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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Solutions for Quiz 7 -- November 19, 2015}
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\noindent
{\it Questions}
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\begin{enumerate}
\item[1)] 
\begin{enumerate} 
\item[(a)] (7) Find $\frac{dy}{dx}$ by implicit differentiation given 
that $x^2y^3 - 5xy + x = 1$.    
\item[] {\it Solution:}  Using implicit differentiation means we have to think of 
$y$ \emph{as an implicitly defined function of $x$}.  This means that the $x^2 y^3$ and 
$-5xy$ terms are products of functions of $x$ and must be differentiated by the product rule (and 
chain rule for the first):
$$x^2\cdot 3y^2 \frac{dy}{dx} + 2x y^3 - 5x \frac{dy}{dx} - 5y + 1 = 0.$$
Then we take the terms without $\frac{dy}{dx}$ to the other side:
$$x^2\cdot 3y^2 \frac{dy}{dx} - 5x \frac{dy}{dx} = 5y - 2xy^3 - 1,$$
Then factor out the $\frac{dy}{dx}$ on the left:
$$\left(3x^2y^2 - 5x\right) \frac{dy}{dx} = 5y - 2xy^3 - 1,$$
and finally divide by the $3x^2y^2 - 5x$ to solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{5y - 2xy^3 - 1}{3x^2y^2 - 5x}.$$
\item[(b)] (3) Find the equation of the tangent line to the curve with the equation
$x^2y^3 - 5xy + x = 1$ at $(x,y) = (0,1)$ using your answer from (a).
\item[] {\it Solution:}  I goofed on this part.  The point I meant was $(1,0)$, not $(0,1)$.
As some of you noticed $\frac{dy}{dx}$ is not even defined at $(0,1)$ (and even worse, that
point doesn't satisfy the equation of the curve).  So I gave everyone full credit for this part
since it was my mistake.  Here's the way it would work with the correct point $(x,y) = (1,0)$:
The slope is found by substituting $x = 1$ and $y = 0$ into the equation for $\frac{dy}{dx}$ from 
part (a):
$$m = \frac{5\cdot 0 - 2\cdot 1\cdot 0^3 - 1}{3\cdot 1^2\cdot 0^2 - 5\cdot 1} = \frac{1}{5}.$$
Then the tangent line is
$$y  = \frac{1}{5}(x - 1) = \frac{1}{5} x - \frac{1}{5}$$ 
\end{enumerate}
\item[2)]  Differentiate the following, but don't simplify:
\begin{enumerate}
\item[(a)] (5) $f(x) = \ln(\cos(x) + \sin(3x))$
\item[] {\it Solution:}  By the rule $\frac{d}{dx} \ln(u) = \frac{1}{u}\cdot \frac{du}{dx}$, 
which incorporates the chain rule (and then the chain rule again on the $\sin(3x)$):
$$f'(x) = \frac{1}{\cos(x) + \sin(3x)}\cdot (-\sin(x) + \cos(3x)\cdot 3)$$
\item[(b)] (5) $g(x) = \tan^{-1}(e^{5x}) + \sin^{-1}(x^2)$
\item[] {\it Solution:}  By the derivative rules for the inverse tangent
$$\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2}\cdot \frac{du}{dx}$$
with $u = e^{5x}$ and the inverse sine:
$$\frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \frac{du}{dx}$$
with $u = x^2$ (note that these incorporate the chain rule):
$$g'(x) = \frac{5e^{5x}}{1 + e^{10x}} + \frac{2x}{\sqrt{1 - x^4}}.$$
\item[] {\it General Comment:}  If you lost a lot of points on this one, 
carefully review the formulas used here.
See the class handout sheet from November 13 for the clearest statements.  These are also 
included (along with a lot of other formulas you will not be responsible for) in Sections 3.8
and 3.9 of the book.
\end{enumerate}
\item[3)] (10)  Water is being poured into a circular cylinder tank with constant radius $r = 5$ 
meters.  If the height of the water in the tank is increasing at a rate of $1$ meter 
per minute, what is the rate of change of the volume of the water in the tank?  
(The volume of a circular cylinder of radius $r$ and height $h$ is $V = \pi r^2 h$.)
\item[] {\it Solution:}  The volume of water inside the tank is also a cylinder with 
constant radius $r = 5$ and height increasing as the water is poured in.
We have $V = \pi r^2 h = 25\pi h$, so taking derivatives with respect to time:
$$\frac{dV}{dt} = 25\pi \frac{dh}{dt}.$$
Since the height of the water in the tank is increasing at $1$ meter per minute, $\frac{dh}{dt} = 1$, 
and hence 
$$\frac{dV}{dt} = 25\pi \cdot 1 = 25 \pi {\rm\ cubic\ meters\ per\ minute}.$$
\end{enumerate}
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