2 .$ $f(x,y) = \y2 .$ $f(x,y) = x + y.$ $f(x,y) = x - y.$ $f(x,y) = y - x.$

[6] For each of the six quadric surfaces we studied, give an example using irrational coefficients and sketch the graph.

[3] Find the following limit or show that it does not exist.

$\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),x>$ $\limv{(x,y)},{(0,0)};\dfric<\sin(x + y),y>$

[3] Use the $\varepsilon-\delta$ defintion of limit to prove $\limp{(x,y)},{(1,1)};3x + 2y p = 5.$ $\limp{(x,y)},{(2,1)};3x + 2y p = 8.$ $\limp{(x,y)},{(\tm1,1)};3x + 2y p = \tm1.$ $\limp{(x,y)},{(1,\tm1)};3x + 2y p = 1.$ $\limp{(x,y)},{(2,2)};3x + 2y p = 12.$ $\limp{(x,y)},{(0,1)};3x + 2y p = 1.$ $\limp{(x,y)},{(1,0)};3x + 2y p = 3.$ $\limp{(x,y)},{(2,1)};2x + 3y p = 5.$ $\limp{(x,y)},{(2,1)};x + 3y p = 5.$ $\limp{(x,y)},{(1,1)};5x - 2y p = 3.$ $\limp{(x,y)},{(\tm2,1)};3x + 2y p = \tm4.$ $\limp{(x,y)},{(0,1)};2x + 3y p = 3.$ $\limp{(x,y)},{(2,\tm1)};3x + 2y p = 4.$

[3] Use the $\varepsilon-\delta$ defintion of limit to prove $\limv{(x,y)},{(1,1)};xy = 1.$ $\limv{(x,y)},{(2,1)};xy = 2.$ $\limv{(x,y)},{(1,2)};xy = 2.$ $\limv{(x,y)},{(2,2)};xy = 4.$ $\limv{(x,y)},{(3,1)};xy = 3.$ $\limv{(x,y)},{(1,3)};xy = 3.$ $\limv{(x,y)},{(2,3)};xy = 6.$ $\limv{(x,y)},{(3,2)};xy = 6.$ $\limv{(x,y)},{(4,1)};xy = 4.$ $\limv{(x,y)},{(1,4)};xy = 4.$ $\limv{(x,y)},{(2,4)};xy = 8.$ $\limv{(x,y)},{(4,2)};xy = 8.$ $\limv{(x,y)},{(4,3)};xy = 12.$

[3] Let $f(x,y,z) = \x2 y + 4y + 3\z2 $. Show that $f$ is differentiable at $(a,b,c)$ for all $(a,b,c)\in\Rss3 $ and find $f'_{{}_{(a,b,c)}}(x,y,z).$

Let $f(x,y,z) = 3\x2 y + 3xy + \y2 .$ $f(x,y,z) = \x2 y + xy + \y2 .$ $f(x,y,z) = 2\x2 y + 2xy + \y2 .$ $f(x,y,z) = 4\x2 y + 4xy + \y2 .$ $f(x,y,z) = 5\x2 y + 5xy + \y2 .$ $f(x,y,z) = 3\x2 y + 3xy + \y2 .$ $f(x,y,z) = \x2 y + xy + \y2 .$ $f(x,y,z) = 2\x2 y + 2xy + \y2 .$ $f(x,y,z) = 4\x2 y + 4xy + \y2 .$ $f(x,y,z) = 5\x2 y + 5xy + \y2 .$ $f(x,y,z) = 3\x2 y + 3xy + \y2 .$ $f(x,y,z) = \x2 y + xy + \y2 .$ $f(x,y,z) = 2\x2 y + 2xy + \y2 .$

[3] Find all local extrema and saddle points of $f.$

[3] Find all extrema of $f$ on the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1).$

[2] Let $R = [1,2]\times[0,1]$ $R = [1,2]\times[1,1]$ $R = [\tm1,1]\times[1,2]$ $R = [0,1]\times[1,2]$ $R = [0,1]\times[1,1]$ $R = [\tm2,0]\times[0,1]$ $R = [\tm2,1]\times[0,1]$ $R = [\tm2,1]\times[1,3]$ $R = [0,1]\times[2,3]$ $R = [\tm1,0]\times[1,2]$ $R = [1,2]\times[0,1]$ $R = [1,2]\times[1,1]$ $R = [\tm1,1]\times[1,2]$ and calculate $\diintlp R;\x2 y + \y3 dA.$

[3] Let $C$ be the circle centered at $(0,0)$ with radius $1$ $\sqrt{2}$ $2$ $\sqrt{3}$ $3$ $4$ $5$ $\sqrt{5}$ $6$ $7$ $\sqrt{7}$ $8$ $9$ and calculate $\diintlp C;x\e\x2 + \y2 ;dA.$

[3] Let $T$ be the triangle with vertices $(0,0),$ $4,0),$ and $4,1)$ and calculate $\diintlp T;\x2 y + \y3 dA.$

Let $f(x,y,z) = \x2 y + 3xy + y\z2 $. Let $\R$ be the rectangular prism determined by $\vi,$ $\vj,$ and $\vk$. Let $T$ be the tetrahedron with vertices $(0,0,0),$ $(2,0,0),$ $(0, 1, 0)$, and $(0,0,3)$. Let $C$ be the right circular cylinder whose base is the unit circle and whose height is 5. Finally, let $S$ be the unit sphere.

[2]()

[2] $\diiintl R;f(x,y,z)dV$

[3] $\diiintl R;f(x,y,z)dT$

[2]()

[3] $\diiintl C;f(x,y,z)dT$

[3] $\diiintl S;f(x,y,z)dT$

(Math 2673 Spring 2015)

(Quizzes)

[1] Describe the region in space being defined.

$\tm6 \leq \x2 - 2x + \y2 + 6y + \z2 \leq \tm1$

$\tm6 \leq (x - 1)^{2} - 1 + (y + 3)^{2} - 9 + \z2 \leq \tm1$

$4 \leq (x - 1)^{2} + (y + 3)^{2} + \z2 \leq 9$

This inequality describes the set of all points on or in the sphere centered at $(1,\tm3,0)$ with radius 3 that are not in the sphere centered at $(1,\tm3,0)$ with radius 2.

(Page 815: 43) [1] Find the distance between the spheres $\x2 + \y2 + \z2 $ $=$ $4$ and $\x2 + \y2 + \z2 $ $=$ $4x + 4y + 4z - 11.$

Let $\vv$ $=$ $\langle1,2,\tm1\rangle,$ $\vw$ $=$ $\langle0,1,1\rangle,$ and $\theta$ be the angle between $\vv$ and $\vw$.

[1] Find $\cos\theta.$

$\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<1,\sqrt{6}\cdot\sqrt{2}>$ $=$ $\fric<1,2\sqrt{3}>$

[1] Find $\vv\times\vw.$

$\vv\times\vw$ $=$ $\langle3,\tm1,1\rangle$

[1] Give the equation of the plane that contains the points $(1,1,0),$ $(2,\tm1,1),$ and $(2,\tm1,3).$

Let $\vv = \left<0,0,2\right>,$ $\vw = \left<1,\tm2,1\right>,$ and $\vn = \vv\times\vw = \left<4,2,0\right>.$

The equation of the plane containing the point $(1,1,0)$ with normal vector $\vn = \langle2,1,0\rangle$ is $2(x - 1) + 1(y - 1) = 0.$

[2]( Let $l_{{}_{1}}$ and $l_{{}_{2}}$ be the lines with parametric equations given below.)

$x = 3t - 1$

$y = t + 6$

$z = 2t + 2$

$x = 4s + 1$

$y = 3s + 5$

$z = \tm s + 7$

[2]( [1] Do $l_{{}_{1}}$ and $l_{{}_{2}}$ intersect? If so, where?)

$3t$ $-$ $1$ $=$ $4s$ $+$ $1$
$t$ $+$ $6$ $=$ $3s$ $+$ $5$

$3t$ $-$ $4s$ $=$ $2$    
$t$ $-$ $3s$ $=$ $\tm1$    

$5s = 5$

$s = 1$

$t = 2$

The lines intersect at $(5,8,6).$

[2] Find the equation of the plane containing $l_{{}_{1}}$ and $l_{{}_{2}}.$

A vector in the direction of $l_{{}_{1}}$ is $\vv = \left\langle3,1,2\right\rangle$ and a vector in the direction of $l_{{}_{2}}$ is $\vw = \left\langle4,3,\tm1\right\rangle$. So a normal vector to the plane is $\vn$ $=$ $\vv\times\vw$ $=$ $\left\langle\tm7,11,5\right\rangle.$ The equation of the plane with normal vector $\left\langle\tm7,11,5\right\rangle$ and containing the point $(5,8,6)$ is $\tm7(x + 1) + 11(y - 6) + 5(z - 2) = 0.$

(Page 895: 21) [2] A ball is thrown eastward into the air from the origin (in the direction of the positive $x$-axis). The initial velocity is $50\vi + 80\vk,$ with speed measured in feet per second. The spin of the ball results in a southward acceleration of 4 ft/ sec$^{2},$ so the acceleration vector is $\va$ $=$ $\tm4\vj - 32 \vk$. Where does the ball land?

[1] Give the domain and range. Sketch or describe the graph.

$f(x,y)$ $=$ $1 - \x2 - \y2 $

D: $\Rss2 $

R: $(\tm\infty,1]$

$z$ $=$ $1 - \x2 - \y2 $

$z - 1$ $=$ $\tm\x2 - \y2 $

$\tm(z - 1)$ $=$ $\x2 + \y2 $

Paraboloid opening downward.

[1] Sketch or describe the graph.

$f(x,y)$ $=$ $\sqrt{1 - \x2 - \y2 }$

D: $\{(x,y)\in\Rss2 :\x2 + \y2 \leq 1\}$

R: $[0,1]$

$z$ $=$ $\sqrt{1 - \x2 - \y2 }$

$\z2 $ $=$ $1 - \x2 - \y2 $

$\x2 + \y2 + \z2 $ $=$ $1$

Top half of the unit sphere.

For each of the following, find the limit or prove that it does not exist.


\begin{multicols}{2}
\par
\prob[1]
\(\limv {(x,y)},(0,0); \dfric<4xy,3\x2 + \y2 ...
...limv{(x,y)},(0,0);\dfric<\x2 y + 3,x + y + 1>\)
\(=\)
\(3\)
\par
\end{multicols}

[1] Find the limit and use the definition of limit to prove your answer.

$\limp{(x,y)},{(1,1)};2x + y p = 3$

Let $\varepsilon > 0$ and choose $\delta > 0$ such that $\delta < \fric<\varepsilon,3>$. Suppose that $d((x,y),(1,1)) < \delta$. Then

[2]()

$\sqrt{(x - 1)^{2} + (y - 1)^{2}} < \delta$

which implies

$\sqrt{(x - 1)^{2}}$ $=$ $\vert x - 1\vert$ $<$ $\delta$

and

$\sqrt{(y - 1)^{2}}$ $=$ $\vert y - 1\vert$ $<$ $\delta.$

So

$2\vert x - 1\vert + \vert y - 1\vert < 3\delta$

$2\vert x - 1\vert + \vert y - 1\vert < \varepsilon$

$\vert 2(x - 1) + y - 1\vert < \varepsilon$

$\vert 2x - 2 + y - 1\vert < \varepsilon$

$\vert 2x + y - 3\vert < \varepsilon$

as desired.

For each of the following, find all local extrema and saddle points.


\begin{multicols}{2}
\par
\prob[1]
\(f(x,y) = \x3 + x\y2 + 2\x2 + \y2 \)
\par
\(...
...al min: \(\tm\fric<1,108>\)
at \((\fric<1,3>,\fric<1,12>)\)
\par
\end{multicols}

[2] Find the maximum and minimum values of $f(x,y) = \e x; + \y2 $ on the region bounded by the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1).$

Since $f_{{}_{x}}(x,y) = \e x; \geq 0$ for all $x\in\R,$ $f$ has no critical points. Since $g(x)$ $=$ $\e x;$ and $h(y)$ $=$ $\y2 $ are increasing on $[0,\infty),$ $f$ has a local minimum of $1$ at $(0,0)$ and a local maximum of $e + 1$ at $(1,1).$

Let $f(x,y,z) = \x2 y + \e x + z;.$

[2]( [1] Find the gradient of $f$ at ${(1,2,\tm1)}.$)

$f_{{}_{x}}(x,y,z)$ $=$ $2xy + \e x + z;$

$f_{{}_{y}}(x,y,z)$ $=$ $\x2 $

$f_{{}_{z}}(x,y,z)$ $=$ $\e x + z;$

$\nabla f(1,2,\tm1)$ $=$ $\left\langle5,1,1\right\rangle$

[1] Find the directional derivative of $f$ at ${(1,2,\tm1)}$ in the direction of $\left\langle\fric<\sqrt{3},3>,\fric<\sqrt{3},3>,\fric<\sqrt{3},3>
\right\rangle.$

$\nabla f(1,2,\tm1)
\cdot
\left\langle\fric<\sqrt{3},3>,\fric<\sqrt{3},3>,\fric<\sqrt{3},3>\right\rangle$ $=$ $\fric<5\sqrt{3},3> + \fric<\sqrt{3},3> + \fric<\sqrt{3},3>$ $=$ $\fric<7\sqrt{3},3>$

[1] In what direction does the maximum rate of change of $f$ occur at $(1,2,\tm1)$?

$\nabla f(1,2,\tm1)$ $=$ $\left\langle5,1,1\right\rangle$

[1] What is the maximum rate of change of $f$ at $(1,2,\tm1)$?

$\Vert\nabla f(1,2,\tm1)\Vert$ $=$ $\sqrt{27}$ $=$ $3\sqrt{3}$

[2] Maximize and minimize $f(x,y,z) = \x2 + y + yz$ subject to $\x2 + \y2 + \z2 = 1$ and $y + z = 0.$


\begin{multicols}{2}
\par
\(g(x,y,z) = \x2 + \y2 + \z2 - 1\)
\par
\(h(x,y,z) = y...
...t>
= \lambda\left<2x,2y,2z\right>
+ \mu\left<0,1,1\right>\)
\par
\end{multicols}


\begin{multicols}{3}
\par
\(2x = 2\lambda x\)
\par
\(z + 1 = 2\lambda y + \mu\)
\par
\(y = 2\lambda z + \mu\)
\par
\end{multicols}

From the first equation, $x = 0$ or $\lambda = 1.$

Note that $y$ $=$ $\tm z$ from the second constraint equation.

So if $x = 0,$ then either $y = \tm\fric<\sqrt{2},2>$ and $z = \fric<\sqrt{2},2>$ or $y = \fric<\sqrt{2},2>$ and $z = \tm\fric<\sqrt{2},2>$.

If $\lambda$ $=$ $1,$ then the second and third equations above yield the following.

[2]()

$z + 1 = 2\lambda y + \mu$

$z + 1 = 2y + \mu$

$z + 1 = \tm2z + \mu$

$3z = \mu - 1$

$y = 2\lambda z + \mu$

$y = 2z + \mu$

$\tm z = 2z + \mu$

$3z = \tm\mu$

So $\mu - 1$ $=$ $\tm\mu$ which means that $\mu$ $=$ $\fric<1,2>$. In this case $z$ $=$ $\tm\fric<1,6>,$ $y$ $=$ $\fric<1,6>,$ and $x$ $=$ $\pms\sqrt{\fric<17,18>}$.

[2]()

$f\left(0,\tm\fric<\sqrt{2},2>,\fric<\sqrt{2},2>\right)$ $=$ $\tm\fric<\sqrt{2} + 1,2>$ (min)

$f\left(0,\fric<\sqrt{2},2>,\tm\fric<\sqrt{2},2>\right)$ $=$ $\fric<\sqrt{2} - 1,2>$

$f\left(\tm\sqrt{\fric<17,18>},\fric<1,6>,\tm\fric<1,6>\right)$ $=$ $\fric<13,12>$ (max)

$f\left(\sqrt{\fric<17,18>},\fric<1,6>,\tm\fric<1,6>\right)$ $=$ $\fric<13,12>$ (max)

[1] Let $R = \{(x,y): 1\leq x \leq 3, \tm2\leq y \leq 1\}$ and calculate $\diintlp R; 3\x2 y - 4xy + 1dA.$


\begin{multicols}{2}
\par
\(\diintlp R; 3\x2 y - 4xy + 1dA\)
\par
\(=\)
\(\dintl...
... 2y p\)
\par
\(=\) \(5 + 2 - (20 - 4)\)
\par
\(=\) \(\tm9\)
\par
\end{multicols}

(Page 1020: 49) [2] Evaluate the following integral. Hint: Reverse the order of integration.

$\dintl0,1;{\dintl {3y},3;\e x^{2}; dx}dy$

[2] Let $R = \{(x,y,z): \x2 + \y2 \leq 1, 0\leq z \leq 3\}$ and calculate $\diiintl R; z\e\x2 + \y2 + \z2 ; dV.$


\begin{multicols}{2}
\par
\(\diiintl R; z\e\x2 + \y2 + \z2 ; dV\)
\par
\(=\)
\(\...
...ar
\(=\)
\(\fric<\pi,2>\left(\e10; - \e9; - e + 1 \right)\)
\par
\end{multicols}

[1] Let $R = \left\{(x,y): \tm2\leq x\leq 2,
\tm\sqrt{4 - \x2 }\leq y\leq\sqrt{4 - \x2 }\right\}$ and calculate $\diintl R;2\e\x2 + \y2 ; dA.$

Note that $\R$ is the circle centered at (0,0) with radius 2. Use polar coordinates.


\begin{multicols}{2}
\par
\(\diintl R;2\e\x2 + \y2 ; dA\)
\par
\(=\)
\(\dintl 0,...
...\e4; - 1\right)p\)
\par
\(=\)
\(2\pi\left(\e4; - 1\right)\)
\par
\end{multicols}

[2] Let $\R$ be the unit sphere and calculate $\diiintl R;
\dfric<\sin(\x2 + \y2 + \z2 ),\sqrt{\x2 + \y2 + \z2 }>dV.$


\begin{multicols}{2}
\par
\(\diiintl R;
\dfric<\sin(\x2 + \y2 + \z2 ),\sqrt{\x2 ...
...>\cos 1\right)\)
\par
\(=\)
\(2\pi\left(1 - \cos 1\right)\)
\par
\end{multicols}

Due 3:30 Friday May 1, 2015

Directions:

Show all of your work.

Write on only one side of the paper.

Use separate paper for each problem.

Put the problems in numerical order.

Staple your pages.

[1] Page 1096: 9

[1] Page 1107: 17

[1] Page 1114: 11

[1] Page 1121: 9

[1] Page 1121: 15

[1] Page 1132: 49

[1] Page 1144: 7

[1] Page 1144: 21

(Exam 1)

Exam 1 Math 2673 Spring 2015

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $\vv$ $=$ $\langle0,\tm1,2\rangle$ and $\vw$ $=$ $\langle1,\tm1,3\rangle.$

[10] Find $\cos\theta$ where $\theta$ is the angle between $\vv$ and $\vw.$

$\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<7,\sqrt{5}\cdot\sqrt{11}>$ $=$ $\fric<7,\sqrt{55}>$

[10] Find proj$_{{}_{\vv}}\vw.$

proj$_{{}_{\vv}}\vw$ $=$ $\fricp<\vv\cdot\vw,\Vert\vv\Vert^{2}> \vv$ $=$ $\fric<7,5>\langle0,\tm1,2\rangle$

[2]( [10] Give the equations of the lines through the point $(1,1,1)$ in the directions of $\vv$ and $\vw.$)

$x$ $=$ $1$

$y$ $=$ $1 - s$

$z$ $=$ $1 + 2s$

$x$ $=$ $1 + t$

$y$ $=$ $1 - t$

$z$ $=$ $1 + 3t$

[10] Give the equation of the plane containing the lines from the previous part. You need not to have done the previous part to do this part.

$\vn$ $=$ $\vv\times\vw$ $=$ $\langle\tm1,2,1\rangle$

$\tm(x - 1) + 2(y - 1) + (z - 1)$ $=$ 0

[10] Give the equation of the sphere with center $(\tm1,2,6)$ and radius $3.$

$(x + 1)^{2} + (y - 2)^{2} + (z - 6)^{2}$ $=$ $9$

(Page 831: 51) [10] A sled is pulled along a level path through snow by a rope. A 30-lb force acting at an angle of $40^{\circ}$ above the horizontal moves the sled 80 ft. Find the work done by the force.

