\magnification=\magstep1
\centerline{Mathematics 241 -- Multivariable Calculus}
\centerline{Sample Exam Questions for Exam 2 -- Solutions}
\centerline{October 20, 2007}
\bigskip
\noindent
I.
\item{A)}  By direct computation,
$$\eqalign{
{\partial w\over \partial s} &= -t \sin^2(st) + t \cos^2(st) \quad {\rm and}\cr
{\partial w\over \partial t} &= -s \sin^2(st) + s \cos^2(st) \cr}$$
Hence
$$s  {\partial w\over \partial s} =  -st \sin^2(st) + st \cos^2(st) = t {\partial w\over \partial t}.
$$
\item{B)}  By the Chain Rule, 
$${\partial w\over \partial s} = {dw\over dx}{\partial x\over \partial s} = f'(st) t$$
and 
$${\partial w\over \partial t} = {dw\over dx}{\partial x\over \partial t} = f'(st) s$$
Hence
$$s  {\partial w\over \partial s} = stf'(st) = t {\partial w\over \partial t}.$$
\bigskip
\noindent
II.
\item{A)}  We have $f_x = {\partial f\over \partial x} = e^{x-y}$ and 
$f_y = {\partial f\over \partial x} = -e^{x-y}$.  These functions are continuous
at all $(x,y)\in {\bf R}^2$.  Hence $f$ is differentiable at all $(x,y) \in {\bf R}^2$.
(See Theorem 3.5 on page 115 of the text.)
\item{B)}  By the definition,
$$D(f) = \pmatrix{f_x & f_y\cr}
= \pmatrix{e^{x-y} & -e^{x-y}\cr}$$
and 
$$D(g) = \pmatrix{ (g_1)_s & (g_1)_t \cr (g_2)_s & (g_2)_t\cr} = \pmatrix{ 2s & - 2t\cr t & s + 2\cr}.$$
\item{C)}  The tangent plane is given by
$$z = f(1,1) + f_x(1,1)(x - 1) + f_y(1,1)(y-1) = 1 + (x - 1) - (y - 1) = 1 + x - y.$$
This meets the $z$-axis when $(x,y) = (0,0)$, and then $z = 1$.  So the intersection
point is $(0,0,1)$.  
\item{D)}  The directional derivative is given as follows.  Let 
$u = v/\|v\| = (1/\sqrt{10}, 3/\sqrt{10})$ be the unit vector in the direction of $v$.  
So then 
$$\eqalign{D_u f(2,1) &= \nabla f(2,1) \cdot (1/\sqrt{10}, 3/\sqrt{10})\cr
&= (f_x(2,1), f_y(2,1))\cdot (1/\sqrt{10}, 3/\sqrt{10})\cr
&= (e,-e) \cdot (1/\sqrt{10}, 3/\sqrt{10})\cr
&= -2e/\sqrt{10}.\cr}$$
\item{E)}  By the Chain Rule, since $g(3,0) = (9,0)$,
$$\eqalign{
D(f\circ g)(3,0) &= D(f)(g(3,0)) D(g)(3,0)\cr
&= \pmatrix{e^{x-y} & -e^{x-y}\cr}|_{(x,y) = (9,0)} \pmatrix{ 2s & - 2t\cr t & s + 2\cr}|_{(s,t) = (3,0)}\cr
&= \pmatrix{e^9 & -e^9\cr} \pmatrix{ 6 & 0\cr 0 & 5\cr}\cr
&= \pmatrix{6 e^9 & -5e^9\cr}\cr}$$
\item{F)}  Substituting and computing the derivatives directly, 
$$(f\circ g)(s,t) = e^{s^2 - t^2 - st - 2t}$$
So 
$$D(f\circ g)(s,t) = \pmatrix{(2s - t)e^{s^2 - t^2 - st - 2t} & (-2t - s - 2)e^{s^2 - t^2 - st - 2t}\cr}$$
Hence substituting $(s,t) = (3,0)$,
$$D(f\circ g)(3,0) = \pmatrix{6e^9 & -5e^9\cr},$$
which is the same as above.
\bigskip
\noindent
III.  
\item{A)}  By our general knowledge of directional derivatives, $f$ 
is increasing the fastest at $(1,1)$ in the direction of $\nabla f(1,1)$.  We have
$\nabla f(x,y) = (-2a x, -2b y)$ so this means we want a unit vector in the direction of 
$(-2a,-2b)$:  
$${\bf u} = \left({-a\over \sqrt{a^2 + b^2}}, {-b\over \sqrt{a^2 + b^2}}\right).$$ 
\item{B)}  The marble would roll in the direction that decreased the altitude the 
fastest, hence in the direction of $-\nabla f(1,1) = (2a,2b)$, so in direction of 
$${\bf v} = \left({a\over \sqrt{a^2 + b^2}}, {b\over \sqrt{a^2 + b^2}}\right).$$
\item{C)}  To remain at a constant altitude, move 
in a direction perpendicular to $\nabla f(1,1)$, so in either of the 
directions specified by 
$${\bf w} = \pm \left({b\over \sqrt{a^2 + b^2}}, {-a\over \sqrt{a^2 + b^2}}\right)$$
(These are chosen to make ${\bf w} \cdot {\bf u} = 0$ using the ${\bf u}$ from part A.)
