\input calc
\def\problemskip{\bigskip}
\def\solnskip{\medskip}
\everymath{\displaystyle}
\parindent=0pt

{\bf Math 225: Calculus III\hfill\bf Exam II Solutions}

\problemskip
1. The force of gravity on a spacecraft located at $(x,y,z)$ is $F(x,y,z) =
3125/(x^2+y^2+z^2)$. Suppose the spacecraft's position and velocity at
time $t=1$ are $\p = 9\j + 12\k$ and $\v = 10\i - 12\j + 90\k$,
respectively. Find ${dF\over dt}$ at time $t=1$.

\solnskip
Solution: ${dF\over dt} = \d{F}/dx {dx\over dt} + \d{F}/dy {dy\over dt}
+ \d{F}/dz {dz\over dt} = {-2(3125)\over r^4}(x{dx\over dt}+y{dy\over dt}
+z{dz\over dt})$ where $r^4 = (x^2 + y^2 + z^2)^2$. At $t=1$,
$(x,y,z) = (0,9,12)$ and ${dx\over dt}\i+ {dy\over dt}\j
+{dz\over dt}\k = 10\i-12\j+90\k$ (given). Plug in to get ${dF\over dt}
=-120$.

\problemskip
2.
Let $f(x,y) = x^2 y - y^2z$. Compute the derivative of $f$
at the point $(1,2,0)$ in the direction $(4,0,3)$.

\solnskip
Solution: $\u = (4\i+ 3\k)/\sqrt{16+9} = (4/5)\i + (3/5)\k$.
$\grad f = 2xy\i + (x^2 - 2yz)\j - y^2\k = 4\i+\j-4\k$ at $(1,2,0)$.
$D_\u f = \grad f \cdot \u = 16/5 -12/5 = 4/5$.


\problemskip
3. Let $f(x,y,z) = e^{-yz}\cos(xy)$. Compute $\grad(f)$ at the point
$(\pi,1,0)$.

\solnskip
Solution: $\grad f = -y\sin(xy)e^{-yz}\i - (x\sin(xy)e^{-yz} +
z\cos(xy)e^{-yz})\j
-ye^{-yz}\cos(xy)\k = -\sin(\pi)\i - \pi\sin(\pi)\j -\cos(\pi)\k = \k$ at
$(\pi,1,0)$.

\problemskip
4. Determine the equation of the plane tangent to the ellipsoid
$x^2 + 2y^2 + 3z^2 = 20$ at the point $(3,2,1)$.

\solnskip
Solution: $\grad f = 2x\i + 4y\j + 6z\k = 6\i + 8\j + 6\k$ at $(3,2,1)$.
The equation of the tangent plane is then $6(x-3)+8(y-2)+6(z-1) = 0$,
or $3x + 4y + 3z = 20$.

\problemskip
5. Find the critical points of the function $f(x,y)
= 2x^3y - 6 x y + 3y^2$.

\solnskip
Solution: (a): $ f_x = 6x^2y - 6y = 0$, (b): $f_y = 2x^3 - 6x + 6y = 0$.
If $y = 0$ then (a) is OK and (b) implies $x = 0$ or $x = \pm\sqrt{3}$,
and we get three critical points, $(0,0)$, $(\pm\sqrt{3},0)$. If $y\ne0$
then
(a) implies that $x = \pm1$; for $x = 1$, (b) implies $y = 2/3$ and
for $x = -1$, (b) implies $y = -2/3$ so we get
two more critical points, $(1,2/3)$, and $(-1,-2/3)$.


\problemskip
6. Choose the statement below that applies to the function
$f(x,y) = x^3 y - x y^2 + 2 x y$.

\solnskip
Solution: The given points to consider are $(\sqrt{2},0)$, $(0,0)$ and
$(0,2)$. $f_x = 3x^2y - y^2 + 2y = 0$, $f_y = x^3 - 2xy + 2x = 0$.
$f_y(\sqrt{2},0) \ne 0$ so $(\sqrt{2},0)$ is not a critical point.
$f_x(0,0) = 0 = f_y(0,0)$ and $f_x(0,2) = 0 = f_y(0,2)$, so we must examine
the discriminant to see what type of critical points they are.
$D = (6xy)(-2x) - (3x^2 - 2y + 2)^2$; $D(0,0) = -(2)^2 < 0$, and
$D(0,2) = -(-2)^2 < 0$, so both are saddle points.


\problemskip
7. Find the maximum value of the function
$f(x,y) = (1-y^2)\log(1+x^2)$
on the rectangle $-1\le x \le 1$, $-1\le y \le 1$.

\solnskip
Solution: Critical points inside: (a): $f_x = {1-y^2\over 1+x^2}(2x) = 0$,
(b): $f_y = -2y\log(1+x^2) = 0$. (a) implies $x = 0$, or $y=\pm1$. When
$x = 0$, (b) implies $y = c$ (arbitrary); when $y=\pm1$ (b) implies $x =
0$.
At these critical points, $f(0,c) = 0$. On the boundary:
$f(x,\pm1) = 0$, and $f(\pm1,y) = (1-y^2)\log(2)$. The latter function has
a maximum of $\log(2) = 0.693$ at $y = 0$.

\problemskip
8. Find the extreme values of the function $f(x,y) = 1 + x^2 y$
on the unit circle $x^2 + y^2 = 1$.