[10] Find the volume of the parallelepiped determined by $\left\langle\tm1,0,4\right\rangle,$ $\left\langle0,2,3\right\rangle,$ and $\left\langle\tm1,3,1\right\rangle.$

$\left\vert\left\langle\tm1,0,4\right\rangle\cdot\left(
\left\langle0,2,3\right\rangle\times
\left\langle\tm1,3,1\right\rangle\right)\right\vert$ $=$
$\tm1$ 0 $4$
   
0 $2$ $3$
   
$\tm1$ $3$ $1$
$=$ $\vert(\tm1)(\tm7) - (0)(3) + (4)(2)\vert$ $=$ $15$

Identify and describe each of the following.

[10] $(x + 1)^{2} + (y - 3)^{2}$ $=$ $4,$ $0\leq z\leq 3$

This is a right circular cylinder. The base is the circle in the $xy$-plane centered at $(\tm1,3)$ with radius 2. The height of the cylinder is 3.

[10] $z$ $=$ $\fric<\x2 ,4> + \fric<\y2 ,9>$

Elliptic paraboloid.

[10] $(x + 1)^{2} + \fric<(y + 4)^{2},16> + \z2 $ $=$ $1$

Ellipsoid.

Let $\vf(t) = \left<\sin t,\cos t,t\right>$

[10] Sketch the graph of $\vf.$

[10] Give the equation of the line that is tangent to the graph of $\vf$ at the point $\left(1,0,\fric<\pi,2>\right).$

$\vf'(t)$ $=$ $\left<\cos t,\tm\sin t,1\right>$

$\vf'\left(\fric<\pi,2>\right)$ $=$ $\left<0,\tm1,1\right>$

$x$ $=$ $1$

$y$ $=$ $\tm t$

$z$ $=$ $1 + t$

(Exam 2)

Exam 2 Math 2673 Spring 2015

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $\vr(t) = \left<t,\fric<2\sqrt{2},3>\t{}^{\fric<3,2>} ,\fric<1,2>\t2 \right>.$

[10] Calculate $\vr'(t)$ and $\Vert\vr'(t)\Vert.$

$\vr'(t)$ $=$ $\left<1,\sqrt{2}\t{}^{\fric<1,2>} ,t\right>$ $=$ $\left<1,\sqrt{2t},t\right>$

$\Vert\vr'(t)\Vert$ $=$ $\sqrt{1 + 2t + \t2 }$ $=$ $\sqrt{(t + 1)^{2}}$ $=$ $\vert t + 1\vert$

[10] Find the arc length of the curve defined by $\vr(t)$ as $t$ ranges from 0 to $1.$

$\dintl0,1;\vert t + 1\vert dt$ $=$ $\dintlp0,1;t + 1dt$ $=$ $\biggmlp0,1;\fric<1,2>\t2 + t p$ $=$ $\fric<3,2>$

[10] Calculate the unit tangent vector at the point $t$ $=$ $2.$

$\vT(t)$ $=$ $\dfric<r'(t),\Vert r'(t)\Vert>$ $=$ $\fric<1,\vert t + 1\vert>\left<1,\sqrt{2t},t\right>$

$\vT(2)$ $=$ $\fric<1,3>\left<1,2,2\right>$

[3]( [10] Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.)

$\va(t)$ $=$ $\vi + 2\vj$

$\vv(0)$ $=$ $\vk$

$\vr(0)$ $=$ $\vi$

$\vv(t)$ $=$ $t\vi + 2t\vj + \vk$ $=$ $\left<t,2t,1\right>$

$\vr(t)$ $=$ $\fric<1,2>\t2 \vi + \t2 \vj + t\vk + \vi$ $=$ $\left<\fric<1,2>\t2 + 1,\t2 ,t\right>$

[2]( Let $f(x,y)$ $=$ $\sqrt{\x2 + \y2 }.$)

[10] Give the domain and range of $f.$

Domain: $\Rss2 $

Range: $[0,\infty)$

[10] \(\text{Sketch or describe the graph of \(f.\)}\)

$z$ $=$ $\sqrt{\x2 + \y2 }$

$\z2 $ $=$ $\x2 + \y2 $

Top half of a cone.

[2]()

Calculate the following limits.

[10] $\limv{(x,y)},{(0,0)};\dfric<\x2 - \y2 ,x + y>$

$=$ $\limv{(x,y)},{(0,0)};\dfric<(x + y)(x - y) ,x + y>$

$=$ $\limp{(x,y)},{(0,0)};x + y p$

$=$ 0

[10] $\limv{(x,y)},{(0,0)};\dfric<\x2 + y,x + \y2 >$ DNE

$\limv{(x,0)},{(0,0)};\dfric<\x2 + y,x + \y2 >$

$=$ $\limv{(x,0)},{(0,0)};\dfric<\x2 ,x>$

$=$ $\limv{(x,0)},{(0,0)};x$

$=$ 0

$=$ $\limv{(0,y)},{(0,0)};\dfric<\x2 + y,x + \y2 >$

$=$ $\limv{(0,y)},{(0,0)};\dfric<y,\y2 >$

$=$ $\limv{(0,y)},{(0,0)};\dfric<1,y>$

DNE

Let $f(x,y)$ $=$ $\x2 \y2 + 2xy.$

[10] Calculate all first order partial derivatives and all second order partial derivatives of $f.$

$f_{{}_{x}}(x,y)$ $=$ $2x\y2 + 2y$

$f_{{}_{y}}(x,y)$ $=$ $2\x2 y + 2x$

$f_{{}_{xx}}(x,y)$ $=$ $2\y2 $

$f_{{}_{yy}}(x,y)$ $=$ $2\x2 $

$f_{{}_{xy}}(x,y)$ $=$ $4xy + 2$

$f_{{}_{yx}}(x,y)$ $=$ $4xy + 2$

[10] Find the critical points of $f.$

$f_{{}_{x}}(x,y)$ $=$ 0

$2x\y2 + 2y$ $=$ 0

$2y(xy + 1)$ $=$ 0

$f_{{}_{y}}(x,y)$ $=$ 0

$2\x2 y + 2x$ $=$ 0

$2x(xy + 1)$ $=$ 0

CP: $(0,0),$ all points on the curve $y$ $=$ $\tm\fric<1,x>$

[10] Find the equation of the plane that is tangent to the graph of $f$ at the point $(2, 1, 8).$

$f_{{}_{x}}(2,1)$ $=$ $6$

$f_{{}_{y}}(2,1)$ $=$ $12$

$z - 8$ $=$ $6(x - 2) + 12(y - 1)$

(Exam 3)

Exam 3 Math 2673 Spring 2015

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $f(x,y,z)$ $=$ $x\e y; + \x3 z.$

[10] Calculate $\nabla f(x,y,z).$

$\nabla f(x,y,z)$ $=$ $\langle\e y; + 3\x2 z, x\e y;,\x3 \rangle$

[10] Calculate the directional derivative of $f$ at the point $(1,0,2)$ in the direction of $\fric<1,\sqrt{3}>\langle1, \tm1, 1\rangle.$

$\nabla f(1,0,2)$ $=$ $\langle7, 1,1\rangle$

$\langle7, 1,1\rangle\cdot \fric<1,\sqrt{3}>\langle1, \tm1, 1\rangle$ $=$ $\fric<1,\sqrt{3}>(7 - 1 + 1)$ $=$ $\fric<7,\sqrt{3}>$

[10] Let $z$ $=$ $\x2 y + \x4 $ where $x$ $=$ $\t3 $ and $y$ $=$ $\sin t$. Calculate $\der t z;.$

$\parti x z;$ $=$ $2xy + 4\x3 $

$\parti y z;$ $=$ $\x2 $

$\der t x;$ $=$ $3\t2 $

$\der t y;$ $=$ $\cos t$

$\der t z;$ $=$ $(2xy + 4\x3 )(3\t2 ) + \x2 (\cos t)$ $=$ $[2(\t3 )(\cos t) + 4\t9 ](3\t2 ) + \t6 (\cos t)$ $=$ $\t6 \cos t + 6\t5 \cos t + 12\t11 $

[10] Let $z$ $=$ $\e xy;$ where $x$ $=$ $t + \s2 $ and $y$ $=$ $\sqrt{t}$. Calculate $\parti s z;.$

$\parti x z;$ $=$ $y\e xy;$

$\parti y z;$ $=$ $x\e xy;$

$\parti s x;$ $=$ $2s$

$\parti s y;$ $=$ 0

$\parti s z;$ $=$ $y\e xy;(2s) + x\e xy;(0)$ $=$ $2s\sqrt{t}\e \sqrt{t}(t + \s2 );$

[10] Maximize and minimize $f(x,y,z)$ $=$ $x + y + z$ subject to $\x2 + \y2 + \z2 = 1.$

[10] Maximize and minimize $f(x,y,z)$ $=$ $\tm8x + 3y + z$ subject to $\x2 + y = 5$ and $y + z = 3.$

[10] Let $R = [0,2]\times[1,3]$ and calculate $\diintlp R;\x2 y + \x3 dA.$

[10] Let $\R$ be the region bounded by the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1)$ and calculate $\diintlp R;\e \x2 ;dA.$

[10] Let $T$ be the solid tetrahedron with vertices $(0,0,0),$ $(1,0,0),$ $(0,1,0),$ and $(0,0,4)$ and calculate $\diintl T;\fric<1,4>dV.$

(Math 2673 Spring 2017)

(Quizzes)

[(01-13-17)]

[1] Give the equation of the sphere with center $(\tm1,3,4)$ and radius $2.$

$(x + 1)^{2} + (y - 3)^{2} + (z - 4)^{2}$ $=$ $4$

[1] Find and simplify an equation for the set of all points in $\Rss3 $ that are equidistant from the points $(1,1,1)$ and $(2, 2, 2).$

$\sqrt{(x - 1)^{2} + (y - 1)^{2} + (z - 1)^{2}}$ $=$ $\sqrt{(x - 2)^{2} + (y - 2)^{2} + (z - 2)^{2}}$

$(x - 1)^{2} + (y - 1)^{2} + (z - 1)^{2}$ $=$ $(x - 2)^{2} + (y - 2)^{2} + (z - 2)^{2}$

$\x2 - 2x + 1 + \y2 - 2y + 1 + \z2 - 2z + 1$ $=$ $\x2 - 4x + 4 + \y2 - 4y + 4 + \z2 - 4z + 4$

$2x + 2y + 2z$ $=$ $9$

[(01-20-17)]

[1] Suppose that vector $\vv$ is parallel to vector $\vw$ $=$ $\left\langle 1,2,2\right\rangle$ and $\Vert\vv\Vert$ $=$ $5$. Find $\vv.$

Note that $\Vert\vw\Vert$ $=$ $3$. So the vector in the direction of $\vw$ with magnitude $5$ is $\fric<5,3>\vw$ $=$ $\left\langle \fric<5,3>,\fric<10,3>,\fric<10,3>\right\rangle$

[1] Answer the following as true or false and justify your answer.

If $\vv$ and $\vw$ are two vectors in $\Rss3 $ such that $\vv\cdot\vw$ $=$ $0,$ then one of $\vv$ or $\vw$ is $\vzero.$

False.

Consider $\vi$ and $\vj$ for example.

Also, recall that two vectors $\vv$ and $\vw$ are orthogonal if and only if $\vv\cdot\vw = 0.$

[1] Let $\vv$ $=$ $\langle1,\tm1,2\rangle$ and $\vw$ $=$ $\langle1,0,3\rangle$. Find $\cos\theta$ where $\theta$ is the angle between $\vv$ and $\vw.$

$\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<7,\sqrt{6}\cdot\sqrt{10}>$ $=$ $\fric<7,2\sqrt{15}>$

(Page 831: 51) [1] A sled is pulled along a level path through snow by a rope. A 30-lb force acting at an angle of $40^{\circ}$ above the horizontal moves the sled 80 ft. Find the work done by the force.

[(01-31-17)]

[2] Find the equation of the plane that contains the points $(1, 1, 4),$ $(\tm1, \tm1, 2),$ and $(2, 0, 1).$

$\vv$ $=$ $\left\langle 1 - \tm1,1 - \tm1,4 - 2\right\rangle$ $=$ $\left\langle 2,2,2\right\rangle$ $=$ $2\left\langle 1,1,1\right\rangle$

$\vw$ $=$ $\left\langle2 - \tm1, 0 - \tm1,1 - 2\right\rangle$ $=$ $\left\langle 3,1,\tm1\right\rangle$

$\vn$ $=$ $\left\langle 1,1,1\right\rangle\times
\left\langle 3,1,\tm1\right\rangle$ $=$ $\left\langle\tm2,4,\tm2\right\rangle$ $=$ $\tm2\left\langle1,\tm2,1\right\rangle$

$(x - 1) -2(y - 1) + (z - 4)$ $=$ 0

$x - 2y + z$ $=$ $3$

[2] Give the equation of the line of intersection of the following planes.

$2x - y - z$ $=$ $2$

$y - z$ $=$ 0

$\vn_{{}_{1}}$ $=$ $\left\langle2,\tm1,\tm1\right\rangle$

$\vn_{{}_{2}}$ $=$ $\left\langle0,1,\tm1\right\rangle$

$\vv$ $=$ $\vn_{{}_{1}}\times \vn_{{}_{2}}$ $=$ $\left\langle 2,2,2\right\rangle$ $=$ $2\left\langle 1,1,1\right\rangle$

Point on both planes: $(1,0,0)$

$x$ $=$ $1 + t$

$y$ $=$ $t$

$z$ $=$ $t$

[(02-08-17)]

[1] Sketch or describe the curve of the following vector function.

$\vr(t)$ $=$ $\left\langle\sin t,\e t;,\cos t\right\rangle$

Let $C$ be the cylinder with base $\x2 + \z2 $ $=$ $1$ (circle centered at $(0,0)$ with radius 1) in the $xz$-plane. As $t\to\infty,$ the curve spirals around $C$ in the direction of the positive $y$-axis. Note that the $y$-coordinate increases exponentially. As $t\to\tm\infty,$ the curve spirals around $C$ toward (but never touching) the $xz$-plane.

[2] Give the vector equation of the curve where the following surfaces intersect.

$\x2 + \y2 $ $=$ $z$

$\x2 - \y2 $ $=$ $1$

Adding the two equations together, yields

$z + 1$ $=$ $2\x2 $

$z$ $=$ $2\x2 - 1$

From the first equation, we have

$y$ $=$ $\sqrt{\x2 - 1}$

$\vr(t)$ $=$ $\left\langle t,\sqrt{\t2 - 1},2\t2 - 1\right\rangle$

Let $\vr(t)$ $=$ $\left\langle \sin(\pi t),\e t - 1;,t\ln t\right\rangle.$

[1] Find the derivative of $\vr.$

$\vr'(t)$ $=$ $\left\langle \pi\cos(\pi t),\e t - 1;,\ln t + 1\right\rangle$

[2] Find the equation of the line that is tangent to the graph of $\vr$ where $t = 1.$

$\vr(1)$ $=$ $\left\langle 0,1,0\right\rangle$

$\vr'(1)$ $=$ $\left\langle \tm\pi,1,1\right\rangle$

$\vf(s)$ $=$ $\left\langle 0,1,0\right\rangle
+ s\left\langle \tm\pi,1,1\right\rangle$

[(02-15-17)]

Let $\vr(t) = \left<\t2 ,\fric<4\sqrt{2},3>\t{}^{\fric<3,2>} ,2t\right>.$

[1] Calculate each of the following.

$\vr'(t)$ $=$ $\left<2t,2\sqrt{2}\t{}^{\fric<1,2>} ,2\right>$ $=$ $2\left<t,\sqrt{2t},1\right>$

$\Vert\vr'(t)\Vert$ $=$ $2\sqrt{\t2 + 2t + 1}$ $=$ $2\sqrt{(t + 1)^{2}}$ $=$ $2\vert t + 1\vert$

$\vr''(t)$ $=$ $\left<2,\sqrt{2}\t{}^{\tm\fric<1,2>} ,0\right>$

[1] Find the arc length of the curve defined by $\vr(t)$ as $t$ ranges from 0 to $1.$

$\dintl0,1;2\vert t + 1\vert dt$ $=$ $\dintl0,1;2(t + 1)dt$ $=$ $\dintlp0,1;2t + 2dt$ $=$ $\biggmlp0,1;\t2 + 2t p$ $=$ $3$

[1] Calculate the curvature at the point $t$ $=$ $2.$

$\kappa(2)$ $=$ $\dfric<\left\Vert\vr'(2)\times\vr''(2)\right\Vert,
\left\Vert\vr'(2)\right\Vert^{3}>$ $=$ $\dfric<\left\Vert{\left\langle4,4,2\right\rangle}\times{\left\langle2,1,0\right\rangle}\right\Vert,6^{3}>$ $=$ $\dfric<\left\Vert{\left\langle\tm2,4,\tm4\right\rangle}\right\Vert,6^{3}>$ $=$ $\dfric<6,6^{3}>$ $=$ $\dfric<1,36>$

[(02-21-17)]

[1] Let $\vr(t) = \left<\t2 ,\fric<4\sqrt{2},3>\t{}^{\fric<3,2>} ,2t\right>$. Calculate the curvature at the point $t$ $=$ $2.$

[(02-22-17)]

[3] A projectile is launched with an initial speed of 100 feet per second at an angle of $\fric<\pi,6>$ to the horizontal. Assume that the only force acting on the object is gravity.)

[2]()

Find the initial velocity vector.


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$\vv_{{}_{0}}$ $=$ $50\sqrt{3}\vi + 50\vj$

Find the vector function that describes velocity.

$\va(t) = \tm32\vj$

$\vv(t) = \tm32t\vj + \vv(0)$

$\vv(t) = \tm32t\vj + 50\sqrt{3}\vi + 50\vj$

$\vv(t) = 50\sqrt{3}\vi + (\tm32t + 50)\vj$

Find the vector function that describes motion.

$\vs(t) = 50\sqrt{3}t\vi + (\tm16\t2 + 50t)\vj + \vs(0)$

$\vs(t) = 50\sqrt{3}t\vi + (\tm16\t2 + 50t)\vj$

Find the maximum height.

$\vv(t) = 50\sqrt{3}\vi + (\tm32t + 50)\vj$

$\tm32t + 50 = 0$

$t = \fric<25,16>$

$\vs(t) = 50\sqrt{3}t\vi + (\tm16\t2 + 50t)\vj$

$\tm16\left(\fric<25,16>\right)^{2}
+ 50\left(\fric<25,16>\right)$ $=$ $\fric<625,16>$

$\fric<625,16>$ ft

Find the horizontal range.

$\vs(t) = 50\sqrt{3}t\vi + (\tm16\t2 + 50t)\vj$

$\tm16\t2 + 50t$ $=$ 0

$t(\tm16t + 50)$ $=$ 0

$t$ $=$ $\fric<25,8>$

$50\sqrt{3}\cdot\fric<25,8>$ $=$ $\fric<625\sqrt{3},4>$

Find the speed of impact.

$\vv(t) = 50\sqrt{3}\vi + (\tm32t + 50)\vj$

$\vv\left(\fric<25,8>\right)$ $=$ $50\sqrt{3}\vi + (\tm32\left(\fric<25,8>\right) + 50)\vj$

$\left\Vert\vv\left(\fric<25,8>\right)\right\Vert$ $=$ $100$

100 ft/sec.

[(03-17-17)]

[3] For the following, find all local extrema and saddle points.


\begin{multicols}{2}
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\(f(x,y) = \fric<2,3>\x3 + x\y2 + 3\x2 + \y2 \)
\par
\...
...m16\)
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Saddle point: \(\left(\tm1,2,\fric<7,3>\right)\)
\par
\end{multicols}

[2] Find the absolute extrema of the function $f(x,y)$ $=$ $\x2 \e y;$ on the region in $\Rss2 $ bounded by the unit circle.

$f_{{}_{x}}(x,y)$ $=$ $2x\e y;$

$f_{{}_{y}}(x,y)$ $=$ $\x2 \e y;$

Critical Points: $(0,y)$ for all $y\in[\tm1,1]$

$f(0,y)$ $=$ 0

Boundary Points: All points on the unit circle.

