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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Problem Set 6B Solutions, November 3, 2017}
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\noindent
Section 3.3
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\noindent
49.  Since $f(x) = x^2e^{-x}$, we can rewrite this as $f(x) = \frac{x^2}{e^x}$ and
apply the quotient rule to get 
$$f'(x) = \frac{e^x\cdot 2x - x^2 e^x}{(e^x)^2} = \frac{x(2 - x)}{e^x}$$
after simplifying.  We are looking for $x = a$ where the tangent line passes
through the origin.  That is clearly true if $a = 0$, but that is not the 
one we are looking for -- there is also one with $a > 0$.  The tangent line at
$x = a$ is 
$$y - \frac{a^2}{e^a} = \frac{a(2 - a)}{e^a}(x - a)$$
This goes through the origin if the equation obtained by setting $x = 0$ and $y = 0$
in the equation of the tangent line is satisfied:
$$-\frac{a^2}{e^a} = \frac{a^3 - 2a^2}{e^a}$$
This implies $-a^2 = a^3 - 2a^2$, so
$$a^3 - a^2 = a^2(a - 1) = 0.$$
That shows $a = 1$ is the other solution.   
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\noindent
Section 3.4
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\noindent
2.  The volume is $v(s) = s^3$.  The rate of change of the volume with respect to
$s$ is the \emph{derivative} $v'(s) = 3s^2$.  When $s = 5$, $v'(5) = 3\cdot 5^2 = 75$.
(The units are not given, but if $s$ was measured in meters, say, then 
the units of the derivative would be cubic meters per meter.)
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\noindent
38.  The height looks like a sinusoid of the form $y = A\sin(Bt)$ for some constants
$A,B$.  The velocity is the derivative of height with respect to $t$.
Its graph should look like a cosine $y = AB\cos(Bt)$.  Since no units are 
given on the axes, any graph that looks like a cosine graph (maximum at 0,
decreasing to zero at one-quarter of the way along, and reaching a minimum
at the half-way point, then increasing back up to zero at three-quarters,
and back up to the maximum at the end of the interval shown) is OK.
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\noindent
Section 3.5
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\noindent
37. 
\begin{enumerate} 
\item[(a)]  $s(t) = 300t - 4t^3$, so $v(t) = s'(t) = 300 - 12 t^2$, and $a(t) = v'(t) = s''(t) = -24t$.
At $t = 5$, $a(5) = -24\cdot 5 = -120$ (units are meters per minute per minute = meters per minute squared).  
\item[(b)]  The graph is a straight line going through the points $(t,a) = (0,0)$ and $(t,a) = (6,-144)$.
$a(t) = v'(t) < 0$ on the whole interval so $v(t)$ is decreasing.  That says the helicopter is
slowing down the whole time.
\end{enumerate}
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\noindent
40.  One of the two graphs (A) and (B) must show the derivative of the other.
The function in graph (B) would have derivative equal to zero at the tops
of the hills and the bottoms of the valleys -- that's three $x$-values in
all, including $x = 0$, and two other $x > 0$.  That matches the graph in (A).
So (B) is $y = f(x)$ and (A) is $y = f'(x)$.  Then (C) is $y = f''(x)$.  
 



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1^+} f(x) = 1$.
This implies $\lim_{x\to 1} f(x)$ \emph{does not exist} (the one-sided limits are not
equal).
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