CARNEGIE MELLON UNIVERSITY
DEPARTMENT OF MATHEMATICAL SCIENCES
21-256 Review Exam 3 Solutions, Spring 2004

Note; below are solutions as I have typed them. at certain points I have skipped the details - you are to fill them in. Please notify me of any typos and/or mistakes by e-mail. Thanks!

1. Find the absolute maximum and minimum values of $f(x)=xe^{-x}$ on the interval $[0,2]$.
   Solution: $f^{\prime}(x)=-xe^{-x}+e^{-x}=e^{-x}(1-x)$. The critical number is $1$. Evaluate: $f(0)=0$, $f(1)=1/e=e/e^2$, $f(2)=2/e^2$. We see that the absolute min. is at 0, value 0, and the abs. max. is at 1, value $1/e$.

2. Find all critical points of $g(x)=x^3-3x-2$. Use the first derivative test to classify these critical points as local maximum, local minimum, or neither.
   Solution: $f^{\prime}(x)=3x^2-3$, so the critical numbers are at $1$ and $-1$. $f^{\prime}(-2)=9>0$, $f^{\prime}(0)=-3<0$, and $f^{\prime}(2)=9>0$, so by the first derivative test, we have a local max. at $x=-1$, and a local min. at $x=1$.

3. Find all critical points of $h(x)=2x+\frac{8}{x}$, $x>0$. Use the second derivative test to classify these critical point(s) as local maximum or local minimum.
   Solution: $h^{\prime}(x)=2-\frac{8}{x^2}$, so the critical number is at $-2$ and $2$. Since the domain does not include negative numbers we discard $-2$. Now $h^{\prime \prime}(x)=\frac{16}{x^3}>0$ for $x>0$. So $h$ is convex, and we have a local minimum at $2$. Actually, this is an absolute min., since $h$ is convex on an interval domain.

4. Let $x=y^2+z^2-2y-4z+5$. Identify the graph of this equation as either an ellipsoid, elliptic paraboloid, hyperbolic paraboloid, cone, hyperboloid of one sheet, hyperboloid of two sheets, or none of the above.
   Solution: $x=y^2+z^2-2y-4z+5=(y-1)^2+(z-2)^2$. This is an elliptic paraboloid, with vertex at $(0,1,2)$, and axis parallel to the $x-$axis.

5. Show that $\lim_{(x,y)\rightarrow (0,0)} \frac{x^2y}{x^4+y^2}$ does not exist.
   Solution: Along the path $x=0$, the limit is $0$. Along the path $y=x^2$, we consider
   
   $$\lim_{x\rightarrow 0} \frac{x^4}{x^4+x^4}=\frac{1}{2}$$
   
   
   . Thus the limit does not exist.

6. Let $f(x,y)=\frac{xy^2}{x^2+y}$. Verify that the conclusion of Clairaut's Theorem holds for this function.
   Solution: Differentiate, and simplify, to obtain $f_x(x,y)=\frac{y^3-x^2y^2}{(x^2+y)^2}$, $f_y(x,y)=\frac{2x^3y+xy^2}{(x^2+y)^2}$, $f_{xy}(x,y)=\frac{3x^2y^2-2x^4y+y^3}{(x^2+y)^3}$, $f_{yx}(x,y)=\frac{3x^2y^2-2x^4y+y^3}{(x^2+y)^3}$, so Clairaut's Theorem holds.

7. Find the linear approximation $L(x,y)$ to the function $f(x,y)=\sqrt{x^2+y}$ at the point $(1,3)$. Use this to approximate $f(\frac{3}{2}, \frac{10}{3} )$.
   Solution: $f_x(x,y)=\frac{x}{\sqrt{x^2+y}}$, $f_y(x,y)=\frac{1}{2\sqrt{x^2+y}}$, $f_x(1,3)=\frac{1}{2}$, $f_y(1,3)=\frac{1}{4}$, thus $L(x,y)=2+\frac{1}{2}(x-1)+\frac{1}{4}(y-3)$. So $f(\frac{3}{2}, \frac{10}{3} ) \approx L(\frac{3}{2}, \frac{10}{3} ) =2+\frac{1}{4}+\frac{1}{12}=\frac{7}{3}$.

8. Suppose $u=xy^2z^2$, $x=p^2q$, $y=pr^2$, and $z=q^2r$. Use the chain rule to find $\frac{\partial u}{\partial p}$ when $p=2$, $q=-1$, and $r=-2$.
   Solution: Determie $x=-4$, $y=8$, and $z=-2$.
   $$\begin{eqnarray*} \frac{\partial u}{\partial p} & = & \frac{\partial u}{\partia... ...2z^2)(2pq)+(2xyz^2)(r^2) \\ & = & -1024-1024 \\ & = & -2048 \end{eqnarray*}$$
   
   
   
   
   

9. Find the directional derivative of $f(x,y)=\ln(x^2+y^2)$ at the point $(1,3)$ in the direction $\vec{v}=\langle 2,-1 \rangle$.
   Solution: $f_x(x,y)=\frac{2x}{x^2+y^2}$, $f_y(x,y)=\frac{2y}{x^2+y^2}$, $f_x(1,3)=\frac{2}{1^2+3^2}=\frac{1}{5}$, $f_y(1,3)=\frac{2(3)}{1^2+3^2}=\frac{3}{5}$,

   $\vert\vert \vec{v} \vert\vert = \sqrt{5}$, so let $\vec{u}= \langle \frac{2}{\sqrt{5}} , \frac{-1}{\sqrt{5}} \rangle$. Then

   $$\begin{eqnarray*} D_{\vec{u}}(1,3) & = & \langle \frac{1}{5} , \frac{3}{5} \rang... ...\sqrt{5}} - \frac{3}{5\sqrt{5}} \\ & = & - \frac{1}{5\sqrt{5}} \end{eqnarray*}$$