$\x2 + \y2 $ $=$ $1$

$x = \sqrt{1 - \y2 }$

$f(\sqrt{1 - \y2 },y)$ $=$ $(1 - \y2 )\e y;$

Consider

$g(y)$ $=$ $(1 - \y2 )\e y;$ on $[\tm1,1]$

$g'(y)$ $=$ $\tm2y\e y; + (1 - \y2 )\e y;$ $=$ $\tm\e y;(\y2 + 2y - 1)$

To find the critical points of $g$, solve:

$\y2 + 2y - 1$ $=$ 0

$y$ $=$ $\fric<\tm2\pmo\sqrt{4 - 4(1)(\tm1)},2>$ $=$ $\fric<\tm2\pmo2\sqrt{2},2>$ $=$ $\tm 1 \pmo \sqrt{2}$

Note that $\tm 1 - \sqrt{2}\notin[\tm1,1]$. So $y$ $=$ $\sqrt{2} -1.$

$\y2 $ $=$ $\left(\sqrt{2} - 1\right)^{2}$ $=$ $3 - 2\sqrt{2}$

$\x2 $ $=$ $1 - \y2 $ $=$ $1 - (3 - 2\sqrt{2})$ $=$ $2\sqrt{2} - 2$ $=$ $2(\sqrt{2} - 1)$

$x$ $=$ $\pms\sqrt{2(\sqrt{2} - 1)}$

Absolute min: 0 at $(0,y)$ for all $y\in[\tm1,1]$

Absolute max: $2(\sqrt{2} - 1)\e\sqrt{2} - 1;$ at $\left(\tm\sqrt{2(\sqrt{2} - 1)},\sqrt{2} - 1 \right)$ and $\left(\sqrt{2(\sqrt{2} - 1)},\sqrt{2} - 1 \right)$

[1] Find the linearization of $f(x,y) = 2\x3 - xy + \y2 ;$ at the point $(1,\tm1,4)$. Use it to approximate $f(.99,\tm1.01)$.

$f_{{}_{x}}(x,y)$ $=$ $6\x2 - y$

$f_{{}_{y}}(x,y) = \tm x + 2y$

$f_{{}_{x}}(1,\tm1)$ $=$ $7$

$f_{{}_{y}}(1,\tm1)$ $=$ $\tm3$

Linearization: $L(x,y) = 7(x - 1) - 3(y + 1) + 4$

$f(.99,\tm1.01)$ $\approx$ $L(.99,\tm1.01)$ $=$ $7(.99 - 1) - 3(\tm1.01 + 1) + 4$ $=$ $\tm.07 + .03 + 4$ $=$ $3.96$

[3] Find the maximum value of the function $f(x,y)$ $=$ $\x2 y$ subject to $\x2 + y$ $=$ $1$.

[2]()

Let $g(x,y)$ $=$ $\x2 + y - 1.$

$\nabla f(x,y) = \left<2xy,\x2 \right>$

$\nabla g(x,y) = \left<2x,1\right>$

$2xy$ $=$ $\lambda 2x$

$x$ $=$ 0 or $\lambda$ $=$ $y$

$f(0,y)$ $=$ 0

Suppose $\lambda$ $=$ $y.$

Then $\x2 $ $=$ $\lambda$ $=$ $y.$

$\x2 + y$ $=$ $1$

$2y$ $=$ $1$

$y$ $=$ $\fric<1,2>$

$y$ $=$ $\lambda$ $=$ $\x2 $ $=$ $\fric<1,2>$

$\x2 $ $=$ $\fric<1,2>$

$x$ $=$ $\pms\sqrt{\fric<1,2>}$

$f\left(\sqrt{\fric<1,2>},\fric<1,2>\right)$ $=$ $\fric<1,4>$ (max)

$f\left(\tm\sqrt{\fric<1,2>},\fric<1,2>\right)$ $=$ $\fric<1,4>$ (max)

Explain why no such minimum exists.

By letting $y$ be a negative number with a large absolute value, $\x2 y$ can be made arbitrarily small. Below is a more precise explanation.

Suppose that $M\leq\tm 1$. Choose $x\in\R$ such that $\tm\x2 < M$ and let $y$ $=$ $1 - \x2 $. Then $\x2 + y$ $=$ $1$ and $f(x,y)$ $=$ $\x2 y$ $<$ $M.$

[2]( For each of the following, give the coordinates of the point in the the other two coordinate systems.)

[1] \(\text{%
Rectangular coordinates:
\(\left(\fric<3,2>,\fric<3\sqrt{3},2>,0\right)\)}\)

$\r2 $ $=$ $\x2 + \y2 $ $=$ $9$

$r$ $=$ $3$

$\rho^{2}$ $=$ $\x2 + \y2 + \z2 $ $=$ $9$

$\rho$ $=$ $3$

$x$ $=$ $r\cos\theta$

$\fric<3,2>$ $=$ $3\cos\theta$

$\cos\theta$ $=$ $\fric<1,2>$

$\theta$ $=$ $\fric<\pi,3>$

$\rho$ $=$ $r\sin\phi$

$3$ $=$ $3\sin\phi$

$\sin\phi$ $=$ $1$

$\phi$ $=$ $\fric<\pi,2>$

Cylindrical: $\left(3,\fric<\pi,3>,0\right)$

Spherical: $\left(3,\fric<\pi,3>,\fric<\pi,2>\right)$

[1] \(\text{%
Cylindrical coordinates:
\(\left(\fric<\sqrt{3},2>,\fric<\pi,6>,\fric<1,2>\right)\)}\)

$x$ $=$ $r\cos\theta$ $=$ $\fric<\sqrt{3},2>\cos\fric<\pi,6>$ $=$ $\fric<3,4>$

$y$ $=$ $r\sin\theta$ $=$ $\fric<\sqrt{3},2>\sin\fric<\pi,6>$ $=$ $\fric<\sqrt{3},4>$

$\rho^{2}$ $=$ $\r2 + \z2 $ $=$ $\fric<3,4> + \fric<1,4>$ $=$ $1$

$\rho$ $=$ $1$

$r$ $=$ $\rho\sin\phi$

$\fric<\sqrt{3},2>$ $=$ $\sin\phi$

$\phi$ $=$ $\fric<\pi,3>$

Rectangular: $\left(\fric<3,4>,\fric<\sqrt{3},4>,\fric<1,2>\right)$

Spherical: $\left(1,\fric<\pi,6>,\fric<\pi,3>\right)$

[1] Spherical cooridinates: $\left(4,\fric<\pi,3>,\fric<\pi,6>\right)$

$\rho$ $=$ $4$

$\theta$ $=$ $\fric<\pi,3>$

$\phi$ $=$ $\fric<\pi,6>$

$r$ $=$ $4\sin\fric<\pi,6>$ $=$ $2$

$z$ $=$ $4\cos\fric<\pi,6>$ $=$ $2\sqrt{3}$

$x$ $=$ $2\cos\fric<\pi,3>$ $=$ $1$

$y$ $=$ $2\sin\fric<\pi,3>$ $=$ $\sqrt{3}$

Rectangular: $\left(1,\sqrt{3},2\sqrt{3}\right)$

Cylindrical: $\left(2,\fric<\pi,3>,2\sqrt{3}\right)$

For the following, $T$ is the solid tetrahedron with vertices $(0,0,0),$ $(1,0,0),$ $(0,1,0),$ and $(0,0,1),$ $\R$ $=$ $\{(x,y,z)\in\Rss3 :\x2 + \y2 \leq 1,0\leq x,0\leq z\leq 3\},$ and $S$ is the region of $\Rss3 $ that lies above the the top half of the cone $\fric<\z2 ,3>$ $=$ $\x2 + \y2 $ and below the sphere \(\text{%
\(\x2 + \y2 + (z - \sqrt{3})^{2}\)
\(=\)
\(1\)}\). Evaluate the following.

[3]()

[1] $\diiintl T; \x2 y dV$

[1] $\diiintl R; \e{}^{\x2 + \y2 };dV$

[1] $\diiintl S; 1 dV$

(Exam 1)

Exam 1 Math 2673 Spring 2017

[10] Give the equation of the sphere with center $(1,\tm1,2)$ and containing the point $(2,1,0).$

$r$ $=$ $\sqrt{(1 - 2)^{2} + (\tm1 - 1)^{2} + (2 - 0)^{2}}$ $=$ $3$

$(x - 1)^{2} + (y + 1)^{2} + (z - 2)^{2}$ $=$ $9$

[10] Describe the region in $\Rss3 $ defined by the following inequality.

$1\leq \y2 + \z2 \leq 4$

Let $C_{{}_{1}}$ be the cylinder with base $\y2 + \z2 $ $=$ $1$ (circle centered at $(0,0)$ with radius 1) in the $yz$-plane and $C_{{}_{2}}$ be the cylinder with base $\y2 + \z2 $ $=$ $4$ (circle centered at $(0,0)$ with radius 2) in the $yz$-plane. The region defined by the inequality is the set of points that are either on the surface of $C_{{}_{2}}$ or in the interior of $C_{{}_{2}}$ but are not in the interior of $C_{{}_{1}}.$

[10] Two force vectors are given below. Find the resultant force vector.


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$(10\sqrt{3}\vi + 10\vj) + (\tm8\vi + 8\sqrt{3}\vj)$ $=$ $(10\sqrt{3} - 8)\vi + (8\sqrt{3} + 10)\vj$ $=$ $\left\langle10\sqrt{3} - 8,8\sqrt{3} + 10\right\rangle$

[10] Determine whether the vectors given below are parallel, perpendicular, or neither.

$\left\langle1,2,\tm2\right\rangle$ $\left\langle2,3,4\right\rangle$

Perpendicular.

$\left\langle1,2,\tm2\right\rangle\cdot\left\langle2,3,4\right\rangle$ $=$ $2 + 6 - 8$ $=$ 0

Let $\vv$ $=$ $\left\langle1,1,2\right\rangle$ and $\vw$ $=$ $\left\langle3,1,3\right\rangle.$

[2]()

[10] Find comp$_{{}_{\vv}}\vw.$

comp$_{{}_{\vv}}\vw$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert>$ $=$ $\fric<10,\sqrt{6}>$

[10] Find proj$_{{}_{\vv}}\vw.$

proj$_{{}_{\vv}}\vw$ $=$ $\fricp<\vv\cdot\vw,\Vert\vv\Vert^{2}> \vv$ $=$ $\fric<5,3> \left\langle1,1,2\right\rangle$

[10] Find the direction cosines of $\left\langle\tm4,2,\sqrt{5}\right\rangle.$

$\left\Vert\left\langle\tm4,2,\sqrt{5}\right\rangle\right\Vert$ $=$ $5$

[3]()

$\cos\alpha$ $=$ $\fric<\tm4,5>$

$\cos\beta$ $=$ $\fric<2,5>$

$\cos\gamma$ $=$ $\fric<\sqrt{5},5>$

[10] Find the volume of the parallelepiped determined by $\left\langle1,2,4\right\rangle,$ $\left\langle0,\tm2,1\right\rangle,$ and $\left\langle1,1,3\right\rangle.$

$\left\vert\left\vert\hspace{-4pt}\begin{tabular}{rrr}
\vspace{-12pt}\\
1 & 2 & 4\\
0 & -2 & 1\\
1 & 1 & 3
\end{tabular}\hspace{-4pt}\right\vert\right\vert$ $=$ $\left\vert 1(\tm6 - 1) - 2(0 - 1) + 4(0 - \tm2)\right\vert$ $=$ $3$

[10] Find the equation of the plane containing the point $(1,\tm1,3)$ with normal vector $\vn = \langle1,\tm2,1\rangle.$

$1(x - 1) - 2(y + 1) + 1(z - 3)$ $=$ 0

[10] Give the equation of the line containing the points $(\tm1,2,0)$ and $(\tm4,1,2).$

[4]()

$\vv$ $=$ $\left\langle3,1,\tm2\right\rangle$

$x$ $=$ $\tm1 + 3t$

$y$ $=$ $2 + t$

$z$ $=$ $0 - 2t$

Identify and describe each of the following.

[10] $\fric<x,2>$ $=$ $\fric<\y2 ,9> + \fric<\z2 ,4>$

This is an elliptic paraboloid whose axis is the positive $x$-axis.

[10] $\x2 - \fric<\y2 ,4> + \fric<\z2 ,3>$ $=$ $2$

This is a hyperboloid of one sheet whose axis is the $y$-axis.

(Exam 2)

Exam 2 Math 2673 Spring 2017

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

[2]( [10] Sketch the curve of the given vector equation. Draw arrows along the curve in the direction of increasing $t.$)

$\vr(t)$ $=$ $\left<\e t;,t\right>$

$x$ $=$ $\e y;$

$y$ $=$ $\ln x$


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[10] A projectile is launched due east with an initial speed of 20 feet per second at an angle of $\fric<\pi,3>$ to the horizontal. Further, suppose that the acceleration due to gravity and wind blowing due south is given by $\va(t)$ $=$ $\left<0,\tm20,\tm32\right>$. Give the vector function that represents the projectile's position at time $t.$

[2]()


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$\vv_{{}_{0}}$ $=$ $\left<10,0,10\sqrt{3}\right>$

$\va(t)$ $=$ $\left<0,\tm20,\tm32\right>$

$\vv(t)$

$=$ $\left<0,\tm20t,\tm32t\right> + \vv_{{}_{0}}$

$=$ $\left<10,\tm20t,\tm32t + 10\sqrt{3}\right>$

$\vs(t)$

$=$ $\left<10t,\tm10\t2 ,\tm16\t2 + 10\sqrt{3}t\right> + \vs_{{}_{0}}$

$=$ $\left<10t,\tm10\t2 ,\tm16\t2 + 10\sqrt{3}t\right> + \vzero$

$=$ $\left<10t,\tm10\t2 ,\tm16\t2 + 10\sqrt{3}t\right>$

Let $\vr(t)$ $=$ $\left<\cos t,\sin t,\sqrt{3}t\right>.$

[10] Find $\vr'(t).$

$\vr'(t)$ $=$ $\left<\tm\sin t,\cos t,\sqrt{3}\right>$

[10] Find the unit tangent vector $\vT(t)$.

$\Vert\vr'(t)\Vert$ $=$ $\sqrt{\sin^{2}t + \cos^{2}t + 3}$ $=$ $2$

$\vT(t)$ $=$ $\fric<1,2>\vr'(t)$ $=$ $\fric<1,2>\left<\tm\sin t,\cos t,\sqrt{3}\right>$

[10] Find the unit normal vector $\vN(t).$

$\vT'(t)$ $=$ $\fric<1,2>\left<\tm\cos t,\tm\sin t,0\right>$

$\Vert\vT'(t)\Vert$ $=$ $\fric<1,2>\sqrt{\cos^{2}t + \sin^{2}t + 0}$ $=$ $\fric<1,2>$

$\vN(t)$ $=$ $2\vT'(t)$ $=$ $=$ $\left<\tm\cos t,\tm\sin t,0\right>$

[10] Find the binormal vector $\vB(t).$

$\vB(t)$ $=$ $\vT(t) \times \vN(t)$ $=$ $\fric<1,2>\left<\tm\cos t,\tm\sin t,0\right>
\times \left<\tm\cos t,\tm\sin t,0\right>$ $=$ $\fric<1,2>\left<\sqrt{3}\sin t,\tm\sqrt{3}\cos t,1\right>$

[10] Find the curvature $\kappa(t).$

$\kappa(t)$ $=$ $\fric<\Vert\vT'(t)\Vert,\Vert\vr'(t)\Vert>$ $=$ $\dfric<{\fric<1,2>},2>$ $=$ $\fric<1,4>$

[10] Reparameterize the curve with respect to arc length beginning at the point $t$ $=$ $0.$

$s$ $=$ $\dintl0,t;\Vert\vr'(u)\Vert du$ $=$ $\dintl0,t;2du$ $=$ $2t$

$t$ $=$ $\fric<1,2>s$

$\vr(t)$ $=$ $\vr\left(\fric<1,2>s\right)$ $=$ $\left<\cos\fric<1,2>s,\sin\fric<1,2>s,\fric<\sqrt{3}s,2>\right>$

[10] Give the line that is tangent to $\vr(t)$ at the point $t$ $=$ $\fric<\pi,2>.$

$\vr\left(\fric<\pi,2>\right)$ $=$ $\left<0,1,\fric<\sqrt{3}\pi,2>\right>$

$\vr'\left(\fric<\pi,2>\right)$ $=$ $\left<\tm1,0,\sqrt{3}\right>$

$\vl(s)$ $=$ $\left<0,1,\fric<\sqrt{3}\pi,2>\right> + s\left<\tm1,0,\sqrt{3}\right>$

[2]( For each of the following, find the limit or prove that it does not exist.)

[10] $\limv{(x,y)},{(0,0)};\dfric<x\y3 ,\x4 + \y4 >$

$\limv{(x,0)},{(0,0)};\dfric<x\cdot0,\x4 + 0>$

$=$ $\limv{(x,0)},{(0,0)};\dfric<0,\x4 >$

$=$ 0

$\limv{(x,x)},{(0,0)};\dfric<x\cdot\x3 ,\x4 + \x4 >$

$=$ $\limv{(x,x)},{(0,0)};\dfric<\x4 ,2\x4 >$

$=$ $\fric<1,2>$

Limit DNE.

[10] $\limv{(x,y)},{(\pi,\pi)};\e{}^{\sin x};\cos y$

$=$ $\e{}^{\sin\pi};\cos\pi$

$\tm1$

[10] $\limv{(x,y)},{(0,0)};\dfric<\sqrt{\x2 + \y2 + 1} - 1,\x2 + \y2 >$

$=$ $\limv{(x,y)},{(0,0)};\dfric<\sqrt{\x2 + \y2 + 1} - 1,\x2 + \y2 >
\cdot
\dfric<\sqrt{\x2 + \y2 + 1} + 1,\sqrt{\x2 + \y2 + 1} + 1>$

$=$ $\limv{(x,y)},{(0,0)};\dfric<\x2 + \y2 + 1 - 1,(\x2 + \y2 )\left(\sqrt{\x2 + \y2 + 1} + 1\right)>$

$=$ $\limv{(x,y)},{(0,0)};\dfric<1,\sqrt{\x2 + \y2 + 1} + 1>$

$=$ $\fric<1,2>$

Let $f(x,y,z) = \x2 \e{}^{xy}; + z\sin y$. Calculate each of the following.

[2]()

[10] $f_{{}_{x}}(x,y,z)$ $=$ $2x \e{}^{xy}; + \x2 y\e{}^{xy};$

[10] $f_{{}_{y}}(x,y,z)$ $=$ $\x3 \e{}^{xy}; + z\cos y$

[10] $f_{{}_{z}}(x,y,z)$ $=$ $\sin y$

[10] $f_{{}_{xyz}}(x,y,z)$

$=$ $f_{{}_{zxy}}(x,y,z)$

$f_{{}_{z}}(x,y,z)$ $=$ $\sin y$

$f_{{}_{zx}}(x,y,z)$ $=$ 0

$f_{{}_{zxy}}(x,y,z)$ $=$ 0

(Exam 3)

Exam 3 Math 2673 Spring 2017

[10] For the function below, find all local extrema and saddle points.

