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\centerline{Mathematics 136, section 4 -- AP Calculus}
\centerline{Solutions for Exam 2}
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\noindent
I.  From the diagram given with the problem, the distance through 
the forest along the path is $2 \sqrt{x^2 +4}$, while the 
distance along the road is $10 - 2x$.  Using the fact that 
$$time = {distance \over rate}$$
if we are traveling at a constant speed, the total time the
trip will take is 
$$T(x) = {2 \sqrt{x^2 +4}\over 3} + {10 - 2x \over 8}$$
We must find the overall minimum of $T(x)$.  To find critical 
points we differentiate using the chain rule:
$$T'(x) = {2\cdot {1\over 2} \cdot 2x\over  3\sqrt{x^2 +4}} - {1\over 4} 
=  {2x\over  3\sqrt{x^2 +4}} - {1\over 4}$$
Set this equal to zero and solve for $x$:
$$\eqalign{{2x\over  3\sqrt{x^2 +4}} - {1\over 4} &= 0\cr
\Rightarrow 8x &= 3\sqrt{x^2 +4}\cr
\Rightarrow 64x^2 &= 9 (x^2 + 4)\cr
\Rightarrow 55x^2 &= 36\cr
\Rightarrow x &= {6\over \sqrt{55}} \doteq .81\ km\cr}$$
(It is not difficult to see that this is a local minimum
for $T(x)$, using the 1st derivative test.)
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\noindent
II. A)  We can approximate the total distance traveled:
$x(5) - x(0) = \int_0^5 v(t)\ dt$ using left- and right-hand 
Riemann sums for the velocity function (with $\Delta t = 1$):
$$LHS = \sum_{i=1}^5 v(t_{i-1}) \Delta t = 0 + 10 + 15 + 40 + 45 = 120$$
and 
$$RHS = \sum_{i=1}^5 v(t_i) \Delta t = 10 + 15 + 40 + 45 + 50 = 170$$
Since $v(t)$ is apparently increasing on the whole interval, we 
expect the actual distance traveled to be somewhere between 120 and 170 ft.
The average 
$${120 + 170\over 2} = 145 $$
is as good an estimate as we can get from the given information.

B)  Since $v(t)$ is apparently increasing the whole time, 
we expect from class (see also pages 223 - 226 in text) that 
the true distance traveled is between 120 and 170 and
$$RHS - LHS = (v(5) - v(0)) \Delta t = 50 \Delta t$$
To get the average of LHS and RHS to be accurate to 
within 1 foot we need to take $\Delta t \le 1/25$.
The idea is that if $\Delta t$ is this small, then
the difference between $LHS$ and $RHS$ is at most 2 ft, while
the average of $LHS$ and $RHS$ is within 1 ft of every point
between them.  (I also gave full credit for $\Delta t = 1/50$, 
which says that $LHS$ and $RHS$ are within 1 ft of each other.)
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\noindent
III.  A)  See class notes.

B)  From the given information in the graph of $g(t)$ and the FTC (first part), 
$$\eqalign{
G(3) - G(0) = \int_0^3 g(t)\ dt = -5\ \Rightarrow\ G(3) = -5 - 1 = -6\cr
G(4) - G(3) = \int_3^4 g(t)\ dt = 1\ \Rightarrow\ G(4) = -6 + 1 = -5\cr
G(5) - G(4) = \int_4^5 g(t)\ dt = -1\ \Rightarrow\ G(5) = -5 - 1 = -6\cr}$$
The graph of $G(t)$ looks like this:
\vglue 1.0truein
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\noindent
V.  A)  Let $u = \ln(x)$, so $du = 1/x\ dx$.  Hence $dv = x^n\ dx$
so $v = x^{n+1}/(n+1)$.  By the integration by parts formula
$$\eqalign{
\int x^n \ln(x)\ dx 
&= {x^{n+1}\over (n+1)}\ln(x) - \int {x^{n+1}\over (n+1)}\cdot {1\over x}\ dx\cr
&= {x^{n+1}\over (n+1)}\ln(x) - {1\over n+1} \int x^n\ dx\cr
&= {x^{n+1}\over (n+1)}\ln(x) - {x^{n+1}\over (n+1)^2} + C\cr}$$
(Note -- we simplify $x^{n+1}/x$ to $x^n$ before integrating on
the right.)

B)  Let $u = \sin(2x)$ (the exponent).  Then $dx = 2\cos(2x)\ dx$, so 
the form of the given integral is 
$${1\over 2} \int e^u\ du = {1\over 2} e^u + C =  {1\over 2} e^{\sin(2x)} + C$$

C)  Complete the square in the quadratic in the denominator to see
that 
$$x^2 + 12 x + 40 = (x + 6)^2 + 4$$
This says that there are no real roots of the quadratic, and hence
we want to use \# 25 in the table.  To set up for this, make
the substitution $x + 6 = u$, so $x = u - 6$, and $du = dx$.  Then
$$\eqalign{\int{5x + 6\over x^2 + 12x + 40}\ dx &=
\int { 5(u-6) + 6 \over u^2 + 4}\ du\cr
&= \int { 5u- 24\over u^2 + 4}\ du\cr
&= {5\over 2} \ln | u^2 + 4 | - {24\over 2} \arctan\left({u\over 2}\right) + C\cr
&= {5\over 2} \ln | x^2 + 6x + 40 | - 12 \arctan\left({x+6\over 2}\right) + C\cr}$$

D)  The most direct way to do this one is to use the trig substitution
$x=3\sin(\theta)$, so $dx = 3\cos(\theta)\ d\theta$.  Substituting in
we get
$$\eqalign{\int {x^2\over \sqrt{9-x^2}}\ dx &=
\int {9\sin^2(\theta)\cdot 3\cos(\theta)\ d\theta\over
      \sqrt{9 - 9\sin^2(\theta)}}\cr
&=\int {9\sin^2(\theta)\cdot 3\cos(\theta)\ d\theta\over
      3 \cos(\theta) }\cr
&=9\int \sin^2(\theta)\ d\theta\cr}$$
Hence, from \# 17 in the table of integrals,
$$\eqalign{
&= 9\left(-{1\over 2}\sin(\theta)\cos(\theta) + {1\over 2}\theta + C\right)\cr
&= {-9\over 2}{x\over 3}{\sqrt{9-x^2}\over 3} + {9\over 2}\arcsin(x/3) + C\cr
&= {-x\sqrt{9-x^2}\over 2} + {9\over 2}\arcsin(x/3) + C\cr}$$
This can also be done using integration by parts if you let
$u = x$, and $dv = x/\sqrt{9-x^2}$ (need the power of $x$ with the 
radical to be able to integrate it by substitution).  
\bye


