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\centerline{Mathematics 136, section 2 -- AP Calculus}
\centerline{Solutions for Final Exam Practice Questions}
\centerline{December 14, 2009}
\bigskip
\noindent
I.  
\item{A)}  The graph $y = 2f(x - 2) + 1$ would be obtained
by shifting the graph $y = f(x)$ to the right by $2$ units, stretching
vertically by a factor of $2$, then shifting the resulting
graph up by $1$ unit. 
\item{B)}  The graph $y = -f(-x)$ would be obtained by reflecting
the graph $y = f(x)$ across the $y$-axis, then reflecting the
resulting graph across the $x$-axis.  
\bigskip
\noindent
II.  
\item{A)}  For problems of this kind, look for where
the given graph has horizontal tangents (zeroes of $f'$),
and positive and negative slopes.  I will look for 
qualitative information rather than any quantitative
estimates of actual slope values.
\item{B)}  These are the reverse of question A.  When
$f(x)$ is $>0$, the antiderivative $g(x)$ is increasing.
When $f(x) < 0$, the antiderivative $g(x)$ is decreasing.  
$g(x)$ will have local maxima at $x$-values where 
$f$ changes from positive to negative, and local minima
where $f$ changes from negative to positive (First Derivative
Test).  
\bigskip
\noindent
III.  
\item{A)}  The derivative $f'(a)$ is:
$$f'(a) = \lim_{h\to 0}{f(a+h) - f(a)\over h},$$
provided that the limit exists.
\item{B)}  The definite integral of $f(x)$ over $[a,b]$
is the limit of the Riemann sums 
$$\int_a^b f(x)\ dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*) \Delta x,$$
provided that the limit exists (independently of the choice of 
the $x_i^*$).
\item{C)}  Let $f(x)$ be continuous on $[a,b]$.  Then
\itemitem{1)}  the function 
$$G(x) = \int_a^x f(t)\ dt$$
is an antiderivative of $f(x)$ on $(a,b)$ (that is $G'(x) = f(x)$
for all $x$ in $(a,b)$).
\itemitem{2)}  If $F(x)$ is any antiderivative of $f(x)$,
then 
$$\int_a^b f(x)\ dx = F(b) - F(a).$$
\bigskip
\noindent  
IV.  
\item{A)}  The plate is a $6\times 4$ rectangle.  Subdivide
it into 6 vertical strips of width $\Delta x = 1$ at the 
given $x$-values.  
Since the density of the metal does not depend on $y$, we can 
estimate the density in any vertical strip by a density value
from the table on that strip times the area of the strip.
Using the left endpoints, for instance, this gives
$$M \doteq 200\cdot 1 \cdot 4 + 190 \cdot 1\cdot 4 + 170\cdot 1\cdot 4 + 140\cdot 1 \cdot 4 + 
100 \cdot 1 \cdot 4 + 90\cdot 1\cdot 4 = 3560
$$
grams.  
\item{B)}  Since we used the left endpoints, and the density
appears to be decreasing left-to-right across the plate,
this estimate is probably {\it too large}.  
\bigskip
\noindent
V.  
\item{A)} Use a $u$-substitution on each term:
$$\eqalign{
\int_0^1 \cos(\pi x) - x^2 e^{x^3}\ dx
&= \left. {1\over \pi} \sin(\pi x) - {1\over 3} e^{x^3}\right|_0^1\cr
&= 0 - {1\over 3} e + {1\over 3}\cr
&= {1\over 3}(1 - e).\cr}$$
\item{B)}  We integrate by parts {\it twice} letting $u$ be
the power of $x$ each time:
$$\eqalign{
\int x^2 \sin(3x)\ dx &= {-x^2\over \cos(3x)} + {2\over 3}\int x\cos(3x)\ dx\cr
&={-x^2\over 3} \cos(3x) + {2\over 3}\left( {x\over 3}\sin(3x) - {1\over 3}\int \sin(3x)\ dx\right)\cr
&={-x^2\over 3}\cos(3x) + {2x\over 9} \sin(3x) + {2\over 27} \cos(3x) + C\cr}$$
\item{C)}  The partial fractions look like 
$${x \over (x + 3)(x^2 + 1)} = {A\over x + 3} + {Bx + C\over x^2 + 1}.$$
Clearing denominators gives:
$$x = A(x^2 + 1) + (Bx + C)(x + 3),$$
or 
$$x = (A + B)x^2 + (3B + C)x + (3C + A).$$
Hence $A+B = 0$, $3B + C = 1$, and $3C + A = 0$.  From the last equation, $A = -3C$. So
the first equation becomes $B - 3C = 0$ and the second is $3B + C = 1$.  Then
$(B-3C) + 3(3B + C) = 10B = 3$, so $B = {3\over 10}$, $C = {1\over 10}$, and $A = {-3\over 10}$.