$f(x,y)$ $=$ $\x2 y + \x2 - y$

$f_{{}_{x}}(x,y)$ $=$ $2x y + 2x$ $=$ $2x(y + 1)$

$f_{{}_{y}}(x,y)$ $=$ $\x2 - 1$ $=$ $(x + 1)(x - 1)$

$f_{{}_{xx}}(x,y)$ $=$ $y$

$f_{{}_{yy}}(x,y)$ $=$ 0

$f_{{}_{xy}}(x,y)$ $2x$

$D(x,y)$ $=$ $\tm4\x2 $

CP: $(\tm1,\tm1)$ $(1,\tm1)$

$D(\tm1,\tm1)$ $=$ $\tm4$

$D(1,1)$ $=$ $\tm4$

Saddle Points: $(\tm1,\tm1,1),(1,\tm1,1)$

Maximize and minimize $f$ subject to the constraint $2\x2 + 3\y2 $ $=$ $1.$

Let $g(x)$ $=$ $2\x2 + 3\y2 - 1$

$\nabla f(x,y)$ $=$ $\langle2x y + 2x,\x2 - 1\rangle$

$\nabla g(x,y)$ $=$ $\langle4x,6y\rangle$

$\nabla f(x,y)$ $=$ $\lambda\nabla g(x,y)$

$2x(y + 1)$ $=$ $4\lambda x$

So $x$ $=$ 0 or $y + 1$ $=$ $2\lambda.$

If $x$ $=$ $0,$ then $y$ $=$ $\pms\fric<\sqrt{3},3>.$

$f\left(0,\fric<\sqrt{3},3>\right)$ $=$ $\tm\fric<\sqrt{3},3>$

$f\left(0,\tm\fric<\sqrt{3},3>\right)$ $=$ $\fric<\sqrt{3},3>$

If $x$ $\ne$ $0,$ then $y + 1$ $=$ $2\lambda.$

$\lambda$ $=$ $\fric<y + 1,2>$

$\x2 - 1$ $=$ $6\lambda y$

$\lambda$ $=$ $\fric<\x2 - 1,6y>$

$\fric<y + 1,2>$ $=$ $\fric<\x2 - 1,6y>$

$6y(y + 1)$ $=$ $2(\x2 - 1)$

$6\y2 + 6y$ $=$ $2\x2 - 2$

$6\y2 + 6y$ $=$ $1 - 3\y2 - 2$

$9\y2 + 6y + 1$ $=$ 0

$(3y + 1)^{2}$ $=$ 0

$y$ $=$ $\tm\fric<1,3>$

$x$ $=$ $\pms\fric<\sqrt{3},3>$

$f\left(\tm\fric<\sqrt{3},3>,\tm\fric<1,3>\right)$ $=$ $\fric<5,9>$

$f\left(\fric<\sqrt{3},3>,\tm\fric<1,3>\right)$ $=$ $\fric<5,9>$

$f\left(0,\fric<\sqrt{3},3>\right)$ $=$ $\tm\fric<\sqrt{3},3>$ (min)

$f\left(0,\tm\fric<\sqrt{3},3>\right)$ $=$ $\fric<\sqrt{3},3>$ (max)

[10] Maximize and minimize $f(x,y)$ $=$ $\e xy;$ subject to the constraint $\x2 + \y2 $ $=$ $2.$

Let $g(x)$ $=$ $\x2 + \y2 - 2$

$\nabla f(x,y)$ $=$ $\langle y\e xy;,x\e xy;\rangle$

$\nabla g(x,y)$ $=$ $\langle2x,2y\rangle$

$\nabla f(x,y)$ $=$ $\lambda\nabla g(x,y)$

$y\e xy;$ $=$ $2\lambda x$

$x\e xy;$ $=$ $2\lambda y$

$\lambda$ $=$ $\fric<x\e xy;,2y>$ $=$ $\fric<y\e xy;,2x>$

$\x2 $ $=$ $\y2 $

$\x2 + \y2 $ $=$ $2$

$2\x2 $ $=$ $2$

$\x2 $ $=$ $1$

$x$ $=$ $\pms1$

$f(\tm1,\tm1)$ $=$ $f(1,1)$ $=$ $e$ (max)

$f(\tm1,1)$ $=$ $f(1,\tm1)$ $=$ $\fric<1,e>$ (min)

[10] Let $z$ $=$ $\x2 + 3\y2 $. Use differentials to approximate the change in $z$ when $(x,y)$ changes from $(2,1)$ to $(2.01,.99).$

$dz$ $=$ $2xdx + 6ydy$

$dz$ $=$ $4(.01) + 6(\tm.01)$ $=$ $\tm.02$

[10] Let $z$ $=$ $\x3 \y2 $ where $x$ $=$ $\sin t$ and $y$ $=$ $\e t;$. Find $\der t z;.$

$\der t z;$ $=$ $\parti x z;\der t x;
+ \parti y z;\der t y;$ $=$ $3\x2 \y2 (\cos t) + 2\x3 y \e t;$ $=$ $3\e2t;\sin^{2}t\cdot\cos t + 2\e2t;\sin^{3}t$

[10] For the function below, find the directional derivative of the function at the indicated point in the direction of the given vector. Recall: A unit vector is required.

$g(x,y)$ $=$ $\e x;\sin y,$ $\left(0,\fric<\pi,6>\right),$ $\vv$ $=$ $\langle3,4\rangle$

Let $\vu$ $=$ $\langle\fric<3,5>,\fric<4,5>\rangle$

$\nabla g(x,y)$ $=$ $\langle\e x;\sin y,\e x;\cos y\rangle$

$\nabla g\left(0,\fric<\pi,6>\right)$ $=$ $\left\langle\fric<1,2>,\fric<\sqrt{3},2>\right\rangle$

$\left\langle\fric<1,2>,\fric<\sqrt{3},2>\right\rangle\cdot
\langle\fric<3,5>,\fric<4,5>\rangle$

$\fric<3,10> + \fric<4\sqrt{3},10>$ $=$ $\fric<3 + 4\sqrt{3},10>$

Evaluate each of the following integrals.

[10] $\diintlp R;x\e y;dA,$ where $\R$ $=$ $[0,1]\times[0,2]$

$\diintlp R;x\e y;dA$ $=$ $\dintlp 0,1;{\dintlp 0,2;x\e y;dy}dx$ $=$ $\dintl 0,1;\left({\biggmlp 0,2;x\e y;p}\right)dx$ $=$ $\dintl0,1;(\e2; - 1)xdx$ $=$ $\biggml0,1;\fric<1,2>(\e2; - 1)\x2 p$ $=$ $\fric<1,2>(\e2; - 1)$

[10] $\diintlp R;\x2 + y dA,$ where $\R$ is the region bounded by the triangle with vertices $(0,0),$ $(1,0),$ and $(1,1)$

$\diintlp R;\x2 + y dA$ $=$ $\dintl0,1;\left({\dintlp0,x;\x2 + ydy}\right)dx$ $=$ $\dintl0,1;\left({\biggmlp0,x;\x2 y+ \fric<1,2>\y2 p}\right)dx$ $=$ $\dintlp0,1; \x3 + \fric<1,2>\x2 dx$ $=$ $\biggmlp0,1; \fric<1,4>\x4 + \fric<1,6>\x3 p$ $=$ $\fric<1,4> + \fric<1,6>$ $=$ $\fric<5,12>$

[10] $\diintlp R;\x2 y dA,$ where $\R$ is the region bounded by the unit circle

$\diintlp R;\x2 y dA$ $=$ $\dintl0,2\pi;\left({\dintl0,1;r\left(r\cos\theta\right)^{2}r\sin\theta dr}\right)d\theta$ $=$ $\dintl0,2\pi;{\dintl0,1;\r4 \cos^{2}\theta\sin\theta dr}d\theta$ $=$ \(\text{\(\dintl0,2\pi;{\biggml0,1;\fric<1,5>\r5 \cos^{2}\theta\sin\theta p}d\theta\)}\) $=$ $\dintl0,2\pi;\fric<1,5>\cos^{2}\theta\sin\theta d\theta$ $=$ $\biggml0,2\pi;\tm\fric<1,15>\cos^{3}\theta p$ $\tm\fric<1,15> - \tm\fric<1,15>$ $=$ 0

(Final)

Final Exam Math 2673 Spring 2017

[10] Find the resultant force of the two vectors pictured below.


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[10] A right triangle has vertices $(1,2,\tm3)$ and $(2,\tm4,1)$. Give a possible third vertex. Note: There is more than one correct answer.

[10] Give the equation of the plane containing the points $(2,1,2), (3,4,8),$ $(1,\tm2,1)$.

Let $\vr(t)$ $=$ $\left\langle\t2 ,2t, \ln t\right\rangle$.

[10] Give an equation of the tangent line (in any form) to the curve at the point $(1,2,0).$

[10] Give the arc length of the curve from $t$ $=$ $1$ to $t$ $=$ $2.$

[10] Find the curvature at the point $(1,2,0)$. Hint: Use the cross product.

[2]( [10] For each of the following, find the limit or prove that it does not exist.)

[10] $\limv{(x,y)},{(0,0)};\dfric<\x2 + \y2 ,x\y2 + 1>$

[10] $\limv{(x,y)},{(0,0)};\dfric<\x2 y + \x2 ,x\y2 - x>$

[10] Find the directional derivative of the given function at the indicated point in the direction of the given vector.

$g(x,y,z)$ $=$ $\e x;y + \cos z,$ $\left(1,2,\fric<\pi,2>\right),$ $\vu$ $=$ $\left\langle\fric<2,7>,\fric<3,7>,\fric<6,7>\right\rangle$

[10] Find all local extrema and saddle points of the function $f(x,y)$ $=$ $\x2 \y2 - \x2 + \y2 $.

[10] Maximize and minimize $f(x,y)$ $=$ $\e xy;$ subject to the constraint $\x2 + \y2 $ $=$ $1.$

[10] Let $T$ be the triangle in $\Rss2 $ with vertices $(0,0), (4,1),$ $(0,3)$ and evaluate $\diintl T;1 dA.$

[10] Let $\R$ $=$ $\{(x,y)\in\Rss2 :\x2 + \y2 \leq 1,0\leq x,0\leq y\}$ and evaluate $\diintl R;\sqrt{\x2 + \y2 }dA.$

Let $S$ be the sphere with center $(0,0,0)$ and radius 1 (the unit sphere) and
$f(x,y,z)$ $=$ $\sqrt[4]{\x2 + \y2 + \z2 }.$

[10] Express (but do not evaluate) the integral $\diiintl S;f(x,y,z)dV$ as an iterated integral using rectangular coordinates.

[10] Express (but do not evaluate) the integral $\diiintl S;f(x,y,z)dV$ as an iterated integral using cylindrical coordinates.

[10] Express (but do not evaluate) the integral $\diiintl S;f(x,y,z)dV$ as an iterated integral using spherical coordinates.

[10] Evaluate $\diiintl S;f(x,y,z)dV$ using one of the iterated integrals above.

(Math 2673 Fall 2017)

(Quiz)

[08-29-17]

[1] Give the equation of the sphere with center $(2,5,\tm1)$ and the following tangent plane.

[2]()

$y$ $=$ 0

This is the $xz$-plane.

The distance between $(2,5,\tm1)$ and the $xz$-plane is $5.$

$(x - 2)^{2} + (y - 5)^{2} + (z + 1)^{2}$ $=$ $25$

$x$ $=$ 0

This is the $yz$-plane.

The distance between $(2,5,\tm1)$ and the $yz$-plane is $2.$

$(x - 2)^{2} + (y - 5)^{2} + (z + 1)^{2}$ $=$ $4$

[1] Describe the region in space being defined.

$\x2 + \y2 \leq 4,$ $0\leq z\leq 5$

This is the solid region bounded by a right circular cylinder. The base of the cylinder is the circle in the $xy$-plane centered at the origin with radius 2. The height of the cylinder is 5.

[09-01-17]

[1] The norm of the vector pictured below is 4. Give the component form of the vector.

[2]()


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$\vv$ $=$ $\left\langle v_{{}_{1}},v_{{}_{2}}\right\rangle$

$\sin30^{\circ}$ $=$ $\fric<v_{{}_{2}},\Vert\vv\Vert>$

$\fric<1,2>$ $=$ $\fric<v_{{}_{2}},4>$

$v_{{}_{2}}$ $=$ $2$

$v_{{}_{1}}^{2} + v_{{}_{2}}^{2}$ $=$ $4^{2}$

$v_{{}_{1}}^{2} + 4$ $=$ $16$

$v_{{}_{1}}$ $=$ $\tm2\sqrt{3}$

$\vv$ $=$ $\left\langle\tm2\sqrt{3},2\right\rangle$


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$\vv$ $=$ $\left\langle v_{{}_{1}},v_{{}_{2}}\right\rangle$

$\sin60^{\circ}$ $=$ $\fric<v_{{}_{2}},\Vert\vv\Vert>$

$\fric<\sqrt{3},2>$ $=$ $\fric<v_{{}_{2}},4>$

$v_{{}_{2}}$ $=$ $2\sqrt{3}$

$v_{{}_{1}}^{2} + v_{{}_{2}}^{2}$ $=$ $4^{2}$

$v_{{}_{1}}^{2} + 12$ $=$ $16$

$v_{{}_{1}}$ $=$ $\tm2$

$\vv$ $=$ $\left\langle\tm2,2\sqrt{3}\right\rangle$

[1] Give an example of three numbers $v_{{}_{1}},$ $v_{{}_{2}},$ and $v_{{}_{3}}$ such that $v_{{}_{1}}$ $=$ $v_{{}_{2}}$ $=$ $v_{{}_{3}}$ and $\Vert\left\langle v_{{}_{1}},v_{{}_{2}},v_{{}_{3}}\right\rangle\Vert$ $=$ $1.$

Note that $\Vert\langle1,1,1\rangle\Vert$ $=$ $\sqrt{3}$. So let $v_{{}_{1}}$ $=$ $v_{{}_{2}}$ $=$ $v_{{}_{3}}$ $=$ $\fric<1,\sqrt{3}>.$

[09-08-17]

[1] Find a vector orthogonal to both $\langle1,1,0\rangle$ and $\langle0,1,1\rangle.$

$\langle1,\tm 1,1\rangle$

To verify, calculate the dot products.

[2] A wagon is pulled 50 ft by exerting a force of 10 lb on the handle at an angle of $\fric<\pi,6>$ with the horizontal. How much work is done?


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$\vF$ $=$ $5\sqrt{3}\vi + 5\vj$

$W$ $=$ $\vF\cdot50\vi$ ft-lb

$W$ $=$ $(5\sqrt{3}\vi + 5\vj)\cdot100\vi$ ft-lb

$W$ $=$ $500\sqrt{3}$ ft-lb

Let $\vu$ $=$ $\left\langle1,\tm1,0\right\rangle,$ $\vv$ $=$ $\left\langle1,0,\tm3\right\rangle,$ and $\vw$ $=$ $\left\langle\tm2,1,3\right\rangle.$

[1] Calculate $\vv\times\vw.$

$\vi$ $\vj$ $\vk$
   
1 0 -3
   
-2 1 3
$=$ $\langle3,3,1\rangle$

[1] Find the volume of the parallelepiped determined by the vectors $\vu,$ $\vv,$ and $\vw.$

$\vert\vu\cdot(\vv\times\vw)\vert$ $=$ $\langle1,\tm1,0\rangle\cdot \langle3,3,1\rangle$ $=$ 0

The vectors are coplanar.

[10-4-17]

Let $\mathcal{C}$ be the curve determined by the vector function $\vr(t)$ $=$ $\left\langle\ln t, \fric<1,2>\t2 ,\sqrt{2}t\right\rangle.$

[2] Give the equations of the normal plane and osculating plane of $\mathcal{C}$ at the point $\left(\ln\sqrt{2},1,2\right).$

$\vr'(t)$ $=$ $\left\langle\fric<1,t>, t,\sqrt{2}\right\rangle$

$\Vert\vr'(t)\Vert$ $=$ $\sqrt{\fric<1,\t2 > + \t2 + 2}$ $=$ $\sqrt{\fric<\t4 + 2\t2 + 1,\t2 >}$ $=$ $\sqrt{\fric<(\t2 + 1)^{2},\t2 >}$ $=$ $\fric<\t2 + 1,t>$ $=$ $t + \fric<1,t>$

$\vT(t)$ $=$ $\fric<t,\t2 + 1>\left\langle\fric<1,t>, t,\sqrt{2}\right\rangle$

$\vT'(t)$ $=$ $\fric<1(\t2 + 1) - t(2t),(\t2 + 1)^{2}>
\left\langle\fric<1,t>, t,\sqrt{2}\right\rangle
+ \fric<t,\t2 + 1>\left\langle\tm\fric<1,\t2 >, 1,0\right\rangle$ $=$ $\fric<\tm\t2 + 1,(\t2 + 1)^{2}>
\left\langle\fric<1,t>, t,\sqrt{2}\right\rangle
+ \fric<t,\t2 + 1>\left\langle\tm\fric<1,\t2 >, 1,0\right\rangle$

$\vr'\left(\sqrt{2}\right)$ $=$ $\left\langle\fric<1,\sqrt{2}>, \sqrt{2},\sqrt{2}\right\rangle$

$\left\Vert\vr'\left(\sqrt{2}\right)\right\Vert$ $=$ $\fric<3,\sqrt{2}>$

$\vT\left(\sqrt{2}\right)$ $=$ $\fric<\sqrt{2},3>\left\langle\fric<1,\sqrt{2}>, \sqrt{2},\sqrt{2}\right\rangle$ $=$ $\fric<1,3>\left\langle1, 2,2\right\rangle$

$\vT'\left(\sqrt{2}\right)$ $=$ $\tm\fric<1,9>\left\langle\fric<1,\sqrt{2}>,\sqrt{2},\sqrt{2}\right\rangle
+ \fric<\sqrt{2},3>\left\langle\tm\fric<1,2>, 1,0\right\rangle$ $=$ $\tm\fric<\sqrt{2},9>\left\langle2,\tm2,1\right\rangle$

$\left\Vert\vT'\left(\sqrt{2}\right)\right\Vert$ $=$ $\fric<\sqrt{2},3>$

$\vN\left(\sqrt{2}\right)$ $=$ $\left(\fric<3,\sqrt{2}>\cdot\tm\fric<\sqrt{2},9>\right)\left\langle2,\tm2,1\right\rangle$ $=$ $\tm\fric<1,3>\left\langle2,\tm2,1\right\rangle$

$\vB\left(\sqrt{2}\right)$ $=$ $\vT\left(\sqrt{2}\right)\times \vN\left(\sqrt{2}\right)$ $=$ $\tm\fric<1,3>\left\langle2,1,\tm2\right\rangle$

Normal Plane:

$\left(x - \ln\sqrt{2}\right) + 2\left(y - 1\right) + 2\left(z - 2\right)$ $=$ 0

Osculating Plane:

$2\left(x - \ln\sqrt{2}\right) + \left(y - 1\right) - 2\left(z - 2\right)$ $=$ 0

[1] Find the curvature of $\mathcal{C}$ at the point $\left(\ln\sqrt{2},1,2\right).$

$\kappa\left(\sqrt{2}\right)$ $=$ $\fric<\left\Vert\vT'\left(\sqrt{2}\right)\right\Vert,\left\Vert\vr'\left(\sqrt{2}\right)\right\Vert>$ $=$ $\dfric<{\fric<\sqrt{2},3>},{\fric<3,\sqrt{2}>}>$ $=$ $\fric<2,9>$

[1] Find the length of the arc from the point $\left(0,\fric<1,2>,\sqrt{2}\right)$ to the point $\left(\ln2\sqrt{2},4,4\right)$ on $\mathcal{C}.$

$\dintl1,2\sqrt{2};\left\Vert\vr'(t)\right\Vert dt$ $=$ $\dintlp1,2\sqrt{2};t + \fric<1,t>dt$ $=$ $\biggmlp1,2\sqrt{2};\fric<1,2>\t2 + \ln\vert t\vert p$ $=$ $4 + \ln2\sqrt{2} - \left(\fric<1,2> + 0\right)$ $=$ $\fric<7,2> + \ln2\sqrt{2}$

[10-6-17]

(Page 895: 21) [2] A ball is thrown eastward into the air from the origin (in the direction of the positive $x$-axis). The initial velocity is $50\vi + 80\vk,$ with speed measured in feet per second. The spin of the ball results in a southward acceleration of 4 ft/ sec$^{2},$ so the acceleration vector is $\va$ $=$ $\tm4\vj - 32 \vk$. Where does the ball land?

[2] Find the linearization of $f(x,y) = 2\x3 \e y;$ at the point $(2,0,16)$. Use it to approximate $f(1.99,.01)$.