\bigskip
\noindent
IV.  Apply the Chain Rule, thinking of the polar coordinates map 
$x = r \cos\theta$, $y = r \sin\theta$ as the ``inside function.'' We have
$$\eqalign{
{\partial z\over \partial r} &= {\partial z\over \partial x}{\partial x\over \partial r} + 
{\partial z\over \partial y}{\partial y\over \partial r}\cr
&= {\partial z\over \partial x}\cos\theta+ 
{\partial z\over \partial y}\sin\theta \quad {\rm and}\cr
{\partial z\over \partial \theta} &= {\partial z\over \partial x}{\partial x\over \partial \theta} + {\partial z\over \partial y}{\partial y\over \partial \theta}\cr
&= -{\partial z\over \partial x}r \sin\theta+ 
{\partial z\over \partial y} r \cos\theta\cr}$$
Hence starting from the right side of the desired equation,
$$\eqalign{
\left({\partial z\over \partial r}\right)^2 + {1\over r^2} \left({\partial z\over \partial \theta}\right)^2
&= \left({\partial z\over \partial x}\cos\theta+ {\partial z\over \partial y}\sin\theta\right)^2 
+ {1\over r^2}\left(-{\partial z\over \partial x}r \sin\theta+ {\partial z\over \partial y} r \cos\theta\right)^2\cr
&= \left({\partial z \over \partial x}\right)^2(\cos^2\theta + \sin^2\theta)
+ 2{\partial z \over \partial x} {\partial z \over \partial y} (\cos\theta\sin\theta - \cos\theta\sin\theta) \cr
& \qquad + \left({\partial z \over \partial y}\right)^2(\cos^2\theta + \sin^2\theta)\cr
&= \left({\partial z \over \partial x}\right)^2 + \left({\partial z \over \partial y}\right)^2,
\cr}$$
which is what we wanted to show(!)
\bigskip
\noindent
V.  
\item{A)}  We have $x'(t) = (3\cos(3t), -3\sin(3t), 3 t^{1/2})$, so 
$$\|x'(t)\| = 3 \sqrt{\cos^2(3t) + \sin^2(3t) + t} = 3\sqrt{1 + t}.$$  
Hence the desired arclength is 
$$L = \int_0^1 3\sqrt{1 + t}\ dt = 2(1 + t)^{3/2}|_0^1 = 2(2\sqrt{2} - 1) = 4\sqrt{2} - 2.$$
\item{B)}  The unit tangent vector is
$$\eqalign{
{\bf T} &= {1\over \|x'(t)\|} x'(t) = {1\over 3\sqrt{1+t}} (3\cos(3t), -3\sin(3t), 3 t^{1/2})\cr
&=\left({\cos(3t)\over \sqrt{1+t}}, {-\sin(3t)\over \sqrt{1+t}}, \sqrt{t\over 1 + t}\right)
.\cr}$$
\item{C)} $x'(t)$ is given above in part A.  Then 
$$x''(t) = (-9\sin(3t),-9\cos(3t), {3\over 2} t^{-1/2}).$$
Using the alternate curvature formula,  we compute 
$$\eqalign{
x'(t) \times x''(t) &=\left(-{9\sin(3t)\over 2\sqrt{t}}+27\sqrt{t}\cos(3t), 
-27 \sqrt{t}\sin(3t)- {9\cos(3t)\over 2\sqrt{t}},\right.\cr
&\qquad  \left.-27(\cos^2(3t)-\sin^2(3t))\right)\cr} $$
Hence $\kappa(t) = \|x'(t) \times x''(t)\|/ \|x'(t)\|^3 = $
$${81\over 4}{36t^2 + 36t + 1\over t} \cdot {1 \over (3\sqrt{1+t})^3} = 
{3\over 4}{36t^2 + 36t + 1\over (1+t)\sqrt{1+t}}$$
\bigskip
\noindent
VI.
\item{A)}  This is clear -- if any two of $x,y,z$ are zero, then all the component
functions of ${\bf F}(x,y,z) = (yz, xz, xy)$ are zero.
\item{B)} This will not be covered on the exam (as announced on Friday, 10/19).
\item{C)} This will not be covered on the exam (as announced on Friday, 10/19).
\item{D)}  $f = xyz + C$ for any constant.
\item{E)}  The curve given by $x = y = z = {1\over 1 - t}$ is a 
flow line of ${\bf F}$ since 
$$x'(t) = y'(t) = z'(t) = (-1)(1 - t)^{-2}(-1) = {1\over (1-t)^2} = y(t)z(t) = x(t)z(t) = x(t) y(t).$$
\bye