Integrating the partial fractions gives:
$$\eqalign{  
\int {{-3\over 10}\over x + 3} + {{3\over 10} x + {1\over 10} \over x^2 + 1}\ dx
&= {-3\over 10} \int {1\over x + 3}\ dx + {3\over 10} \int {x \over x^2 + 1}\ dx + {1\over 10} \int {1 \over x^2 + 1}\ dx\cr
&= {-3\over 10} \ln|x + 3| + {3\over 20} \ln(x^2 + 1) + {1\over 10} \tan^{-1}(x) + C.\cr}$$
\item{D)}  Completing the square gives
$$\int {dx \over \sqrt{(x - 8)^2 + 1}}$$
We let $x - 8 = \tan\theta$, so $dx = \sec^2\theta\ d\theta$, and the integral becomes
$$\int {\sec^2\theta\ d\theta\over \sqrt{\tan^2\theta + 1}} = \int \sec\theta\ d\theta$$
Using the appropriate entry from the integral table, this gives
$$= \ln|\sec\theta + \tan\theta| + C = \ln|\sqrt{(x - 8)^2 + 1} + x - 8| + C.$$
\bigskip
\noindent
VI.  
\item{A)}  The cross-sections by planes $x = $ constant
are washers, so the volume is computed by the integral
$$V = \int_0^2 \pi (4 - x^2)^2\ dx = {256\pi\over 15}.$$
\item{B)}  The cross-sections by planes $y = $ constant
are washers with outer radius 3 and inner radius $3 - \sqrt{4 - y}$.
So the volume is
$$V = \int_0^4 \pi(3)^2 - \pi(3 - \sqrt{4 - y})^2\ dy = 24.$$
\item{C)}  The coordinates of the centroid are
$$\overline{x} = {\int_0^2 x(4 - x^2)\ dx\over \int_0^2 4 - x^2\ dx} = {3\over 4},$$
and
$$\overline{y} = {\int_0^2 {1\over 2}(4 - x^2)^2\ dx\over \int_0^2 4 - x^2\ dx} = {8\over 5}.$$
\bigskip
\noindent
VII.  
\item{A)} Graph omitted -- the curves $xy = c$ are rectangular hyperbolas
with asymptotes along the coordinate axes.
\item{B)}  Separating variables and integrating:
$$\eqalign{\int {dy\over y} &= \int x\ dx\cr
              \ln|y| &= {x^2\over 2} + C\cr
              y &= k e^{x^2/2}\cr}$$
              where $k = \pm e^C$.  
To get $y(0) = 4$, we want $k = 4$, so the solution is
$$y = 4 e^{x^2/2}.$$
\bigskip
\noindent
VIII.
\item{A)}  The differential equation is ${dT\over dt} = k(T - A)$.
\item{B)}  The solution, after separating and integrating is 
$$T = A + k e^{kt},$$
\item{C)}  Using the result from B with $A = 70$, we have $T(0) = 200$, 
so $k = 130$.  Then at $t = 5$, $170 = 70 + 130 e^{5k}$, so 
$$k = {1\over 5}\ln\left({10\over 13}\right) = -.0525.$$
Then we want the time $t$ when
$120 = 70 + 130 e^{(-.0525)t}$, so 
$$t = {\ln\left({50\over 130}\right) \over -.0525} \doteq 18.2$$
minutes.
\bigskip
\noindent
IX.  Put the top of the punch bowl at $y = 0$.  The work done is
$$W = \int_{-0.5}^0 \pi (\sqrt{(.5)^2 - y^2})^2 \cdot 950 \cdot 9.8 (0- y) \ dy \doteq 457$$
Joules. (Note: 1J is 1 $kg\cdot m^2/sec^2$.)
\bigskip
\noindent
X.  
\item{A)}  Since $0 < {e\over \pi} < 1$, the geometric series converges
to 
$$ {1\over 1 - {e\over \pi}} = {\pi \over \pi - e}.$$
\item{B)}  For each $k \ge 0$, the $k$th derivative of $f(x) = e^{2x}$ at $a = 0$
is $f^{(k)}(0) = 2^k$.  So the Taylor series is
$$\sum_{k = 0}^\infty {2^k x^k \over k!} = 1 + 2x + {4x^2\over 2!} + {8x^3 \over 6!} + \cdots .$$
\item{C)}  The fifth degree Taylor polynomial of $\sin(x)$ at $ a = 0$ is
$$p_5(x) = x - {x^3\over 3!} + {x^5\over 5!}$$
At $x = .6$, this gives
$$\sin(.6) \doteq p_5(.6) =.564648.$$
The actual value of $\sin(.6)$ is about $.564642$, so the absolute error is about $.000006$. 
 


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