$f_{{}_{x}}(x,y)$ $=$ $6\x2 \e y;$

$f_{{}_{y}}(x,y)$ $=$ $2\x3 \e y;$

$f_{{}_{x}}(2,0)$ $=$ $24$

$f_{{}_{y}}(2,0)$ $=$ $16$

Linearization: $L(x,y)$ $=$ $24(x - 2) + 16y + 16$

$f(1.99,.01)$ $\approx$ $L(1.99,.01)$ $=$ $24(1.99 - 2) + 16(.01) + 16$ $=$ $\tm.24 + .16 + 16$ $=$ $15.92$

[1] Let $f(x,y)$ $=$ $\e x;\cos y$. Calculate $f'_{{}_{\left(0,\frac{\pi}{6}\right)}}(x,y).$

$f_{{}_{x}}(x,y)$ $=$ $\e x;\cos y$

$f_{{}_{y}}(x,y)$ $=$ $\tm\e x;\sin y$

$f_{{}_{x}}\left(0,\frac{\pi}{6}\right)$ $=$ $\fric<\sqrt{3},2>$

$f_{{}_{y}}\left(0,\frac{\pi}{6}\right)$ $=$ $\tm\fric<1,2>$

$f'_{{}_{\left(0,\frac{\pi}{6}\right)}}(x,y)$ $=$ $\fric<\sqrt{3},2>x - \fric<1,2>y$

Let $f(x,y)$ $=$ $\e y;\sin x$. Calculate $f'_{{}_{\left(\frac{\pi}{6},0\right)}}(x,y).$

$f_{{}_{x}}(x,y)$ $=$ $\e y;\cos x$

$f_{{}_{y}}(x,y)$ $=$ $\e y;\sin x$

$f_{{}_{x}}\left(\frac{\pi}{6},0\right)$ $=$ $\fric<\sqrt{3},2>$

$f_{{}_{y}}\left(\frac{\pi}{6},0\right)$ $=$ $\fric<1,2>$

$f'_{{}_{\left(\frac{\pi}{6},0\right)}}(x,y)$ $=$ $\fric<\sqrt{3},2>x + \fric<1,2>y$

Let $z$ $=$ $xy + y$ where $x$ $=$ $\t2 - 3$ and $y$ $=$ $2t - 6$.

[1] Use the chain rule to find $\der t z;$.

$\der t z;$ $=$ $\parti x z;\der t x;
+ \parti y z;\der t y;$ $=$ $2ty + 2(x + 1)$ $=$ $2t(2t - 6) + 2(\t2 - 3 + 1)$ $=$ $6\t2 -12t - 4$

[1] Express $z$ as function of $t$ and then find $\der t z;$.

$z$ $=$ $xy + y$ $=$ $(\t2 - 3)(2t - 6) + (2t - 6)$ $=$ $2\t3 - 6\t2 - 4t + 12$

$\der t z;$ $=$ $6\t2 -12t - 4$

[1] Let $f(x,y,z)$ $=$ $x\sin y + \e x + z;$. Find the directional derivative of $f$ at ${(1,0,\tm1)}$ in the direction of $\left\langle\fric<\sqrt{2},2>,\fric<1,2>,\fric<1,2>\right\rangle.$

$\nabla f(x,y,z)$ $=$ $\left\langle \sin y + \e x + z;,x\cos y,\e x + z; \right\rangle$

$\nabla f(1,0,\tm1)$ $=$ $\left\langle 1,1,1\right\rangle$

$\left\langle 1,1,1\right\rangle
\cdot
\left\langle\fric<\sqrt{2},2>,\fric<1,2>,\fric<1,2>\right\rangle$

$=$ $\fric<\sqrt{2},2> + \fric<1,2> + \fric<1,2>$ $=$ $\fric<\sqrt{2} + 2,2>$

[2] Let $R = [0,1]\times[0,1]$ and calculate $\diintlp R;2xy\e\x2 ; dA.$

$\dintl 0,1;{\dintl 0,1;2xy\e\x2 ; dx}dy$

$=$ $\dintl 0,1;{\biggml 0,1;y\e\x2 ; p}dy$

$=$ $\dintl 0,1; y(e - 1)dy$

$=$ $\biggml 0,1; \fric<1,2>\y2 (e - 1)p$

$=$ $\fric<1,2>(e - 1)$

[2] Let $T$ be the triangle with vertices $(0,0),$ $\left(\sqrt{\fric<\pi,2>},0\right),$ and $\left(\sqrt{\fric<\pi,2>},\sqrt{\fric<\pi,2>}\right)$ and calculate $\diintlp T;\cos\x2 dA.$

$\dintl 0,\sqrt{\fric<\pi,2>};{\dintl 0,x;\cos\x2 dy}dx$

$=$ $\dintl 0,\sqrt{\fric<\pi,2>};\biggml 0,x;y\cos\x2 pdx$

$=$ $\dintl 0,\sqrt{\fric<\pi,2>};x\cos\x2 dx$

$=$ $\biggml 0,\sqrt{\fric<\pi,2>};\fric<1,2>\sin\x2 p$

$=$ $\fric<1,2>$

[2] Let $\R$ be the region in the $xy$-plane that is inside the circle centered at $(0,0)$ with radius 2 and outside the circle centered at $(0,0)$ with radius 1 and calculate $\diintlp R;\sqrt{\x2 + \y2 + 1}dA.$

[2]()

$\dintl 0,2\pi;{\dintl 1,2;r\sqrt{\r2 + 1}dr}d\theta$

$=$ $\dintl 0,2\pi;{\biggml 1,2;\fric<1,3>\left(\r2 + 1\right)^{{}^{\fric<3,2>}}p}d\theta$

$=$ $\dintl 0,2\pi;\fric<1,3>\left(5\sqrt{5} - 2\sqrt{2}\right)d\theta$

$=$ $\biggml 0,2\pi;\fric<1,3>\left(5\sqrt{5} - 2\sqrt{2}\right)\theta p$

$=$ $\fric<2\pi,3>\left(5\sqrt{5} - 2\sqrt{2}\right)$

[2] Evaluate the following integral. Hint: Use polar coordinates.

$\dintl \tm2,2;{\dintl 0,\sqrt{4 - \y2 };
\left(\x2 + \y2 \right)^{{}^{\fric<3,2>}}dx}dy$

Note that $\dintl \tm2,2;{\dintl 0,\sqrt{4 - \y2 };
\left(\x2 + \y2 \right)^{{}^{\fric<3,2>}}dx}dy$ $=$ $\diintl R;\left(\x2 + \y2 \right)^{{}^{\fric<3,2>}}dA$ where $\R$ is the right half of the circle centered at $(0,0)$ with radius $2.$

[2]()

$\dintl \tm2,2;{\dintl 0,\sqrt{4 - \y2 };
\left(\x2 + \y2 \right)^{{}^{\fric<3,2>}}dx}dy$

$=$ $\dintl\tm{\fric<\pi,2>},{\fric<\pi,2>};{\dintl 0,2;%
r\left(\r2 \right)^{{}^{\fric<3,2>}}dr}d\theta$

$=$ $\dintl\tm{\fric<\pi,2>},{\fric<\pi,2>};{\dintl 0,2;\r4 dr}d\theta$

$=$ $\dintl\tm{\fric<\pi,2>},{\fric<\pi,2>};{\biggml 0,2;\fric<1,5>\r5 p}d\theta$

$=$ $\dintl\tm{\fric<\pi,2>},{\fric<\pi,2>};\fric<32,5>d\theta$

$=$ $\biggml\tm{\fric<\pi,2>},{\fric<\pi,2>};\fric<32,5>\theta p$

$=$ $\fric<32\pi,5>$

(Exam 1)

Exam 1 Math 2673 Fall 2017

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Consider the points $P(2,0,1),$ $Q(5,1,0),$ and $R(1,2,2).$

[10] Give the equation(s) of the line (in any form) that passes through the points $P$ and $Q.$

$\vv$ $=$ $\langle3,1,\tm1\rangle$

$\vr(t)$ $=$ $\langle2,0,1\rangle + t\langle3,1,\tm1\rangle$

[10] Give the equation of the plane that contains the points $P,$ $Q,$ and $R.$

$\vw$ $=$ $\langle\tm4,1,2\rangle$

$\vn$ $=$ $\vv\times\vw$ $=$ $\langle3,\tm2,7\rangle$

[10] Give the equation for the set of all points that are equidistant from the points $(1,0,\tm1)$ and $(2,4,1).$

$(x - 1)^{2} + \y2 + (z + 1)^{2}$ $=$ $(x - 2)^{2} + (y - 4)^{2} + (z - 1)^{2}$

$\x2 - 2x + 1 + \y2 + \z2 + 2z + 1$ $=$ $\x2 - 4x + 4 + \y2 - 8y + 16 + \z2 - 2z + 1$

$2x + 8y + 4z$ $=$ $19$

[10] Are the following two vectors parallel, perpendicular, or neither?

$\vv$ $=$ $\langle1,\tm2,3\rangle$

$\vw$ $=$ $\langle2,\tm4,\tm5\rangle$.

No. If $\vw = \alpha\vv,$ then $2 = \alpha\cdot1$ which implies that $\alpha = 2$ and $\tm5 = \alpha\cdot3$ which implies that $\alpha = \tm\fric<5,3>$. Therefore, no such $\alpha$ exists.

[10] Give the direction cosinses of the following vector.

$\langle\tm3, 0, 4\rangle$

$\Vert\langle\tm3, 0, 4\rangle\Vert$ $=$ $5$

$\cos\alpha$ $=$ $\tm\fric<3,5>$.

$\cos\beta$ $=$ 0

$\cos\gamma$ $=$ $\fric<4,5>$.

[10] Find the work done by a force of $\langle2,1,\tm6\rangle$ that moves an object from $(0,0,0)$ to $(4,5,\tm2)$ along a straight line. Distance is measured in feet and force in pounds.

$\vD$ $=$ $\langle4,5,\tm2\rangle$

$\vF$ $=$ $\langle2,1,\tm6\rangle$

$\vD\cdot\vF$ $=$ $25$

[10] Determine whether the given pairs of lines are parallel, skew, or intersecting. If they intersect, find the point of intersection.

[2]()

$x$ $=$ $1$

$y$ $=$ $1 + 2t$

$z$ $=$ $1 - 3t$

$x$ $=$ $2 + s$

$y$ $=$ $\tm1 - 4s$

$z$ $=$ $3 + 5s$

[2]()

$2 + s$ $=$ $x$

$2 + s$ $=$ $1$

$s$ $=$ $\tm1$

$x$ $=$ $1$

$y$ $=$ $3$

$z$ $=$ $\tm2$

Check that $(1,3,\tm2)$ lies on both lines.

Describe or sketch the given surface.

[10] $\tm\x2 + \y2 + \z2 $ $=$ $1$

This is a hyperboloid of one sheet with $x$-axis as its central axis.

[10] $\x2 + \y2 $ $=$ $1$

This is a circular cylinder with radius 1 and having the $z$-axis as the central axis.

[10] $y$ $=$ $\x2 + 2\z2 $

This is an elliptic paraboloid having the $y$-axis as the central axis.

Let $\vr(t)$ $=$ $\langle\e t;,\cos t,\sin t\rangle.$

[2]()

[10] Sketch the graph of $\vr.$

[10] Calculate the derivative of $\vr.$

$\vr'(t)$ $=$ $\langle\e t;,\tm\sin t,\cos t\rangle.$

(Exam 2)

Exam 2 Math 2673 Fall 2017

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $\mathcal{C}$ be the curve defined by $\vr(t)$ $=$ $\langle\cos t,\sin t,2\sqrt{2}t\rangle.$

[2]()

[10] \(\text{Find the unit tangent vector \(\vT(t).\)}\)

$\vr'(t)$ $=$ $\langle\tm\sin t,\cos t,2\sqrt{2}\rangle$

$\Vert\vr'(t)\Vert$ $=$ $\sqrt{\sin^{2}t + \cos^{2}t + 8}$ $=$ $3$

$\vT(t)$ $=$ $\fric<1,3>\langle\tm\sin t,\cos t,2\sqrt{2}\rangle$

[10] \(\text{Find the unit normal vector \(\vN(t).\)}\)

$\vT'(t)$ $=$ $\fric<1,3>\langle\tm\cos t,\tm\sin t,0\rangle$

$\Vert\vT'(t)\Vert$ $=$ $\fric<1,3>$

\(\text{
\(\vN(t)\)
\(=\)
\(3\cdot\fric<1,3>\langle\tm\cos t,\tm\sin t,0\rangle\)
\(=\)
\(\langle\tm\cos t,\tm\sin t,0\rangle\)}\)

[10] Find the binormal vector $\vB(t).$

$\vB(t)$

$=$ $\vT(t) \times \vN(t)$

$=$ $\fric<1,3>\langle\tm\cos t,\sin t,2\sqrt{2}\rangle\times
\langle\tm\cos t,\tm\sin t,0\rangle$

$=$ $\fric<1,3>\langle2\sqrt{2}\sin t,\tm2\sqrt{2}\cos t,1\rangle$

[10] Find the curvature $\kappa(t).$

$\kappa(t)$ $=$ $\fric<\left\Vert\vT'(t)\right\Vert,\left\vert\vr'(t)\right\Vert>$ $=$ $\fric<{\fric<1,3>},3>$ $=$ $\fric<1,9>$

[10] Reparameterize the curve with respect to arc length beginning at the point $t$ $=$ $0.$

$s$ $=$ $\dintl0,t;3du$ $=$ $\biggml0,t;3u p$ $=$ $3t$

$t$ $=$ $\fric<1,3>s$

$\vr\left(\fric<1,3>s\right)$ $=$ $\left\langle\cos\left(\fric<1,3>s\right),
\sin\left(\fric<1,3>s\right),
2\sqrt{2}\left(\fric<1,3>s\right)\right\rangle$ $=$ $\left\langle\cos\left(\fric<1,3>s\right),
\sin\left(\fric<1,3>s\right),
\fric<2\sqrt{2},3>s\right\rangle$

[10] A ball is thrown into the air from the origin so that its initial velocity is $25\vi + 35\vk$. Length is measured in feet and time is measured in seconds. The acceleration vector is \(\text{%
\(\va\)
\(=\)
\(5\vj - 32 \vk\)}\). Find the vector function that describes motion.

$\va(t)$ $=$ $\left\langle0,5,\tm32\right\rangle$

$\vv(t)$ $=$ $\left\langle0,5t,\tm32t\right\rangle + \langle25,0,35\rangle$ $=$ $\left\langle25,5t,\tm32t + 35\right\rangle$

$\vs(t)$ $=$ $\left\langle25t,\fric<5,2>\t2 ,\tm16\t2 + 35t\right\rangle$

[10] Give the domain and range of the following function.

$f(x,y,z)$ $=$ $\dfric<\x2 - \y2 ,\sqrt{z}>$

Domain: $\{(x,y,z)\in\Rss3 : z\geq 0\}$

Range: $\R$

For $a\geq 0,$ $f(\sqrt{a},0,1)$ $=$ $a.$

For $a < 0,$ $f(0,\sqrt{\tm a},1)$ $=$ $a.$

[10] Sketch or describe the graph of the following function.

$f(x,y)$ $=$ $\sqrt{\x2 + \y2 }$

$z$ $=$ $\sqrt{\x2 + \y2 }$

$\z2 $ $=$ $\x2 + \y2 $

Top half of the cone $\z2 $ $=$ $\x2 + \y2 .$

[2]( Calculate the following limits.)

[10] $\limv{(x,y)},{(0,1)};\dfric<y,\x2 - \y2 >$ $=$ $\tm1$

[10] $\limv{(x,y)},{(0,0)};\dfric<y,\x2 - \y2 >$

$\limv{(x,0)},{(0,0)};\dfric<y,\x2 - \y2 >$ $=$ $\limv{(x,0)},{(0,0)};\dfric<0,\x2 - 0>$ $=$ 0

\(\text{%
\(\limv{(0,y)},{(0,0)};\dfric<y,\x2 - \y2 >\)
\(=\)
\(\limv{(0,y)},{(0,0)};\dfric<y,\tm\y2 >\)
\(=\)
\(\limv{(0,y)},{(0,0)};\tm\dfric<1,y>\)}\)

$\limv{(0,y)},{(0,0)};\tm\dfric<1,y>$ does not exist

$\limv{(x,y)},{(0,0)};\dfric<y,\x2 - \y2 >$ does not exist

Let $f(x,y)$ $=$ $\x2 y + \x2 - y.$

[2]( [10] Find all local extrema and saddle points.)

$f_{{}_{x}}(x,y)$ $=$ $2x y + 2x$ $=$ $2x(y + 1)$

$f_{{}_{y}}(x,y)$ $=$ $\x2 - 1$ $=$ $(x + 1)(x - 1)$

Critical points: $(\tm1,\tm1),$ $(1,\tm1)$

$f_{{}_{xx}}(x,y)$ $=$ $2y + 2$

$f_{{}_{xy}}(x,y)$ $=$ $2x$

$f_{{}_{yy}}(x,y)$ $=$ 0

$D(x,y)$ $=$ $\tm4\x2 $

$D(\tm1,\tm1)$ $=$ $\tm4$

$D(1,\tm1)$ $=$ $\tm4$

Saddle Points: $(\tm1,\tm1,1),$ $(1,\tm1,1)$

[10] Find the absolute extrema of $f$ on the region in $\Rss2 $ bounded by the unit circle.

The critical points $(\tm1,\tm1)$ and $(1,\tm1)$ are not in the region.

Boundary points:

$\x2 + \y2 $ $=$ $1$

$\x2 $ $=$ $1 - \y2 $

$x$ $=$ $\pms\sqrt{1 - \y2 }$

$f\left(\sqrt{1 - \y2 },y\right)$ $=$ $f\left(\tm\sqrt{1 - \y2 },y\right)$ $=$ $(1 - \y2 )y + 1 - \y2 - y$ $=$ $\tm\y3 - \y2 + 1$

Consider $g(y)$ $=$ $\tm\y3 - \y2 + 1$ on $[\tm1,1].$

$g(\tm1)$ $=$ $f(0,\tm1)$ $=$ $1$

$g(1)$ $=$ $f(0,1)$ $=$ $\tm1$

$g'(y)$ $=$ $\tm3\y2 - 2y$ $=$ $\tm y(3y + 2)$

$g(0)$ $=$ $f(1,0)$ $=$ $f(\tm1,0)$ $=$ $1$

$g\left(\tm\fric<2,3>\right)$ $=$ $f\left(\fric<\sqrt{5},3>,\tm\fric<2,3>\right)$ $=$ $f\left(\tm\fric<\sqrt{5},3>,\tm\fric<2,3>\right)$ $=$ $\fric<23,27>$

So $f$ attains its maximum value of $1$ at the points $(0,\tm1),$ $(\tm1,0),$ and $(1,0)$ and its minimum value of $\tm1$ at the point $(0,1).$

Total Points:

(Exam 3)

Exam 3 Math 2673 Fall 2017

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $f(x,y,z)$ $=$ $\x2 \y3 + xz.$

[10] Calculate the gradient of $f$ at the point $(1,\tm1,3).$

$\nabla f(x,y,z)$ $=$ $\langle2x\y3 + z,3\x2 \y2 ,x\rangle$

$\nabla f(1,\tm1,3)$ $=$ $\langle1,3,1\rangle$

[10] Give the linear approximation of $f$ at the point $(1,\tm1,3)$ and use it to approximate $f(1.01,\tm1.01,2.99).$

$L(x,y,z)$ $=$ $1(x - 1) + 3(y + 1) + 1(z - 3) + 2$

$L(1.01,\tm1.01,2.99)$ $=$ $.01 - .03 - .01 + 2$ $=$ $1.97$

[10] Give the directional derivative of $f$ at the point $(1,\tm1,3)$ in the direction of $\left\langle\fric<\sqrt{2},4>,\fric<\sqrt{2},4>,\fric<\sqrt{3},2>\right\rangle.$

$\langle1,3,1\rangle\cdot
\left\langle\fric<\sqrt{2},4>,\fric<\sqrt{2},4>,\fric<\sqrt{3},2>\right\rangle$ $=$ $\fric<\sqrt{2},4> + \fric<3\sqrt{2},4> + \fric<\sqrt{3},2>$ $=$ $\sqrt{2} + \fric<\sqrt{3},2>$

[10] Find an equation for the normal line to the level surface $f(x,y,z)$ $=$ $2$ at the point $(1,\tm1,3).$

$\vr(t)$ $=$ $\langle1,\tm1,3\rangle + t\langle1,3,1\rangle$

[10] Give the derivative of $f$ at the point $(a,b,c).$

$f'_{{}_{(a,b,c)}}(x,y,z)$ $=$ $\nabla f(a,b,c)\cdot\langle x,y,z\rangle$ $=$ $\langle2a\b3 + c,3\a2 \b2 ,a\rangle\cdot\langle x,y,z\rangle$ $=$ $(2a\b3 + c)x + 3\a2 \b2 y + az$

[10] Let $z$ $=$ $x\e y;$ where $x$ $=$ $\sin t$ and $y$ $=$ $\sqrt{t}$. Find $\der t z;.$

$\der t z;$ $=$ $\parti x z;\der t x;
+ \parti y z;\der t y;$ $=$ $\e y;\cos t + x\e y;\cdot\fric<1,2>\t{}^{\tm\fric<1,2>} $ $=$ $\e\sqrt{t};\cos t + \fric<\e\sqrt{t};\sin t,2\sqrt{t}>$

Let $z$ $=$ $x\e y;$ where $x$ $=$ $\s2 + t$ and $y$ $=$ $st$.

