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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Problem Set 2B Solutions, September 15, 2017}
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\noindent
Note:  Because of the confusion in class on Wednesday, credit was given for
either 1.3/39 {\it or} 1.4/39.  Solutions for {\it both} are shown below.
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\noindent
1.3/30.  With $f(x) = |x|$ and $g(\theta) = \sin(\theta)$,
$$(f\circ g)(\theta) = f(g(\theta)) = |\sin(\theta)|.$$
The domain is the set of all real $\theta$ because any such $\theta$
is in the domain of $g$ and then $g(\theta)$ is in the domain of $f$.
On the other hand,
$$(g\circ f)(x) = \sin(|x|)$$
The domain is the set of all real $x$ because any such $x$
is in the domain of $f$ and then $f(x)$ is in the domain of $g$.
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\noindent
1.3/36.  When $x < 0$, we get a portion of a line of slope $1$ and $y$-intercept $+1$;
when $x \ge 0$, we get a portion of a line of slope $-1$ and $y$-intercept $+1$.
The graph is shown in Figure 1.
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\noindent
1.3/39.  Starting at $t = t_0$, the time 10 years later is $t = t_0 + 10$.
We have, using rules for exponents:
\begin{align*}
P(t_0+10) &= 30 \cdot 2^{(0.1)(t_0 + 10)}\\ 
&= 30 \cdot 2^{(0.1)t_0 + 1}\\ 
&= 30 \cdot 2^{(0.1)t_0}\cdot 2\\
&= 2 P(t_0).
\end{align*}
This shows the population doubles in the time from $t_0$ to $t_0 + 10$.  
\begin{align*}
P(t_0+1/k) &= a \cdot 2^{k(t_0 + 1/k)}\\ 
&= a \cdot 2^{kt_0 + 1}\\ 
&= a \cdot 2^{kt_0}\cdot 2\\
&= 2 P(t_0).
\end{align*}
So the population doubles after $1/k$ years.
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\caption{The graph $y = f(x)$ in 1.3/36}
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\noindent
1.4/39.  This looks like the $\cos$ graph, but with amplitude $3$ and 
period $4\pi$, so an equation would be 
$$y = 3 \cos(x/2)$$
(there is no vertical shift in this case).  
There are other correct ways to write this of course.  For instance
if you think of the part of the graph between $x = -\pi$ and $x = 3\pi$
as the period, then its a horizontally shifted and vertically stretched
sine graph with equation
$$y = 3 \sin((x + \pi)/2).$$
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\noindent
1.4/40.  I saw this first as another cosine graph, but reflected across the 
$x$-axis.  The amplitude is $2$ and the period is $4\pi/3$, so 
$$y = -2\cos(3x/2)$$
is one possible equation.  There are others too as in the solution
for 39 above.  
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\noindent
1.4/59.  By the Law of Cosines, 
$$PQ^2 = 10^2 + 8^2 - 2\cdot 10 \cdot 8 \cdot \cos(7\pi/9) \doteq
100 + 64 - 160 \cdot (-.76604) \doteq 286.57$$
So $PQ \doteq \sqrt{286.57} \doteq 16.928$.



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