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\centerline{Mathematics 136 -- Advanced Placement Calculus}
\centerline{Solutions to Ungraded Problems on Problem Set 3}
\centerline{September 29, 2009}
\vskip 10pt
\noindent
2.8/10 
\begin{enumerate}
\item[(a)]  The rate of increase of the population is small at first, then
much larger around $t = 8$, then smaller again (tending to zero, apparently)
as $t$ increases farther.
\item[(b)]  The rate of increase is largest at $t = 8$.
\item[(c)]  The graph of the population function is concave up for $0 < t < 8$
and concave down for $8 < t < 18$.
\item[(d)]  At $t = 8$, the population is $350$.  The inflection point on the graph is 
at $(8, 350)$. 
\end{enumerate} 
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\noindent 
2.8/16.  
\begin{enumerate}
\item[(a)]  $f$ is increasing where $f'$ is positive, so on $(1,6)$ and for $x > 8$.  
$f$ is decreasing where $f'$ is negative, so on $(0,1)$ and $(6,8)$.
\item[(b)]  $f$ has local minima at $x = 1$ and $x = 8$ and a local maximum at $x = 6$
(using what we stated as the First Derivative Test in class on 9/29).
\item[(c)]  $f$ is concave up where $f'$ is increasing, so on $(0,2), (3,5), (7,\infty)$.
$f$ is concave down where $f'$ is decreasing, so on $(2,3)$ and $(5,7)$.
\item[(d) and (e)]  The graph should follow parts (a), (b), (c).  Note that $f$ is increasing
all the way from $x = 1$ to $x = 6$; what changes on that interval is the concavity (three changes!).
There are inflection points at $x = 2,3,5$ and $x = 7$.  
\end{enumerate}
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\noindent
3.1/7.  $f'(x) = 3x^2 - 4$.
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\noindent
3.1/25.  $x = Ay^{-10} + Be^y$ (where $A,B$ are constants).  So $\frac{dx}{dy} = -10A y^{-11} + Be^y$.
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\noindent
3.2/11.  By the quotient rule:
$$y' = \frac{(1-x^2)(3x^2) - x^3\cdot(-2x)}{(1-x^2)^2} = \frac{3x^2-x^4}{(1-x^2)^2}.$$
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\noindent
3.2/22.  By the quotient rule:
$$
f'(x) = \frac{(x+e^x)(-xe^x-e^x) - (1 - xe^x)(1+e^x)}{(x + e^x)^2} = 
\frac{-x^2 e^x - e^{2x} - e^x - 1}{(x + e^x)^2} \text{ after simplifying}.
$$
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\noindent
3.2/45.  The idea of this problem is to use the product or quotient rule as appropriate, 
and read the values of $f,g$, and the derivatives of $f,g$ from the graph, recalling that
$f'(a),g'(a)$ are slopes of tangent lines.
\begin{enumerate}
\item[(a)] 
$$u'(1) = f(1)g'(1) + f'(1)g(1) = 2\cdot (-1) + 2\cdot 1 = 0.$$
\item[(b)] 
$$v'(5) = \frac{g(5)f'(5) - f(5)g'(5)}{(g(5))^2} = \frac{2\cdot (-1/3) - 3\cdot (2/3)}{2^2} = -\frac{2}{3}.$$
\end{enumerate}
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\noindent
3.3/5.  By the product rule, $y' = \sec\theta\cdot\sec^2\theta + \tan\theta\cdot\sec\theta\tan\theta = 
\sec\theta(\sec^2\theta + \tan^2\theta)$.  This can also be rewritten in various ways by 
using trig identities, recalling that $1 + \tan^2\theta = \sec^2\theta$.
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\noindent
3.3/35.  
\begin{enumerate}
\item[(a)] Since $x(t) = 8\sin(t)$, $v(t) = x'(t) = 8\cos(t)$ and $a(t) = x''(t) = -8\sin(t)$.
\item[(b)] At $t = \frac{2\pi}{3}$, $v = 8\cos\left(\frac{2\pi}{3}\right) = -4$ and 
$a = -8\sin\left(\frac{2\pi}{3}\right) = -4\sqrt{3}$.  This means that the mass is moving to 
the left ($v < 0$).
\end{enumerate}
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\noindent
3.4/15.  The general derivative rule for the exponential functions is 
$$\frac{d}{dx} a^x = a^x \ln(a).$$
(This can be seen by rewriting $a^x = e^{x \ln(a)}$ and using the chain rule.)
So $h'(t) = 3t^2 - 3^t \ln(3)$.
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\noindent
3.4/30.  Think of the function $f(t)$ in the form $f(t) = \left(\frac{t}{t^2+4}\right)^{1/2}$ and 
use the chain rule and quotient rule:
$$
f'(t) = \frac{1}{2}\left(\frac{t}{t^2+4}\right)^{-1/2}\left(\frac{(t^2+4) - t\cdot(2t)}{(t^2 + 4)^2}\right).
$$
This can be simplied to:
$$
f'(t) = \frac{4 - t^2}{2 t^{1/2}(t^2+4)^{3/2}}.
$$
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\noindent
3.5/15.  By implicit differentiation:
$$
-e^y \sin(x) + \cos(x)\cdot e^y y' = \cos(xy)\cdot(y + xy').
$$
Solving for $y'$:
$$
(\cos(x)e^y - x\cos(xy))y' = y\cos(xy) + e^y\sin(x).
$$
So 
$$
y' = \frac{y\cos(xy) + e^y\sin(x)}{\cos(x)e^y - x\cos(xy)}.
$$
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