[10] Find $\parti s z;.$

$\parti s z;$ $=$ $\parti x z;\parti s x; + \parti y z;\parti s y;$ $=$ $\e y;\cdot 2s + x\e y;\cdot t$ $=$ $2s\e st; + t(\s2 + t)\e st;$

[10] Find $\parti t z; .$

$\parti t z;$ $=$ $\parti x z;\parti t x; + \parti y z;\parti t y;$ $=$ $\e y;\cdot 1 + x\e y;\cdot s$ $=$ $\e st; + s(\s2 + t)\e st;$

[10] Maximize and minimize $f(x,y)$ $=$ $\x2 + \y2 $ subject to the constraint $x + \y2 $ $=$ $\fric<9,2>.$

[2]()

Let $g(x)$ $=$ $x + \y2 - \fric<9,2>.$

$\nabla f(x,y)$ $=$ $\langle2x,2y\rangle$

$\nabla g(x,y)$ $=$ $\langle 1,2y\rangle$

$\nabla f(x,y)$ $=$ $\lambda\nabla g(x,y)$

$\langle2x,2y\rangle$ $=$ $\lambda\langle 1,2y\rangle$

$2x$ $=$ $\lambda$ and $2y$ $=$ $2y\lambda$

So $y$ $=$ 0 or $\lambda$ $=$ $1.$

If $y$ $=$ $0,$ then $x$ $=$ $\fric<9,2>.$

$f\left(\fric<9,2>,0\right)$ $=$ $\fric<81,4>$

If $\lambda$ $=$ $1,$ then $x$ $=$ $\fric<1,2>$ and $y$ $=$ $\pms 2.$

$f\left(\fric<1,2>,\tm 2\right)$ $=$ $\fric<17,4>$ (min)

$f\left(\fric<1,2>,2\right)$ $=$ $\fric<17,4>$ (min)

$f\left(\fric<9,2>,0\right)$ $=$ $\fric<81,4>$ (max)

[10] Maximize and minimize $f(x,y,z)$ $=$ $xy + yz$ subject to the constraints \(\text{\(\x2 + \y2 \)
\(=\)
\(1\)}\) and $yz$ $=$ $1.$

[2]()

Let $g(x,y,z)$ $=$ $\x2 + \y2 - 1$

and

$h(x,y,z)$ $=$ $xy - 1.$

$\nabla f(x,y,z)$ $=$ $\langle y,x + z,y\rangle$

$\nabla g(x,y,z)$ $=$ $\langle 2x,2y,0\rangle$

$\nabla h(x,y,z)$ $=$ $\langle 0,z,y\rangle$

$\nabla f(x,y,z)$ $=$ $\lambda\nabla g(x,y,z) + \mu\nabla h(x,y,z)$

$\langle y,x + z,y\rangle$ $=$ $\lambda\langle 2x,2y,0\rangle + \mu\langle 0,z,y\rangle$

$y$ $=$ $2x\lambda,$ $x + z$ $=$ $2y\lambda + z\mu,$ $y$ $=$ $y\mu$

Since $y$ $=$ $y\mu,$ $y$ $=$ 0 or $\mu$ $=$ $1.$

However, $yz$ $=$ $1,$ so $y$ $\ne$ $0.$

Therefore, $\mu$ $=$ $1.$

$x + z$ $=$ $2y\lambda + z\mu$

$x + z$ $=$ $2y\lambda + z\cdot 1$

$x$ $=$ $2y\lambda$

$\lambda$ $=$ $\fric<x,2y>$

$y$ $=$ $2x\lambda$

$\lambda$ $=$ $\fric<y,2x>$

$\fric<x,2y>$ $=$ $\fric<y,2x>$

$\x2 $ $=$ $\y2 $

$\x2 + \y2 $ $=$ $1$

$2\x2 $ $=$ $1$

$x$ $=$ $\pms\fric<\sqrt{2},2>$

$y$ $=$ $\pms\fric<\sqrt{2},2>$

$yz$ $=$ $1$

$f\left(\tm\fric<\sqrt{2},2>,\tm\fric<\sqrt{2},2>,\tm\sqrt{2}\right)$ $=$ $\fric<3,2>$ (max)

$f\left(\tm\fric<\sqrt{2},2>,\fric<\sqrt{2},2>,\sqrt{2}\right)$ $=$ $\fric<1,2>$ (min)

$f\left(\fric<\sqrt{2},2>,\tm\fric<\sqrt{2},2>,\tm\sqrt{2}\right)$ $=$ $\fric<1,2>$ (min)

$f\left(\fric<\sqrt{2},2>,\fric<\sqrt{2},2>,\sqrt{2}\right)$ $=$ $\fric<3,2>$ (max)

Alternatively:

Since $xy$ $=$ $1,$ $f(x,y,z)$ $=$ $xy + 1$. So let $g(x,y)$ $=$ $xy + 1$ and maximize and minimize $g$ subject to the constraint $\x2 + \y2 $ $=$ $1.$

[2]()

Let $h(x)$ $=$ $\x2 + \y2 - 1.$

$\nabla g(x,y)$ $=$ $\langle y,x\rangle$

$\nabla h(x,y,z)$ $=$ $\langle2x,2y\rangle$

$\nabla g(x,y)$ $=$ $\lambda\nabla h(x,y)$

$y$ $=$ $2x\lambda,$ $x$ $=$ $2y\lambda$

As above, $x$ $=$ $\pms\fric<\sqrt{2},2>$ and $y$ $=$ $\pms\fric<\sqrt{2},2>.$

$g\left(\tm\fric<\sqrt{2},2>,\tm\fric<\sqrt{2},2>\right)$ $=$ $\fric<3,2>$ (max)

$g\left(\tm\fric<\sqrt{2},2>,\fric<\sqrt{2},2>\right)$ $=$ $\fric<1,2>$ (min)

$g\left(\fric<\sqrt{2},2>,\tm\fric<\sqrt{2},2>\right)$ $=$ $\fric<1,2>$ (min)

$g\left(\fric<\sqrt{2},2>,\fric<\sqrt{2},2>\right)$ $=$ $\fric<3,2>$ (max)

(Final)

Final Exam Math 2673 Fall 2017

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $\vu$ $=$ $\langle1,1,\tm1\rangle,$ $\vv$ $=$ $\langle1,0,\tm1\rangle,$ and $P$ $=$ $(3,\tm2,1).$

[10] Find the angle between $\vu$ and $\vv.$

[10] Give the equation of the line through $P$ in the direction of $\vu.$

[10] Give the equation of the plane that contains $P,$ $\vu,$ and $\vv.$

Let $\vr(t)$ $=$ $\left\langle t ,2\sqrt{2t}, \ln t\right\rangle$.

[10] Give an equation of the tangent line (in any form) to the curve at the point $\left(\e 2;,2\sqrt{2}e,2\right).$

[10] Give the arc length of the curve from $t$ $=$ $1$ to $t$ $=$ $e.$

[10] A ball is thrown into the air from the origin so that its initial velocity is $20\vi + 15\vk$. Length is measured in feet and time is measured in seconds. The acceleration vector is \(\text{%
\(\va\)
\(=\)
\(10\vj - 32 \vk\)}\). Find the vector function that describes motion.

Sketch or describe the following surfaces.

[10] $\x2 + \fric<\y2 ,9> + \fric<\z2 ,4>$ $=$ $1$

[10] $\z2 $ $=$ $\x2 + \y2 $

[10] $\tm\x2 + \y2 - \z2 $ $=$ $1$

Let $f(x,y)$ $=$ $\x2 + xy$.

[10] Show that $f$ is differentiable at $(a,b)$ for all $(a,b)\in\Rss2 .$

[10] Find $f'_{{}_{(a,b)}}(x,y).$







[10] Maximize and minimize $f(x,y,z)$ $=$ $\x2 + 2yz$ subject to the constraint \(\text{\(2x + 2y + 2z\)
\(=\)
\(1.\)}\)

Calculate $\diintlp R;4xy\sqrt{\x2 + \y2 }dA$ for each of the following.

[2]()

[10] $\R$ $=$ $[0,1]\times[0,1]$

[10] $\R$ is the unit circle

[10] Let $\R$ be the region bounded by the planes $x$ $=$ $0,$ $y$ $=$ $0,$ $z$ $=$ $0,$ and \(\text{\(x + y + z\)
\(=\)
\(1\)}\) and calculate $\diiintl R; \x2 y dV.$

[10] Let $\R$ be the region in the first octant bounded by the unit sphere and calculate $\diiintl R; e^{{}^{\left(\x2 + \y2 + \z2 \right)^{{}^{\frac{3}{2}}}}}dV.$

Final Exam Math 2673 Fall 2017

Name:

Directions: Show all of your work and justify all of your answers. An answer without justification will receive a zero.

Let $\vu$ $=$ $\langle1,1,\tm1\rangle,$ $\vv$ $=$ $\langle1,0,1\rangle,$ and $P$ $=$ $(1,\tm2,3).$

[10] Find the angle between $\vu$ and $\vv.$

[10] Give the equation of the line through $P$ in the direction of $\vu.$

[10] Give the equation of the plane that contains $P,$ $\vu,$ and $\vv.$

Let $\vr(t)$ $=$ $\left\langle t ,2\sqrt{2t}, \ln t\right\rangle$.

[10] Give an equation of the tangent line (in any form) to the curve at the point $\left(\e 2;,2\sqrt{2}e,2\right).$

[10] Give the arc length of the curve from $t$ $=$ $1$ to $t$ $=$ $e.$

[10] Calculate the curvature at the point $\left(\e 2;,2\sqrt{2}e,2\right).$

Sketch or describe the following surfaces.

[10] $\fric<\y2 ,9> + \fric<\z2 ,4>$ $=$ $1 - \x2 $

[10] $\z2 $ $=$ $\x2 - \y2 $

[10] $\x2 - \y2 - \z2 $ $=$ $1$

Let $f(x,y)$ $=$ $x + x\y2 $.

[10] Show that $f$ is differentiable at $(a,b)$ for all $(a,b)\in\Rss2 .$

[10] Find $f'_{{}_{(a,b)}}(x,y).$







[10] Maximize and minimize $f(x,y,z)$ $=$ $\x2 + 2yz$ subject to the constraint \(\text{\(2x + 2y + 2z\)
\(=\)
\(1.\)}\)

Calculate $\diintlp R;4xy\sqrt{\x2 + \y2 }dA$ for each of the following.

[2]()

[10] $\R$ $=$ $[0,1]\times[0,1]$

[10] $\R$ is the unit circle

[10] Let $\R$ be the region bounded by the planes $x$ $=$ $0,$ $y$ $=$ $0,$ $z$ $=$ $0,$ and \(\text{\(x + y + z\)
\(=\)
\(1\)}\) and calculate $\diiintl R; \x2 y dV.$

[10] Let $\R$ be the region in the first octant bounded by the unit sphere and calculate $\diiintl R;\sqrt{\left(\x2 + \y2 + \z2 \right)^{3}}dV.$

(Math 2673 Fall 2018)

(Quizzes)

[2] The bottom of the rectangular prism below lies in the $xy$-plane and its height is $7$. Give the coordinates of the eight corners.

[2]()


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\put(100,100){%
\begin{picture}(0,0)
\axestenbytenbyten...
...hed,arrows=*-*]%
(0,0)(-35.535533906,-35.535533906)
\end{picture}}
\end{picture}

$(0,0,0)$

$(5,0,0)$

$(0,2,0)$

$(5,2,0)$

$(0,0,7)$

$(5,0,7)$

$(0,2,7)$

$(5,2,7)$

The top of the rectangular prism below lies in the $xy$-plane and its height is $4$. Give the coordinates of the eight corners.

[2]()


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\put(100,100){%
\begin{picture}(0,0)
\axestenbytenbyten...
...d,arrows=*-*]%
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$(0,0,0)$

$(5,0,0)$

$(0,2,0)$

$(5,2,0)$

$(0,0,\tm4)$

$(5,0,\tm4)$

$(0,2,\tm4)$

$(5,2,\tm4)$

[1] Give the equation of the sphere with center $(\tm1,4,2)$ and radius $\sqrt{3}.$

$(x + 1)^{2} + (y - 4)^{2} + (z - 2)^{2}$ $=$ $3$

Give the equation of the sphere with center $(\tm1,4,3)$ and radius $\sqrt{2}.$

$(x + 1)^{2} + (y - 4)^{2} + (z - 3)^{2}$ $=$ $2$

[1]

[2]()

In the figure below, draw the vector $\vv - \vw$.


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...{\vector(3,-1){60}}
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In the figure below, draw the vector $\vw - \vv$.


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...){\vector(3,-1){60}}
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[2] A spider is in the southwest corner of a room that is $10\sqrt{3}$ ft by $10\sqrt{3}$ ft. The spider walks along a straight line that makes an angle of $\fric<\pi,6>$ radians with the south wall. The spider is walking at a constant rate of 2 ft/min. Where and when will the spider reach another wall? Hint: Give the position of the spider as a vector determined by time.


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\end{picture}}
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$\Vert\vv\Vert$ $=$ $2t$

$\vv$ $=$ $\left\langle\sqrt{3}t,t\right\rangle$

The spider reaches the east wall when $\sqrt{3}t$ $=$ $10\sqrt{3}$.

So after 10 minutes the spider's position relative to the southwest corner is $\langle10\sqrt{3},10\rangle$ which is on the east wall and 10 ft from the south wall.

[4] Let $\vx$ $=$ $\langle1, \tm1, 1\rangle,$ $\vy$ $=$ $\langle1, 1, 2\rangle,$ and $\vv$ $=$ $\langle3, 1, \tm2\rangle$. Also, let $\theta$ be the angle between $\vx$ and $\vv$. Find each of the following.

$\Vert\vx\Vert$ $=$ $\sqrt{3}$

$\Vert\vv\Vert$ $=$ $\sqrt{14}$

comp$_{{}_{\vv}}\vx$ $=$ $\fric<0,\sqrt{14}>$ $=$ 0

proj$_{{}_{\vv}}\vx$ $=$ $\vzero$

$\vx\times\vv$ $=$
$\vi$ $\vj$ $\vk$
   
$1$ $\tm1$ $1$
   
$3$ $1$ $\tm2$
$=$ $\langle1,5,4\rangle$

$\cos\theta$ $=$ $\fric<0,\sqrt{3}\sqrt{14}>$ $=$ 0

$\sin\theta$ $=$ $\fric<{\Vert\langle1,5,4\rangle\Vert},\sqrt{3}\sqrt{14}>$ $=$ $\fric<\sqrt{42},\sqrt{42}>$ $=$ $1$

The volume of the parallelpiped determined by the vectors $\vx,$ $\vy,$ and $\vv.$

$\left\Vert\hspace{-4pt}
\begin{tabular}{rrr}
\vspace{-12pt}\\
1 & -1 & 1\\
1 & 1 & 2\\
3 & 1 & -2
\end{tabular}\hspace{-4pt}\right\Vert$ $=$ $\vert 1(\tm2 - 2) + 1(\tm2 - 6) + 1(1 - 3)\vert$ $=$ $14$

[4] Let $\vx$ $=$ $\langle1, \tm1, 1\rangle,$ $\vy$ $=$ $\langle1, 1, 2\rangle,$ and $\vv$ $=$ $\langle3, 1, \tm2\rangle$. Also, let $\theta$ be the angle between $\vy$ and $\vv$. Find each of the following.

$\Vert\vy\Vert$ $=$ $\sqrt{6}$

$\Vert\vv\Vert$ $=$ $\sqrt{14}$

comp$_{{}_{\vv}}\vy$ $=$ $\fric<0,\sqrt{14}>$ $=$ 0

proj$_{{}_{\vv}}\vy$ $=$ $\vzero$

$\vy\times\vv$ $=$
$\vi$ $\vj$ $\vk$
   
$1$ $1$ $2$
   
$3$ $1$ $\tm2$
$=$ $\langle\tm4,8,\tm2\rangle$

$\cos\theta$ $\fric<0,\sqrt{6}\sqrt{14}>$ $=$ 0

$\sin\theta$ $=$ $\fric<{\Vert\langle\tm4,8,\tm2\rangle\Vert},\sqrt{6}\sqrt{14}>$ $=$ $\fric<\sqrt{84},\sqrt{84}>$ $=$ $1$

The volume of the parallelpiped determined by the vectors $\vx,$ $\vy,$ and $\vv.$

$\left\Vert\hspace{-4pt}
\begin{tabular}{rrr}
\vspace{-12pt}\\
1 & -1 & 1\\
1 & 1 & 2\\
3 & 1 & -2
\end{tabular}\hspace{-4pt}\right\Vert$ $=$ $\vert 1(\tm2 - 2) + 1(\tm2 - 6) + 1(1 - 3)\vert$ $=$ $14$

[2]()

Let $\vr(t)$ $=$ $\left\langle\sqrt{t} + 1,
\sin\left(\pi t\right),\fric<t,\e t;>\right\rangle.$

Let $\vr(t)$ $=$ $\left\langle\sqrt{t} + 1,
\tm\cos\left(\pi t\right),\fric<t,\e t;>\right\rangle.$

[1] Calculate $\limv t,0;\vr(t).$

[2]()

$\limv t,0;\vr(t)$

$=$ $\limv t,0;
\left\langle\sqrt{t} + 1,
\sin\left(\pi t\right),\fric<t,\e t;>\right\rangle$

$=$ $\langle1,0,0\rangle$

$\limv t,0;\vr(t)$

$=$ $\limv t,0;
\left\langle\sqrt{t} + 1,
\tm\cos\left(\pi t\right),\fric<t,\e t;>\right\rangle$

$=$ $\langle1,\tm1,0\rangle$

[1] Find $\vr'(t).$

[2]()

$\vr'(t)$

$=$ $\left\langle\fric<1,2\sqrt{t}>,
\pi\cos\left(\pi t\right),
\fric<\e t; - t\e t;,\e2t;>
\right\rangle$

$=$ $\left\langle\fric<1,2\sqrt{t}>,
\pi\cos\left(\pi t\right),
\fric<1 - t,\e t;>
\right\rangle$

$\vr'(t)$

$=$ $\left\langle\fric<1,2\sqrt{t}>,
\pi\sin\left(\pi t\right),
\fric<\e t; - t\e t;,\e2t;>
\right\rangle$

$=$ $\left\langle\fric<1,2\sqrt{t}>,
\pi\sin\left(\pi t\right),
\fric<1 - t,\e t;>
\right\rangle$

[1] Give the equation(s) of the line that is tangent to the graph of $\vr$ at the point where $t$ $=$ $4.$

[2]()

$\vr(4)$ $=$ $\left\langle3,0,\fric<4,\e4;>\right\rangle$

$\vr'(4)$ $=$ $\left\langle\fric<1,4>,\pi,\fric<\tm3,\e4;>\right\rangle$

$\vl(t)$ $=$ $\left\langle3,0,\fric<4,\e4;>\right\rangle
+ t \left\langle\fric<1,4>,\pi,\fric<\tm3,\e4;>\right\rangle$

$\vr(4)$ $=$ $\left\langle3,\tm1,\fric<4,\e4;>\right\rangle$

$\vr'(4)$ $=$ $\left\langle\fric<1,4>,0,\fric<\tm3,\e4;>\right\rangle$

$\vl(t)$ $=$ $\left\langle3,\tm1,\fric<4,\e4;>\right\rangle
+ t \left\langle\fric<1,4>,0,\fric<\tm3,\e4;>\right\rangle$

[1] Calculate the following integral.

[2]()

$\dintl0,1;\left\langle\cos(\pi t),\e t;,\fric<1,\t2 + 1>\right\rangle dt$

$=$ $\biggml0,1;\left\langle\fric<1,\pi>\sin(\pi t),\e t;,\inv\tan;t\right\rangle p$

$=$ $\left\langle0,e,\fric<\pi,4>\right\rangle - \left\langle0,1,0\right\rangle$

$=$ $\left\langle0,e - 1,\fric<\pi,4>\right\rangle$

Consider the vector function $\vr(t)$ $=$ $\left\langle\e t;,\sqrt{2}t,\e\tm t;\right\rangle$.

