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Key for Test 2

Remember we are switching rooms on Monday:
\begin{tabular}{lll}
Section 2 & 10:00-11:00 & JTB 110 \\
Section 4 & 11:15-12:15 & JWB 333 \\
\end{tabular}

JTB is on presidents circle, 
just west of the path that goes up to the bookstore, union, and library.

JWB is on presidents circle, just east of Kingsbury hall.

\begin{enumerate}

\item  See page 111.  
I was looking for functions with corners(like $|x|$) and functions with 
vertical tangent lines(like $x^{\frac{1}{3}}$).
A vertical line does have a vertical tangent line but is not a function
so is not correct.

\item See page 144.  
Let $h$ be the height of the baby and $t$ stand for time.
Then the height of the baby is increasing means that $\frac{dh}{dt}>0$
since this is the rate that the baby is growing
(the baby is growing not shrinking).
The baby grows at a slower rate means that $\frac{d^2h}{dt^2}<0$
since this says that $\frac{dh}{dt}$ is shrinking.

\item  See sections 3.1 and 3.2.  
$f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim_{h\rightarrow 0} \frac{(x+h)^2-1-(x^2-1)}{h}$
$=\lim_{h\rightarrow 0} \frac{x^2+2xh+h^2-1-x^2+1}{h}$
$=\lim_{h\rightarrow 0} \frac{2xh+h^2}{h}$
$=\lim_{h\rightarrow 0} 2x+h$
$=2x$

\item See section 3.8.
Check first that the point satisfies the equation(it does).
The slope of the tangent line is $\frac{dy}{dx}$.
So take the derivative with respect to $x$ ($\frac{d}{dx}$):
This is going to be a product rule (with chain rule (implicit))
$(\frac{d}{dx}(x^2))\cos y + x^2 (\frac{d}{dx}(\cos y)) = \frac{d}{dx}2$.
$2x(\frac{dx}{dx})\cos y + x^2(-\sin y)\frac{dy}{dx} = 0$.
$2x\cos y + x^2(-\sin y)\frac{dy}{dx} = 0$.
Now solve for $\frac{dy}{dx}$ to get:
$\frac{dy}{dx}=\frac{2x\cos y}{x^2\sin y}$
Plug in the point we are interested in: $(2,\frac{\pi}{3})$ which gives:
$\frac{dy}{dx}=\frac{1}{\sqrt{3}}$
($\cos (\frac{\pi}{3})=\frac{1}{2}$, $\sin (\frac{\pi}{3})=\frac{\sqrt{3}}{2}$)

\item  
\begin{enumerate}
\item  Chain rule:
$(\cos(\cos(\sin(\pi x))))(\cos(\sin(\pi x)))'$
$=(\cos(\cos(\sin(\pi x))))(-\sin(\sin(\pi x)))(\sin(\pi x))'$
$=(\cos(\cos(\sin(\pi x))))(-\sin(\sin(\pi x)))(\cos(\pi x))(\pi x)'$
$=(\cos(\cos(\sin(\pi x))))(-\sin(\sin(\pi x)))(\cos(\pi x))(\pi)$
$=-\pi \cos(\cos(\sin(\pi x)))\sin(\sin(\pi x))\cos(\pi x)$
\item  Quotient rule.
$\frac{(3t^2-1)'(2t-7)-(2t-7)'(3t^2-1)}{(2t-7)^2}$
$=\frac{6t(2t-7)-2(3t^2-1)}{(2t-7)^2}$
$\frac{6t^2-42t+2}{(2t-7)^2}$
\item  Product rule with chain rule.
$(\sin y)' (\cos y^2)+(\sin y) (\cos y^2)'$  
$=\cos y \cos y^2 + \sin y (-sin y^2)(y^2)'$
$=\cos y \cos y^2 + \sin y (-\sin y^2)(2y)$
$=\cos y \cos y^2 - 2y\sin y \sin y^2$
\end{enumerate}

\item $v=\frac{ds}{dt}=-6t^2+6=-6(t^2-1)=-6(t-1)(t+1)$
So $v=0$ when $t=1$ or $t=-1$.
This gives three regions on a number line to test.
Testing gives that:
$v>0$ (moving to the right) when $-1<t<1$.
$v<0$ (moving to the left) when $t<-1$ or $t>1$.
$a=\frac{dv}{dt}=-12t$.
$a<0$ when $t>0$.
$a>0$ when $t<0$.
To draw the diagram, note that $s(1)=-9$,$s(0)=-5$,$s(1)=-1$.
Now piece together the info in the previous parts.

\item
What is changing?:
$V$ (volume), $h$ (height), and $r$ (radius of the pile)
What do we know?:
$\frac{dV}{dt}=10$ and $h=\frac{1}{7}r$ (which means $r=7h$).
What do we want to find?: $\frac{dh}{dt}$.
So we find an equation that relates $V$ and $h$.
$V=\frac{1}{3}\pi r^2h$.
We need to get rid of the $r$, so use $r=7h$.
$V=\frac{1}{3}\pi (7h)^2h$.
$V=\frac{49\pi}{3}h^3$.
Now take the derivative with respect to $t$ ($\frac{d}{dt}$).
$\frac{dV}{dt}=49\pi h^2 \frac{dh}{dt}$.
Solve for $\frac{dh}{dt}$.
$\frac{dh}{dt}=\frac{\frac{dV}{dt}}{49\pi h^2}$.
Now plug in for $\frac{dV}{dt}$ and $h$ to get the answers.

\end{enumerate}



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