\solnskip
Solution: Using Lagrange multipliers, solve (a): $2xy = \lambda 2 x$,
(b): $x^2 = \lambda 2 y$, and (c): $x^2 + y^2 = 1$. If $x = 0$, then
(a) is OK, and (c) implies $y = \pm1$ ((b) implies $\lambda = 0$), so
we get critical points $(0,\pm1)$ and $f(0,\pm1) = 1$. If $x\ne 0$
then (a) implies $\lambda = y$, (b) implies $x = \pm\sqrt{2} y$, and
(c) implies $x = \pm\sqrt{1/3}$, so we get the critical points
$(\pm\sqrt{2/3},\pm\sqrt{1/3})$ and $f(\pm\sqrt{2/3},\pm\sqrt{1/3}) =
1 \pm{2\over 3\sqrt{3}} = 1.385$, $0,615$.

You could also plug in the constraint equation into $f$ and find the
maximum of
$g(y) = f(\pm\sqrt{1-y^2},y) = 1 + (1-y^2)y$: $g'(y) = 0$ when $3y^2 = 1$,
etc.

\problemskip
9. Let $R$ be the region between the $x$-axis and $y = x$ for $0 \le x \le
1$.
Compute $\iint_R 6 y e^{x^3} \, dA$.

\solnskip
Solution: $\int_0^1\int_0^x 6y e^{x^3}\,dy\,dx =
\int_0^13x^2 e^{x^3}\,dx = e^{x^3}|_0^1 = e - 1 = 1.728$.

\problemskip
10. Reverse the order of integration of the integral
$\int_1^5\int_{\sqrt{y-1}}^2 f(x,y)\,dx\,dy$.

\solnskip
Solution: The region (described horizontally) is: $\sqrt{y-1}\le x \le 2$,
$1 \le y \le 5$. Rewriting this vertically gives $1\le y\le 1+x^2$,
$0\le x\le 2$ (draw the picture) and the integral is
$\int_0^2\int_1^{1+x^2} f(x,y)\,dy\,dx$.

\problemskip
11. Find the area of the region inside the cardiod $r = 1 + \sin(\theta)$
in
the first quadrant.

\solnskip
Solution: 
$\int_0^{\pi/2}\int_0^{1+\sin(\theta)} r\,dr\,d\theta =
\int_0^{\pi/2} {1\over2}(1+\sin(\theta))^2\,d\theta =
\int_0^{\pi/2} {1\over2} + \sin(\theta)+ {1\over2}\sin(\theta)^2\,d\theta =
\int_0^{\pi/2} {1\over2} + \sin(\theta)+ {1\over4}-
{1\over4}\cos(2\theta)\,d\theta =
{3\over4}\theta - \cos(\theta) - {1\over8}\sin(2\theta)|_0^{\pi/2} = 1 +
{3\pi\over8}$.


\problemskip
12. Compute the volume of the solid region under the graph of $f(x,y) = 4 -
x^2 - y^2$
over the triangular region defined by $x + y \le 1$ in the first quadrant.

\solnskip
Solution: The region is $0\le y \le 1-x$, $0\le x\le 1$ (draw the picture).
So
the volume is given by
$\int_0^1\int_0^{1-x} 4 - x^2 - y^2\,dy\,dx =
\int_0^1 4(1-x) - x^2(1-x) - {1\over3}(1-x)^3\,dx =
4x - 2x^2 -{1\over3}x^3 + {1\over4}x^4 + {1\over12}(1-x)^4|_0^1 =
4 - 2 - {1\over3} + {1\over4} - {1\over12} = {22\over12} = {11\over6}$.


\problemskip
13. Find the average value of the function $f(x,y) = y\sin(x y)$
over the region $0 \le x \le \sqrt{\pi}$, $0 \le y \le \sqrt{\pi}$.

\solnskip
Solution: The region is a square with area $(\sqrt{\pi})^2 = \pi$,
so the average is $\bar{f} =
{1\over\pi}\int_0^{\sqrt{\pi}}\int_0^{\sqrt{\pi}}y\sin(x y)\,dx\,dy =
{1\over\pi}\int_0^{\sqrt{\pi}}-\cos(x y)|_0^{\sqrt{\pi}}\,dy =
{1\over\pi}\int_0^{\sqrt{\pi}}-\cos(\sqrt{\pi}y) + 1\,dy =
{1\over\pi}(y - {1\over\sqrt{\pi}}\sin(\sqrt{\pi}y))|_0^{\sqrt{\pi}} =
{1\over\pi}\sqrt{\pi} = {1\over\sqrt{\pi}}$.


\problemskip
14. Suppose $z$ is a function of $u$ and $v$ and that $u = x^2 - y^2$ and
$v = \log(x - y)$. If $z_u(3,0) = -3$ and $z_v(3,0) = 5$, compute 
$\d{z}/dx$ when $x = 2$ and $y = 1$.

\solnskip
Solution: $z_x = z_u u_x + z_v v_x = 2x z_u + {1\over x-y}z_v$. Note that
when
$(x,y) = (2,1)$, we get $(u,v) = (3,0)$. Plug in the given values
for $x$, $y$, $z_u$ and $z_v$ to get $z_x = 4(-3) + 1(5) = -7$.

\problemskip
15.
Choose the function below that has the following graph.

\solnskip
Solution: The graph clearly has more than one critical point
(you should be able to see a local minimum and two saddle points).
All of the given functions have only one critical point except
$f(x,y)  = y^2 + x^2 y - y$ which has three critical points: $(0,1/2)$
and $(\pm1,0)$.
 
You could also examine the edges: When $y \approx 1$,
the graph looks like a parabola opening up and when $y \approx -1$,
the graph looks like a parabola opening down. Only two of the given
answers give something like $x^2 + c$ when $y=1$ and only one of these
(the above answer) looks like $-x^2 + c$ when $ y =-1$.



\bye