[2]()

[2] Calculate and simplify $\Vert\vr'(t)\Vert.$

$\vr'(t)$ $=$ $\left\langle\e t;,\sqrt{2},\tm\e\tm t;\right\rangle$

$\Vert\vr'(t)\Vert$

$=$ $\sqrt{\e2t; + 2 + \e\tm2t;}$

$=$ $\sqrt{\fric<\e4t; + 2\e2t; + 1,\e2t;>}$

$=$ $\sqrt{\fric<(\e2t; + 1)^{2},\e2t;>}$

$=$ $\fric<\e2t; + 1,\e t;>$

$=$ $\e t; + \e\tm t;$

[1] Give the arc length of the curve from $t$ $=$ 0 to $t$ $=$ $1.$

$\dintlp0,1;\e t; + \e\tm t;dt$

$=$ $\biggmlp0,1;\e t; - \e\tm t;p$

$=$ $e - \fric<1,e> - (1 - 1)$

$=$ $e - \fric<1,e>$

Let $\mathcal{C}$ be the curve defined by $\vr(t)$ $=$ $\left\langle\t2 , t, 1\right\rangle.$

[1] Find the unit tangent vector.

$\vr'(t)$ $=$ $\left\langle2t,1,0\right\rangle$

$\Vert\vr'(t)\Vert = \sqrt{4\t2 + 1}$

$\vT(t)$ $=$ $\fric<1,\sqrt{4\t2 + 1}>\left\langle2t , 1, 0\right\rangle$

[1] Find the derivative of the unit tangent vector.

$\vT(t)$ $=$ $\left(4\t2 + 1\right)^{{}^{\tm\fric<1,2>}}\left\langle2t,1,0\right\rangle$

$\vT'(t)$

$=$ $\tm\fric<1,2>\left(4\t2 + 1\right)^{{}^{\tm\fric<3,2>}}(8t)\left\langle2t,1,0\r...
...ngle
+ \left(4\t2 + 1\right)^{{}^{\tm\fric<1,2>}}\left\langle2,0,0\right\rangle$

$=$ $\fric<1,\sqrt{\left(4\t2 + 1\right)^{3}}>\left[
(\tm4t)\left\langle2t,1,0\right\rangle + (4\t2 + 1)\left\langle2,0,0\right\rangle\right]$

$=$ $\fric<2,\sqrt{\left(4\t2 + 1\right)^{3}}>\left[
\left\langle\tm4\t2 ,\tm2t,0\right\rangle + \left\langle4\t2 + 1,0,0\right\rangle\right]$

$=$ $\fric<2,\sqrt{\left(4\t2 + 1\right)^{3}}>\left\langle1,\tm2t,0\right\rangle$

[1] Find the curvature at the point $(4,2,1).$

$\vr'(t)$ $=$ $\left\langle2t,1,0\right\rangle$

$\vr''(t)$ $=$ $\left\langle2,0,0\right\rangle$

$\vr'(2)\times \vr''(2)$ $=$ $\langle4,1,0\rangle\times\langle2,0,0\rangle$ $=$ $\langle0,0,\tm2\rangle$

$\Vert\vr'(2)\times \vr''(2)\Vert$ $=$ $2$

$\Vert\vr'(2)\Vert$ $=$ $\sqrt{17}$

$\kappa(2)$ $=$ $\fric<2,17\sqrt{17}>$

A ball is thrown with an initial speed of 30 feet per second at an angle of $\fric<\pi,6>$ to the horizontal. Assume that the only force acting on the object is gravity.


\begin{multicols}{2}
\par
\subprob[1]
Find the initial velocity vector.
\par\s...
...m32\cdot\fric<15,16> + 15\vj\right\Vert\)
\par
\(=\)
\(30\)
\par
\end{multicols}

Let $f(x,y)$ $=$ $\x2 \sqrt[3]{y}.$

[1] Find the linearization of $f$ at the point $(10,125).$

$f_{{}_{x}}(x,y)$ $=$ $2x\sqrt[3]{y}$

$f_{{}_{y}}(x,y)$ $=$ $\x2 \fric<1,3>\y{}^{\tm\fric<2,3>} $ $=$ $\fric<\x2 ,3\sqrt[3]{\y2 }>$

$f(10,125)$ $=$ $500$

$f_{{}_{x}}(10,125)$ $=$ $100$

$f_{{}_{y}}(10,125)$ $=$ $\fric<4,3>$

$L(x,y)$ $=$ $500 + 100(x - 10) + \fric<4,3>(y - 125)$

[1] Use your answer from the previous part to estimate $(9.9)^{2}\sqrt[3]{126}.$

$f(9.9,126)$

$\approx$ $L(9.9,126)$

$=$ $500 + 100(9.9 - 10) + \fric<4,3>(126 - 125)$

$=$ $500 - 10 + \fric<4,3>$

$=$ $\fric<1474,3>$

Let $f(x,y)$ $=$ $x\y2 + y.$

[1] Assuming that $f$ is differentiable at $(a,b)$ for all $(a,b)\in\Rss2 ,$ find $f'_{{}_{(a,b)}}(x,y).$

$f_{{}_{x}}(x,y)$ $=$ $\y2 $

$f_{{}_{y}}(x,y)$ $=$ $2xy + 1$

$f'_{{}_{(a,b)}}(x,y)$ $=$ $\b2 x + (2ab + 1)y$

[2] Show that $f$ is differentiable at $(a,b)$ for all $(a,b)\in\Rss2 .$

$\limv{(h,k)},(0,0);\dfric<{f(a + h, b + k) - f(a,b)
- \left[f_{{}_{x}}(a,b)\cdot h + f_{{}_{y}}(a,b)\cdot k\right]},
\Vert(h,k)\Vert>$

$=$ $\limv{(h,k)},(0,0);\dfric<(a + h)(b + k)^{2} + (b + k) - (a\b2 + b)
- [\b2 h + (2ab + 1)k],\sqrt{\h2 + \k2 }>$

$=$ $\limv{(h,k)},(0,0);\dfric<(a + h)(\b2 + 2bk + \k2 ) + b + k - a\b2 - b
- \b2 h - 2abk - k,\sqrt{\h2 + \k2 }>$

$=$ $\limv{(h,k)},(0,0);\dfric<
a\b2 + 2abk + a\k2
+ \b2 h + 2bhk + h\k2
+ b + k - a\b2 - b
- \b2 h -2abk - k,\sqrt{\h2 + \k2 }>$

$=$ $\limv{(h,k)},(0,0);\dfric<
a\k2 + 2bhk + h\k2 ,\sqrt{\h2 + \k2 }>$

$\limv{(h,k)},(0,0);\left\vert\dfric<a\k2 + 2bhk + h\k2 ,\sqrt{\h2 + \k2 }>\right\vert$ $=$ $\limv{(h,k)},(0,0);\dfric<\vert a\k2 + 2bhk + h\k2 \vert,\sqrt{\h2 + \k2 }>$ $=$ $\limv{(h,k)},(0,0);\dfric<\vert k\vert\vert ak + 2bh + hk\vert,\sqrt{\h2 + \k2 }>$

\(\text{%
\(\leq\)
\(\limv{(h,k)},(0,0);\dfric<\vert k\vert\vert ak + 2bh + hk\v...
...t k\vert>\)
\(=\)
\(\limv{(h,k)},(0,0);\vert ak + 2bh + hk\vert\)
\(=\)
\(0\)}\)

Let $z$ $=$ $x\e y;.$

[1] Calculate $\parti x z;$ and $\parti y z;.$

$\parti x z;$ $=$ $\e y;$

$\parti y z;$ $=$ $x\e y;$

[1] Suppose that $x$ $=$ $\t2 $ and $y$ $=$ $\sin t$. Calculate $\der t z;.$

$\der t z;$ $=$ $\parti x z;\cdot\der t x; + \parti y z;\cdot\der t y;$ $=$ $\e y;(2t) + x\e y;\cos t$ $=$ $2t\e\sin t; + \t2 \e\sin t;\cos t$

[1] Suppose that $x$ $=$ $\s2 + t$ and $y$ $=$ $st$. Calculate $\parti s z;$ and $\parti t z; .$

$\parti s z;$ $=$ $\parti x z;\cdot\parti s x; + \parti y z;\cdot\parti s y;$ $=$ $\e y;(2s) + x\e y;(t)$ $=$ $2s\e st; + t(\s2 + t)\e st;$

$\parti t z;$ $=$ $\parti x z;\cdot\parti t x; + \parti y z;\cdot\parti t y;$ $=$ $\e y;(1) + x\e y;(s)$ $=$ $\e st; + s(\s2 + t)\e st;$

Let $f(x,y,z)$ $=$ $\x2 y + x\e z;,$ $c$ $=$ $(1,\tm1,0)$ $[(\tm1,1,0)],$ $\vv$ $=$ $\langle2,2,1\rangle,$ and $\vu$ $=$ $\fric<\vv,\Vert\vv\Vert>.$

[1] Calculate each of the following.

[2]()

$\nabla f(c)$

$\nabla f(x,y,z)$ $=$ $\left\langle2xy + \e z;,\x2 ,x\e z;\right\rangle$

$\nabla f(1,\tm1,0)$ $=$ $\left\langle\tm1,1,1\right\rangle$

$\nabla f(\tm1,1,0)$ $=$ $\left\langle\tm1,1,\tm1\right\rangle$

$D_{{}_{\vu}}f(c)$

$\vu$ $=$ $\fric<1,3>\vv$ $=$ $\fric<1,3>\langle2,2,1\rangle$

$\left\langle\tm1,1,1\right\rangle\cdot\fric<1,3>\langle2,2,1\rangle$ $=$ $\fric<1,3>$

$\left\langle\tm1,1,\tm1\right\rangle\cdot\fric<1,3>\langle2,2,1\rangle$ $=$ $\tm\fric<1,3>$

[1] Let $\mathcal{S}$ be the surface $f(x,y,z)$ $=$ 0. Give the equation of each of the following at the point $c.$

The tangent plane to $\mathcal{S}.$

[2]()

$\tm1(x - 1) + 1(y + 1) + 1(z - 0)$ $=$ 0

$\tm x + y + z$ $=$ $\tm2$

$\tm1(x + 1) + 1(y - 1) - 1(z - 0)$ $=$ 0

$\tm x + y - z$ $=$ $2$

The normal line to $\mathcal{S}.$

$\vl(t)$ $=$ $\langle1,\tm1,0\rangle + t\langle\tm1,1,1\rangle$

$\vl(t)$ $=$ $\langle\tm1,1,0\rangle + t\langle\tm1,1,\tm1\rangle$

[2] Let $\R$ be the the region bounded by the triangle with vertices $(0,0),$ $(0,1),$ and $(1,1)$. Express $\diintl R;x\sin\y3 dA$ as an iterated integral for each order of integration. Evaluate one of the iterated integrals. Hint: One iterated integral is more difficult than the other.


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[2]()

Type I

$\dintl0,1;{\dintl x,1;x\sin\y3 dy}dx$

Type II

$\dintl0,1;{\dintl 0,y;x\sin\y3 dx}dy$

Type II

$\dintl0,1;{\dintl 0,y;x\sin\y3 dx}dy$

$=$ $\dintl0,1;{\biggml 0,y;\fric<1,2>\x2 \sin\y3 p}dy$

$=$ $\dintl0,1;\fric<1,2>\y2 \sin\y3 dy$

$=$ $\biggml0,1;\tm\fric<1,6>\cos\y3 p$

$=$ $\tm\fric<1,6>(\cos 1 - 1)$

$=$ $\fric<1,6>(1 - \cos 1)$

[2] Calculate the following integral. Hint: Sketch the region of integration and convert to polar coordinates.

$\dintl\tm1,1;{\dintl0,\sqrt{1 - \x2 };x\y2 dy}dx$

$=$ $\dintl0,\pi;{\dintl0,1;r\cos\theta(r\sin\theta)^{2}rdr}d\theta$

$=$ $\dintl0,\pi;{\dintl0,1;\r4 \cos\theta\sin^{2}\theta dr}d\theta$

$=$ $\dintl0,\pi;{\biggml0,1;\fric<1,5>\r5 \cos\theta\sin^{2}\theta p}d\theta$

$=$ $\dintl0,\pi;\fric<1,5>\cos\theta\sin^{2}\theta d\theta$

$=$ $\biggml0,\pi;\fric<1,15>\sin^{3}\theta p$

$=$ 0

[2] Let $\R$ be the region bounded by the planes $x$ $=$ $1,$ $y$ $=$ $0,$ $y$ $=$ $x,$ $z$ $=$ $0,$ and $z$ $=$ $y$. Calculate the following integral.

$\diiintl R;72\x2 z\e\y3 ;dV$

$=$ $\dintl0,1;{\dintl 0,x;{\dintl 0,y;72\x2 z\e\y3 ;dz}dy}dx$

$=$ $\dintl0,1;{\dintl 0,x;{\biggml 0,y;36\x2 \z2 \e\y3 ;p}dy}dx$

$=$ $\dintl0,1;{\dintl 0,x;36\x2 \y2 \e\y3 ;dy}dx$

$=$ $\dintl0,1;{\biggml 0,x;12\x2 \e\y3 ;p}dx$

$=$ $\dintlp0,1;12\x2 \e\x3 ;- 12\x2 dx$

$=$ $\biggmlp0,1;4\e\x3 ; - 4\x3 p$

$=$ $4e - 4 - 4$

$=$ $4(e - 2)$

[3] Let $\R$ be the region between the planes $z$ $=$ 0 and $z$ $=$ $1$ that lies above the cone $\z2 $ $=$ $\x2 + \y2 $. Express $\diiintl R;xyz dV$ in terms of rectangular, cylindrical, and spherical coordinates. Evaluate all of the integrals. Follow the directions given in class.

[2]()

Rectangular:

$\diiintl R;xyz dV$

$=$ $\dintl\tm1,1;{\dintl{\tm\sqrt{1 - \x2 }},{\sqrt{1 - \x2 }};{\dintl \sqrt{\x2 + \y2 },1;xyz dz}dy}dx$

$=$ $\dintl\tm1,1;{\dintl{\tm\sqrt{1 - \x2 }},{\sqrt{1 - \x2 }};{\biggml \sqrt{\x2 + \y2 },1;xy\cdot\fric<1,2>\z2 p}dy}dx$

$=$ $\dintl\tm1,1;{\dintl{\tm\sqrt{1 - \x2 }},{\sqrt{1 - \x2 }};\fric<1,2>xy[1 - (\x2 + \y2 )]dy}dx$

$=$ $\fric<1,2>\dintl\tm1,1;{\dintlp{\tm\sqrt{1 - \x2 }},{\sqrt{1 - \x2 }};xy - \x3 y - x\y3 dy}dx$

$=$ $\fric<1,2>\dintl\tm1,1;{\biggmlp{\tm\sqrt{1 - \x2 }},{\sqrt{1 - \x2 }};\fric<1,2>x\y2 - \fric<1,2>\x3 \y2 - \fric<1,4>x\y4 p}dx$

$=$ $\fric<1,2>\dintl\tm1,1;0dx$

$=$ 0

Cylindrical:

$\diiintl R;xyz dV$

$=$ $\dintl0,2\pi;{\dintl 0,1;{\dintl r,1;r(r\cos\theta)(r\sin\theta)z dz}dr}d\theta$

$=$ $\dintl0,2\pi;{\dintl 0,1;{\dintl 4,1;\r3 (\cos\theta\sin\theta)z dz}dr}d\theta$

$=$ $\dintl0,2\pi;{\dintl 0,1;{\biggml r,1;\fric<1,2>\r3 (\cos\theta\sin\theta)\z2 p}dr}d\theta$

$=$ $\fric<1,2>\dintl0,2\pi;{\dintl 0,1;(\r3 - \r5 )(\cos\theta\sin\theta)dr}d\theta$

$=$ $\fric<1,2>\dintl0,2\pi;(\cos\theta\sin\theta)\biggmlp 0,1;\fric<1,4>\r4 - \fric<1,6>\r6 p d\theta$

$=$ $\fric<1,2>\dintl0,2\pi;\fric<1,12>(\cos\theta\sin\theta) d\theta$

$=$ $\fric<1,24>\dintl0,2\pi;\cos\theta\sin\theta d\theta$

$=$ $\fric<1,48>\biggml0,2\pi;\sin^{2}\theta p$

$=$ 0

Spherical:

$\diiintl R;xyz dV$

$=$ $\dintl0,{\fric<\pi,2>};{\dintl 0,2\pi;{\dintl0,\sec\phi;(\rho\sin\phi\cos\theta)(\rho\sin\phi\sin\theta)
(\rho\cos\phi)\rho^{2}\sin\phi d\rho}d\theta}d\phi$

$=$ $\dintl0,{\fric<\pi,2>};{\dintl 0,2\pi;{\dintl0,\sec\phi;\rho^{5}\sin^{3}\phi\cos\phi\sin\theta\cos\theta d\rho}d\theta}d\phi$

$=$ $\dintl0,{\fric<\pi,2>};{\dintl 0,2\pi;{\biggml0,\sec\phi;
\fric<1,6>\rho^{6}\sin^{3}\phi\cos\phi\sin\theta\cos\theta p}d\theta}d\phi$

$=$ $\dintl0,{\fric<\pi,2>};{\dintl 0,2\pi;
{\fric<1,6>\sec^{6}\phi\sin^{3}\phi\cos\phi\sin\theta\cos\theta}d\theta}d\phi$

$=$ $\dintl0,{\fric<\pi,2>};{\dintl 0,2\pi;
{\fric<1,6>\sec^{2}\phi\tan^{3}\phi\sin\theta\cos\theta}d\theta}d\phi$

$=$ $\dintl0,{\fric<\pi,2>};{\biggml 0,2\pi;\fric<1,12>\sec^{2}\phi\tan^{3}\phi\sin^{2}\theta p}d\phi$

$=$ $\dintl0,{\fric<\pi,2>};0 d\phi$

$=$ 0

(Exam 1)

Exam 1 Math 2673 Fall 2018

[2]()

[10] Give the center and radius of the following sphere.

$\x2 - 2x + \y2 + \z2 + 6z + 6$ $=$ 0

$(x - 1)^{2} + \y2 + (z + 3)^{2}$ $=$ $\tm6 + 1 + 9$ $=$ $4$

Center: $(1, 0, \tm3)$

Radius: 2

[10] Graph the point $(4,3,\tm2)$.


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[10] Suppose $\vv\in\vV_{{}_{2}},$ $\Vert\vv\Vert$ $=$ $10,$ and the angle between $\vv$ and the positive $x$-axis is $\fric<\pi,6>$ radians. Give the component form of $\vv.$

$\vv$ $=$ $10\left(\cos\fric<\pi,6>\right)\vi + 10\left(\sin\fric<\pi,6>\right)\vj$ $=$ $5\sqrt{3}\vi + 5\vj$ $=$ $\left\langle5\sqrt{3},5\right\rangle$

Let $\vv$ $=$ $\langle1, \tm2, 2\rangle,$ $\vw$ $=$ $\langle1, 4, 1\rangle,$ and $\theta$ be the angle between $\vv$ and $\vw$. Give each of the following.

