\magnification=\magstep1
\centerline{Mathematics 134 -- Intensive Calculus for Science 2}
\centerline{Solutions for Exam 3}
\centerline{April 27, 2006}
\bigskip
\noindent
I.  
\item{A)}  Series A is {\it not geometric} because there is not a constant
ratio between successive terms.  For instance ${1/2!\over 1} = 1/2$, but
${1/3!\over 1/2!} = 2/6 = 1/3$.  Series B is geometric with $a = 3/2, r = 1/2$.
\item{B)}  From the sum formula for infinite geometric series
$$\eqalign{3 &= {a\over 1-r}\cr
             &= {b/2 \over 1 - b/2}\cr
\Rightarrow 3 &= {b\over 2 - b}\cr
           6-3b &= b\cr
            6 &= 4b\cr
\Rightarrow b &= 6/4 = 3/2\cr}$$
\bigskip
\noindent
II.  
\item{A)} 
For $f(x) = \cos(2x)$, by the chain rule:
$$\eqalign{
              &\quad f(0) = 1\cr
f'(x) = -2\sin(2x) &\quad f'(0) = 0\cr
f''(x) = -4\cos(2x) &\quad f''(0) = -4\cr
f'''(x) = +8\sin(2x) &\quad f'''(0) = 0\cr
f^{(4)}(x) = 16\cos(2x) &\quad f^{(4)}(0) = 16\cr}$$
So the Taylor polynomial is
$$1 + 0\cdot x + {-4\over 2!}x^2 + {0\over 3!}x^3 + {16\over 4!}x^4 =  
1 - 2x^2 + {2\over 3}x^4$$
\item{B)} Using the general formula for $(1 + x)^p$, the first three terms
are all nonzero for $p = 1/3$:
$$1 + {1\over 3} x + {\left({1\over 3}\right)\left({-2\over 3}\right)\over 2} x^2 =
1 + {1\over 3}x - {1\over 9}x^2.$$
(You would get the same thing by computing the Taylor polynomial of degree
$2$ using the method of part A too.)
\bigskip 
\noindent
III.  
\item{A)} It's slope field 2 for the following reason.  
${dy \over dx} = (y-1)(y-5)$ has zero slopes (equilibrium
solutions) along the lines $y = 1$ and $y = 5$.  Slope field
1 doesn't do that (slope is not zero along $y = 1$).
There are other ways to tell too, from the signs slopes at 
various $y$-values.
\item{B)}  For slope field 1, the solution with $y(0) = 3$
increases up to a horizontal asymptote at $y = 5$.  For
slope field 2, the solution with $y(0) = 3$ decreases 
to a horizontal asymptote at $y = 1$.
\item{C)}  Separating variables we have
$$\int {dy\over y} = \int {dx\over x(x+1)}$$
The integral on the right can be done with \#26 from
the table or by the partial fraction method:
$${1\over x(x+1)} = {1\over x} - {1\over x+1}$$
so
$$\ln|y| = \ln|x| - \ln|x+1| + c$$
Exponentiating both sides, 
$$y = k\left({x\over x+1}\right)$$
where $k = \pm e^c$.  
\bigskip
\noindent
IV.  
\item{A)} The given information says that if $Q$
is the number of bacteria as a function of time, 
then ${dQ\over dt} = kQ$.  This is exponential
growth, so 
$$Q(t) = Q(0) e^{kt} = 200 e^{kt}$$
\item{B)}  We know 
$$360 = Q(.5) = 200 e^{(.5)k} \Rightarrow k = {1\over .5}\ln(360/200) \doteq 1.1756$$
So then we want to solve
$$10000 = 200 e^{(1.1756)t} \Rightarrow t = {1\over 1.1756}\ln(10000/200) \doteq 3.33$$
(hours).
\bye