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\centerline{Mathematics 136 -- AP Calculus}
\centerline{Makeup Exam 2 Solutions -- November 14, 2003}
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I. (20)  In an alternate universe, the ``super-attractive'' force
exerted by two objects of mass $m_1,m_2$ 
is {\it directly} proportional to the square of the distance between
them: $F = S m_1m_2 r^2$, where $S$ is a positive constant and $r$ is the 
distance between the masses.  Four 1kg masses are fixed 
at points $x = -12,-4,2,7$ respectively along a straight line
in this alternate universe.  At what location $x$ along that line should
another 2kg mass be placed to {\it minimize} the sum of the super-attractive
forces exerted on it by the four unit masses?
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{\it Solution:}
The distance between $a$ and $b$ along the number line is 
$|b - a| = |a - b|$.  When we square, the absolute value 
can be disregarded, since the sign does not matter.  Hence the 
total force exerted by the four masses on the 2kg mass at $x$ is
$$\eqalign{F(x) &= 2S((x-(-12))^2 + (x - (-4))^2 + (x - 2)^2 + (x - 7)^2)\cr
&= 2S((x+12)^2 + (x+4)^2 + (x-2)^2 + (x-7)^2)\cr}$$
To minimize $F$ we differentiate and set to zero:
$$0 = F'(x) = 2S(2(x+12) + 2(x+4) + 2(x-2) + 2(x-7)) = 4S(4x + 7)$$
This is zero if and only if $x = -7/4$.   (This is clearly a 
local minimum for $F$, since $F'' = 16S > 0$.)
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\noindent
II.  In a car moving at 90 ft/sec, the driver 
suddenly saw an obstacle 400 feet ahead and braked
to a stop in 10 seconds.  The car's velocity was recorded
by a sensor in its onboard computer every two seconds:
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$$\matrix{t\ (sec) & 0 & 2 & 4 & 6 & 8 & 10\cr
     v(t)\ (ft/sec) & 90 & 75 & 50 & 25 & 5 & 0 \cr}$$
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\item{A)} (10)  Can you say for sure from the information
here whether the car hit the obstacle or not?  Explain, using 
the left- and right-hand Riemann sums for $v(t)$.
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{\it Solution:}  The left-hand Riemann sum gives
$$LHS = 2\cdot 90 + 2\cdot 75 + 2\cdot 50 + 2\cdot 25 + 2\cdot 5 = 490$$
The right-hand Riemann sum gives
$$RHS= 2\cdot 75 + 2\cdot 50 + 2\cdot 25 + 2\cdot 5 + 2\cdot 0 = 310$$
Since the velocity is apparently decreasing on the whole interval,
we expect the LHS is an {\it overestimate} of the actual 
distance travelled and the RHS is an 
{\it underestimate} of the actual distance.  This says 
$310 \le $ actual distance $\le 490$.  But it is not 
possible to say for sure from this information whether
a collision occurred.
\item{B)} (10)  From the given information, what is your
best estimate about whether the car hit the obstacle?
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{\it Solution:}  The average $(RHS + LHS)/2 = 400$ exactly.
There is a good chance the car did hit or just brush the obstacle.
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III. 
\item{A)} (10) State the first and second parts of the 
Fundamental Theorem of Calculus.
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{\it Solution:}  See text or class notes.
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\item{B)} (10)  The following graph shows $y = f(x)$ for 
some function.  Sketch the graph of the antiderivative
$F(x)$ of $g(x)$ with $F(x) = 0$.  What is $F(4)$?
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{\it Solution:}  The graph of $F$ should have
a linear segment from $x=0$ to $x=1$ of slope 1, so $F(1) = 1$.
Then between $x = 1$ and $x = 2$, the graph is part of a
parabola opening down.  The areas above and below the $x$-axis
between $x=1$ and $x = 2$ exactly balance out so $F(2) = F(1) = 1$.
Between $x =2$ and $x = 3$, the graph is part of a parabola opening
up; the area below the axis is $-1/2$ (a triangle of base 1 and height 1).
So $F(3) - F(2) = -1/2$ and $F(3) = 1/2$.  Then
the rest of the graph to $x = 4$ is part of a parabola opening
down and $F(4) - F(3) =-1/2$, so $F(4) = 0$. 
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IV.  Methods of integration. In B,C,D you may use the table of integrals.
\item{A)} (10) Using integration by parts, show that
$$\int x^n \cos(ax)\ dx = {x^n\over a} \sin(ax) - {n\over a}\int x^{n-1} \sin(ax)\ dx.$$ 
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{\it Solution:}  Let $u = x^n$, $dv = \cos(ax)\ dx$.  Then 
$du = nx^{n-1}$ and $v = {1\over a}\sin(ax)$, so using the integration
by parts formula,
$$\int x^n \cos(ax)\ dx = \int u\ dv = uv - \int v\ du  = {x^n\over a} \sin(ax) - {n\over a}\int x^{n-1}\sin(ax)\ dx$$
which is what we had to show.
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\item{B)} (10)  $\int \tan(\sqrt{x})/\sqrt{x}\ dx$
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{\it Solution:}  (This is a $u$-substitution form.)  
Let $u = \sqrt{x} = x^{1/2}$.  Then
$du = {1\over 2\sqrt{x}}$ so the $\sqrt{x}$ in the denominator
of the integrand is part of $du$.   We have, using \# 7 in the 
table (or by another substitution)
$$\eqalign{\int \tan(\sqrt{x})/\sqrt{x}\ dx &= 2\int\tan(u)\ du\cr
&= -2 \ln(\cos(u)) + C\cr
&= -2 \ln(\cos(\sqrt{x})) + C\cr}$$
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\item{C)} (10)  $\int (3x + 2)/(x^2 + 8x + 7)\ dx$
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{\it Solution:}  The quadratic in the denominator factors
as $x^2 + 8x + 7 = (x+1)(x+7)$ (or, completing square,
get $(x+4)^2 - 9$ and {\it the negative sign} says there
are real roots).  Hence we want to use \#27 in the table
with $a = -1, b = -7, c = 3, d = 2$.  The result is 
$${1\over 6}\left(((-1)(3)+2)\ln|x+1| - ((-7)(3)+2)\ln|x+7|\right)+C$$
which simplifies to
$${1\over 6}\left(-\ln|x+1| + 19\ln|x+7|\right)+C$$
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\item{D)} (10) $\int (4 - x^2)^{-3/2}\ dx$
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{\it Solution:}  (This is a trigonometric substitution form.)
Let $x = 2\sin(\theta)$, so $dx = 2\cos(\theta)d \theta$.
Then the integral becomes
$$\eqalign{
\int (4 - x^2)^{-3/2}\ dx &= 2\int (4\cos^2(\theta))^{-3/2}\cos(\theta)\ d\theta\cr
&= {1\over 4}\int \cos^{-3}(\theta) \cos(\theta)\ d\theta\cr
&= {1\over 4}\int \cos^{-2}(\theta)\ d\theta\cr
&= {1\over 4}\int \sec^2(\theta)\ d\theta\cr
&= {1\over 4} \tan(\theta) + C\cr}$$
To convert back to $x$, we use $\sin(\theta) = x/2$, 
so $\cos(\theta) = \sqrt{4-x^2}/2$, and 
$\tan(\theta) = \sin(\theta)/\cos(\theta) = x/\sqrt{4-x^2}$.
The final answer is 
$$\int (4 - x^2)^{-3/2}\ dx = {1\over 4} {x\over \sqrt{4-x^2}} + C$$
\bye