[2]()

[10] $\vv\cdot\vw$ $=$ $1 - 8 + 2$ $=$ $\tm5$

[10] $\cos\theta$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert\Vert\vw\Vert>$ $=$ $\fric<\tm5,3\sqrt{18}>$ $=$ $\tm\fric<5,9\sqrt{2}>$

[10] comp$_{{}_{\vv}}\vw$ $=$ $\fric<\vv\cdot\vw,\Vert\vv\Vert>$ $=$ $\tm\fric<5,3>$

\(\text{%
\subprob[10]
\(\text{proj}_{{}_{\vv}}\vw\)
\(=\)
\(\fricp<\vv\cdot\vw,...
...\)
\(=\)
\(\tm\fric<5,9>\vv\)
\(=\)
\(\tm\fric<5,9>\langle1, \tm2, 2\rangle\)}\)

Consider the points $P(1,\tm2,0),$ $Q(\tm4,3,8),$ and $R(2,1,3).$

[2]([10] Give the equation of the plane containing the points $P,$ $Q,$ and $R.$)

$\overset{\longrightarrow}{PQ}$ $=$ $\langle\tm5,5,8\rangle$

$\overset{\longrightarrow}{PR}$ $=$ $\langle1,3,3\rangle$

$\vn$ $=$ $\overset{\longrightarrow}{PQ}\times\overset{\longrightarrow}{PR}$ $=$ $\langle\tm9,23,\tm20\rangle$

$\tm9(x - 1) + 23(y + 2) - 20(z - 0)$ $=$ 0

$\tm9x + 23y - 20z$ $=$ $\tm55$

[10] Give the equation(s) of the line containing the points $P$ and $Q.$

[3]()

$\overset{\longrightarrow}{PQ}$ $=$ $\langle\tm5,5,8\rangle$

Parametric

$x$ $=$ $1 - 5t$

$y$ $=$ $\tm2 + 5t$

$z$ $=$ $8t$

Symmetric

$\fric<x - 1,\tm5>$ $=$ $\fric<y + 2,5>$ $=$ $\fric<z,8>$

Vector

\(\text{%
\(\vl(t)\)
\(=\)
\(\langle1,\tm2,0\rangle + t\langle\tm5,5,8\rangle\)}\)

[2]()

For each of the following, describe and sketch the surface.

[10] $\x2 + \z2 $ $=$ $1$

Circular cylinder whose axis is the $y$-axis.


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[10] $\x2 + \z2 $ $=$ $y$

Circular paraboloid whose axis is the $y$-axis.


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[10] $\x2 + \z2 $ $=$ $\y2 $

Cone whose axis is the $y$-axis.


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[10] $\x2 - \y2 + \z2 $ $=$ $1$

Hyperboloid of one sheet whose axis is the $y$-axis.


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[10] $\x2 + \y2 + \z2 $ $=$ $1$

The unit sphere.


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[10] Two chipmunks are sitting at the same place at the bottom of a hill. They begin moving at the same time. The trajectory of the first chipmunk is $\vr_{{}_{1}}(t)$ $=$ $\left\langle5t,\t2 ,t\right\rangle$. The trajectory of the second chipmunk is $\vr_{{}_{2}}(t)$ $=$ $\left\langle\t2 + t,4t,t\right\rangle$. Will the chipmunks meet again? If so, when and where?

Consider the second component of each function and solve the following.

$\t2 $ $=$ $4t$

$t$ $=$ $0,$ $t$ $=$ $4$

Note that $\vr_{{}_{1}}(4)$ $=$ $\left\langle20,16,4\right\rangle$ and $\vr_{{}_{2}}(4)$ $=$ $\left\langle20,16,4\right\rangle$.

So the chipmunks will meet again at time $t$ $=$ $4$ at the point $(20,16,4).$

(Exam 2)

Exam 2 Math 2673 Fall 2018

A projectile is launched E $30^{\circ}$N at an angle of elevation of $45^{\circ}$ at a speed of 50 ft/sec. The acceleration vector is $\left\langle5,4,\tm32\right\rangle$.

[10] Find and simplify the initial velocity vector.

$\vv_{{}_{0}}$ $=$ $\left\langle50\cos120^{\circ},50\cos30^{\circ},50\cos45^{\circ},\right\rangle$ $=$ $\left\langle\tm25,25\sqrt{3},25\sqrt{2}\right\rangle$

[10] Find the position of the object as a vector-valued function of time.

$\va(t)$ $=$ $\left\langle5,4,\tm32\right\rangle$

$\vv(t)$ $=$ $\left\langle5,4,\tm32\right\rangle$ $\left\langle5t,4t,\tm32t\right\rangle
+ \left\langle\tm25,25\sqrt{3},25\sqrt{2}\right\rangle$ $=$ $\left\langle5t - 25,4t + 25\sqrt{3},\tm32t + 25\sqrt{2}\right\rangle$

$\vs(t)$ $=$ $\left\langle\fric<5,2>\t2 - 25t,2\t2 + 25\sqrt{3}t,\tm16\t2 + 25\sqrt{2}t\right\rangle$

Let $\vr(t)$ $=$ $\left\langle\sqrt{3}t,\sin t, \cos t\right\rangle.$

[2]()

[10] Find the equation of the line that is tangent to the graph of $\vr$ at the point $t$ $=$ $\fric<\pi,6>.$

$\vr'(t)$ $=$ $\left\langle\sqrt{3},\cos t, \tm\sin t\right\rangle$

$\vr\left(\fric<\pi,6>\right)$ $=$ $\left\langle\fric<\sqrt{3}\pi,6>,\fric<1,2>,\fric<\sqrt{3},2>\right\rangle$

$\vr'\left(\fric<\pi,6>\right)$ $=$ $\left\langle\sqrt{3},\fric<\sqrt{3},2>, \tm\fric<1,2>\right\rangle$

$\vl(t)$ $=$ $\left\langle\fric<\sqrt{3}\pi,6>,\fric<1,2>,\fric<\sqrt{3},2>\right\rangle
+ t\left\langle\sqrt{3},\fric<\sqrt{3},2>, \tm\fric<1,2>\right\rangle$

[10] Find the unit tangent vector $\vT(t).$

$\Vert\vr'(t)\Vert$ $=$ $2$

$\vT(t)$ $=$ $\fric<1,2>\left\langle\sqrt{3},\cos t, \tm\sin t\right\rangle$

[10] Find the unit normal vector $\vN(t).$

$\vT'(t)$ $=$ $\fric<1,2>\left\langle0,\tm\sin t, \tm\cos t\right\rangle$

$\Vert\vT'(t)\Vert$ $=$ $\fric<1,2>$

$\vN(t)$ $=$ $\left\langle0,\tm\sin t, \tm\cos t\right\rangle$

[10] Find the curvature $\kappa(t).$

$\kappa(t)$ $=$ $\dfrac{\fric<1,2>}{2}$ $=$ $\fric<1,4>$

[10] Find the length of the curve from $t$ $=$ 0 to $t$ $=$ $4.$

$\dintl0,4;2dt$ $=$ $\biggml0,4;2t p$ $=$ $8$

[10] Let $f(x,y)$ $=$ $\sqrt{\x2 + \y2 }$.

[2]()

[10] Find the domain and range of $f.$

D: $\Rss2 $

R: $[0,\infty)$

[10] Sketch or describe the graph of $f.$

Top half of the cone $\x2 + \y2 $ $=$ $\z2 .$

[2]( For each of the following, find the limit or show that it does not exist.)

[10] $\limv{(x,y)},(0,0);\dfric<\x3 + \y2 ,x + \y2 >$

$\limv{(x,0)},(0,0);\dfric<\x3 + \y2 ,x + \y2 >$

$=$ $\limv x,0;\dfric<\x3 ,x>$

$=$ $\limv x,0;\x2 $

$=$ 0

$\limv{(0,y)},(0,0);\dfric<\x3 + \y2 ,x + \y2 >$

$=$ $\limv y,0;\dfric<\y2 ,\y2 >$

$=$ $1$

DNE

[10] $\limv{(x,y)},(0,0);\dfric<\x2 - \y2 ,x + y>$

$=$ $\limv{(x,y)},(0,0);\dfric<(x + y)(x - y),x + y>$

$=$ $\limp{(x,y)},(0,0);x - y p$

$=$ 0

[10] $\limv{(x,y)},(0,0);\dfric<\ln\vert y\vert,\e x;>$

$\limv y,0;\ln\vert y\vert$ $=$ $\tm\infty$

$\limv x,0;\e x;$ $=$ $1$

$\limv{(x,y)},(0,0);\dfric<\ln\vert y\vert,\e x;>$ $=$ $\tm\infty$

Let $f(x,y)$ $=$ $\x2 \e y; + \x2 - 3\y2 .$

[2]()

[10] Calculate all first order and second order partial derivatives.

$f_{{}_{x}}(x,y)$ $=$ $2x\e y; + 2x$ $=$ $2x(\e y; + 1)$

$f_{{}_{y}}(x,y)$ $=$ $\x2 \e y; - 6y$

$f_{{}_{xy}}(x,y)$ $=$ $2x\e y;$

$f_{{}_{yx}}(x,y)$ $=$ $2x\e y;$

$f_{{}_{xx}}(x,y)$ $=$ $2\e y; + 2$

$f_{{}_{yy}}(x,y)$ $=$ $\x2 \e y; - 6$

[10] Find the local extrema and saddle points.

$f_{{}_{x}}(x,y)$ $=$ $2x(\e y; + 1)$ $=$ 0

$x$ $=$ 0

$f_{{}_{y}}(0,y)$ $=$ $\tm6y$ $=$ 0

$y$ $=$ 0

CP: $(0,0)$

$D(0,0)$ $=$ $\tm24$

Saddle Point: $(0,0,0)$

[10] Let $f(x,y)$ $=$ $xy - 2x$. Find the absolute extrema of $f$ on the region bounded by the curves \(\text{\(y\)
\(=\)
\(\tm\x2 + 1\)}\) and $y$ $=$ $\x2 - 1.$

[3]()

$f_{{}_{x}}(x,y)$ $=$ $y - 2$

$f_{{}_{y}}(x,y)$ $=$ $x$

\(\text{%
CP: \((0,2)\)
(not in region)}\)

$f\left(x,\tm\x2 + 1\right)$ $=$ $\tm\x3 - x$

Let $g(x)$ $=$ $\tm\x3 - x.$

$g'(x)$ $=$ $\tm3\x2 - 1$

$\tm3\x2 - 1$ $=$ 0

$\tm3\x2 $ $=$ $1$

No solution.

$f(\tm1,0)$ $=$ $2$

$f(1,0)$ $=$ $\tm2$

$f\left(x,\x2 - 1\right)$ $=$ $\x3 - 3x$

Let $g(x)$ $=$ $\x3 - 3x.$

$g'(x)$ $=$ $3\x2 - 3$

$3\x2 - 3$ $=$ 0

$x$ $=$ $\pms1$

$f(\tm1,0)$ $=$ $2$

$f(1,0)$ $=$ $\tm2$

Min: $\tm2$

Max: $2$

(Exam 3)

Exam 3 Math 2673 Fall 2018

Name:

Let $f(x,y,z)$ $=$ $3xy + \y2 \e z;$

[10] Give the linear approximation of $f$ at the point $(1,2,0)$ and use it to approximate $f(.99,2.1,.001).$

[2]()

$f_{{}_{x}}(x,y,z)$ $=$ $3y$

$f_{{}_{y}}(x,y,z)$ $=$ $3x + 2y \e z;$

$f_{{}_{z}}(x,y,z)$ $=$ $\y2 \e z;$

$f_{{}_{x}}(1,2,0)$ $=$ $6$

$f_{{}_{y}}(1,2,0)$ $=$ $7$

$f_{{}_{z}}(1,2,0)$ $=$ $4$

$f(1,2,0)$ $=$ $10$

$L(x,y,z)$ $=$ $10 + 6(x - 1) + 7(y - 2) + 4z$

$L(.99,2.1,.001)$ $=$ $10 + 6(\tm.01) + 7(.1) + 4(.001)$ $=$ $10.644$

[10] Given that $f$ is differentiable, find $f_{{}_{(a,b,c)}}(x,y,z)$ for all $(a,b,c)\in\Rss 3 .$

$f_{{}_{x}}(a,b,c)$ $=$ $3b$

$f_{{}_{y}}(a,b,c)$ $=$ $3a + 2b \e c;$

$f_{{}_{z}}(a,b,c)$ $=$ $\b2 \e c;$

$f'_{{}_{(a,b,c)}}(x,y,z)$ $=$ $3bx + (3a + 2b \e c;)y + (\b2 \e c;)z$

[10] Find the gradient of $f.$

$\nabla f(x,y,z)$ $=$ $\left\langle3y,3x + 2y \e z;,\y2 \e z;\right\rangle$

[10] Find the directional derivative of $f$ at $(1,2,0)$ in the direction of $\vu$ $=$ $\left\langle\fric<1,2>,\fric<1,2>,\fric<1,\sqrt{2}>\right\rangle.$

$\nabla f(1,2,0)$ $=$ $\left\langle6,7,4\right\rangle$

$\left\langle6,7,4\right\rangle\cdot
\left\langle\fric<1,2>,\fric<1,2>,\fric<1,\sqrt{2}>\right\rangle$

$=$ $3 + \fric<7,2> + 2\sqrt{2}$

$=$ $\fric<13 + 4\sqrt{2},2>$

[10] In what direction does the maximum rate of change of $f$ occur at (1,2,0)? What is the maximum rate of change?

$\nabla f(1,2,0)$ $=$ $\left\langle6,7,4\right\rangle$

$\Vert\nabla f(1,2,0)\Vert$ $=$ $\Vert\left\langle6,7,4\right\rangle\Vert$ $=$ $\sqrt{101}$

[10] Find an equation for the tangent plane to the level surface $f(x,y,z)$ $=$ $10$ at the point $(1,2,0).$

$\nabla f(1,2,0)$ $=$ $\left\langle6,7,4\right\rangle$

$6(x - 1) + 7(y - 2) + 4z$ $=$ 0

[10] Find an equation for the normal line to the level surface $f(x,y,z)$ $=$ $10$ at the point $(1,2,0).$

$\vl(t)$ $=$ $\langle1,2,0\rangle + t\langle6,7,4\rangle$

Let $\R$ $=$ $[0,8]\times[0,4].$

[10] Use the midpoint rule with $m$ $=$ $n$ $=$ $2$ to estimate $\diintl R;\x2 \e y;dA.$


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$8f(2,1) + 8f(6,1) + 8f(2,3) + 8f(6,3)$ $=$ $8(4e) + 8(36e) + 8(4\e3;) + 8(36\e3;)$ $=$ $320e + 320\e3;$

[10] Evaluate $\diintl R;\x2 \e y;dA.$

[3]()

$\diintl R;\x2 \e y;dA$

$=$ $\dintl0,8;\dintl0,4;\x2 \e y;dy dx$

$=$ $\dintl0,8;\biggml0,4;\x2 \e y;p dx$

$=$ $\dintl0,8;\x2 (\e4; - 1) dx$

$=$ $\biggml0,8;\fric<1,3>(\e4; - 1)\x3 p$

$=$ $\fric<512,3>(\e4; - 1)$

$\diintl R;\x2 \e y;dA$

$=$ $\dintl0,4;\dintl0,8;\x2 \e y;dx dy$

$=$ $\dintl0,4;\biggml0,8;\fric<1,3>\x3 \e y;p dy$

$=$ $\dintl0,4;\fric<512,3>\e y;dy$

$=$ $\biggml0,4;\fric<512,3>\e y;p$

$=$ $\fric<512,3>(\e4; - 1)$

[10] Maximize and minimize $f(x,y,z)$ $=$ $xz + yz$ subject to the constraint $\x2 + \y2 + \z2 $ $=$ $1.$

Let $g(x,yz,)$ $=$ $\x2 + \y2 + \z2 -1.$

$\nabla f(x,y,z)$ $=$ $\lambda\nabla g(x,y,z)$

$\left\langle z,z,x + y\right\rangle$ $=$ $\lambda\left\langle2x,2y,2z\right\rangle$

If $\lambda$ $=$ $0,$ then $z$ $=$ 0 and $y$ $=$ $\tm x$.

$f(x,y,0)$ $=$ 0

Now suppose that $\lambda\ne 0$.

From $\left\langle z,z,x + y\right\rangle$ $=$ $\lambda\left\langle2x,2y,2z\right\rangle$ we have the following.

[3]()

$z$ $=$ $2\lambda x$

$z$ $=$ $2\lambda y$

$x + y$ $=$ $2\lambda z$

From the first two equations, we see that $x$ $=$ $y.$

If $x$ $=$ $0,$ then $y$ $=$ 0 and $z$ $=$ 0. This is not possible since $\x2 + \y2 + \z2 $ $=$ $1$. So $x$ $\ne$ 0. Also, from the first equation we now have $\lambda$ $=$ $\fric<z,2x>.$

[2]()

Considering the third equation, yields the following.

$x + y$ $=$ $2\lambda z$

$2x$ $=$ $2\lambda z$

$x$ $=$ $\lambda z$

$\lambda$ $=$ $\fric<x,z>$

So now we have the following.

$\fric<z,2x>$ $=$ $\fric<x,z>$

$2\x2 $ $=$ $\z2 $

$\x2 + \y2 + \z2 $ $=$ $1$

$2\x2 + \z2 $ $=$ $1$

$2\z2 $ $=$ $1$

$z$ $=$ $\pms\fric<1,\sqrt{2}>$

$x$ $=$ $\pms\fric<1,2>$

$y$ $=$ $\pms\fric<1,2>$

$f\left(\tm\fric<1,2>,\tm\fric<1,2>\tm\fric<1,\sqrt{2}>\right)$ $=$ $\fric<1,\sqrt{2}>$ (max)

$f\left(\tm\fric<1,2>,\tm\fric<1,2>\fric<1,\sqrt{2}>\right)$ $=$ $\tm\fric<1,\sqrt{2}>$ (min)

$f\left(\fric<1,2>,\fric<1,2>\tm\fric<1,\sqrt{2}>\right)$ $=$ $\tm\fric<1,\sqrt{2}>$ (min)

$f\left(\fric<1,2>,\fric<1,2>\fric<1,\sqrt{2}>\right)$ $=$ $\fric<1,\sqrt{2}>$ (max)

(Final)

Final Exam Math 2673 Fall 2018

[10] Give the equation of the sphere with center $(2,\tm1,5)$ and tangent plane $x$ $=$ $\tm3.$

[10] A 10 lb ornament is hung from a ceiling with two wires. One wire makes an angle of $30^{\circ}$ with the ceiling. The other wire makes an angle of $60^{\circ}$ with the ceiling. Find the tension on each wire. Hint: See example 37.

[10] Find the volume of the parallelepiped determined by the vectors $\vu$ $=$ $\langle1,\tm2,1\rangle,$ $\vv$ $=$ $\langle0,\tm1,3\rangle,$ and \(\text{%
\(\vw\)
\(=\)
\(\langle2,2,0\rangle.\)}\)

Sketch or describe each of the following.

[10] $\fric<\x2 ,4> - \fric<\y2 ,9>$ $=$ $\z2 $

[10] $\fric<\x2 ,4> + \fric<\y2 ,9>$ $=$ $\z2 + 1$

[10] $z - \x2 $ $=$ $\y2 $

Consider the vector function $\vr(t)$ $=$ $\langle\sin t,\cos t, t\rangle$.

[10] Give the equation(s) of the line that is tangent to the graph of $\vr$ at the point $(0,\tm1,\pi).$

[10] Give the length of the curve from the point $(0, 1, 0)$ to the point $(0,\tm1,\pi).$

[2]( For each of the following, find the limit or prove that it does not exist.)

[10] $\limv{(x,y)},{(0,0)};\dfric<\x2 - \y2 ,\sqrt{x - y}>$

[10] $\limv{(x,y)},{(0,0)};\dfric<\x2 + \y2 ,x + \y2 >$

[10] Find all local extrema and saddle points of the following function.

$f(x,y)$ $=$ $\x2 y + xy$

[10] Maximize and minimize $f(x,y,z)$ $=$ $\x2 + y - z$ subject to the constraints $\x2 + \y2 $ $=$ $1$ and \(\text{
\(y + z\)
\(=\)
\(0.\)}\)

[10] Let $\R$ be the region in the plane bounded by the curves $y$ $=$ $0,$ $x$ $=$ $1,$ and $y$ $=$ $\sqrt{x}$. Calculate the following integral.

$\diintl R;\dfric<\e y;,\sqrt{x}>dA$

[10] Evaluate the following integral.

$\dintl0,1;{\dintl\tm\sqrt{1 - \x2 },\sqrt{1 - \x2 };\e \x2 + \y2 ; dy}dx$

[10] Let $\R$ be the region bounded by the planes $x$ $=$ $0,$ $x$ $=$ $1,$ $y$ $=$ $0,$ $z$ $=$ $0,$ and $z$ $=$ $x + y$.

Calculate the following integral.

$\diiintl R;x dV$

[10] Let $\R$ be the region bounded by the top half of the cone $\x2 + \y2 $ $=$ $\z2 $ and the plane $z$ $=$ $1.$ Express $\diiintl R;x dV$ in terms of rectangular, cylindrical, and spherical coordinates. Evaluate one of the integrals.