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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Problem Set 7B Solutions, November 17, 2017}
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\noindent
Section 3.8
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\noindent
70.  To find $\frac{dy}{dx}$, we use implicit differentiation:
$$2(x^2 + y^2 - 4x)(2x + 2y\frac{dy}{dx} - 4) = 4x + 4y\frac{dy}{dx}$$
So solving for $\frac{dy}{dx}$,
$$\frac{dy}{dx} = \frac{4x + 2(4 - 2x)(x^2 + y^2 - 4x)}{4y(x^2 + y^2 - 4x) - 4y}.$$
The four $y$-coordinates for the intersections with the line $x = 1$ are found by 
substituting into the original equation:
$$(y^2 - 3)^2 = 2(1 + y^2)$$
and then rearranging and solving for $y$:
$$y^4 - 6y^2 + 9 = 2 + 2y^2, \quad {\rm so} \quad y^4 - 8y^2 + 7 = 0.$$
This factors as 
$$(y^2 - 1)(y^2 - 7) = 0$$
so the solutions are $y = \pm 1$, $y = \pm \sqrt{7}$.  Then at $(x,y) = (1,1)$ for 
instance, we have
$$\frac{dy}{dx} = \frac{4 + 2(2)(-2)}{4(-2) - 4} = \frac{1}{3}$$
so the tangent line at $(1,1)$ is 
$$y - 1 = \frac{1}{3}(x - 1).$$
The other three are found similarly:
\begin{itemize}
\item  At $(x,y) = (1,-1)$, $\frac{dy}{dx} = \frac{-1}{3}$ and the tangent line is
$$y + 1 = \frac{-1}{3}(x - 1)$$
\item At $(x,y) = (1,\sqrt{7})$, $\frac{dy}{dx} = \frac{5\sqrt{7}}{21}$ (after simplifying),
so the tangent line is 
$$y - \sqrt{7} =  \frac{5\sqrt{7}}{21}(x - 1)$$
\item Finally, at $(x,y) = (1,-\sqrt{7})$, $\frac{dy}{dx} = -\frac{5\sqrt{7}}{21}$, and 
the tangent line is 
$$y + \sqrt{7} = -\frac{5\sqrt{7}}{21}(x - 1).$$
\end{itemize}
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\noindent
Section 3.10
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\noindent
16.  In the figure given with the problem, let $x$ be the distance from the intersection
to the car as a function of $t$ and let $z$ be the distance from the car to the house.
By the Pythagorean theorem, 
$$2^2 + x^2 = z^2.$$
So taking the derivative with respect to $t$ (implicitly)
$$2x\frac{dx}{dt} = 2z\frac{dz}{dt}.$$
At the instant of time the problem is asking about, $x = 6$ and $\frac{dx}{dt} = 80$.
From the Pythagorean Theorem, $4 + 36 = 40 = z^2$, so $z = \sqrt{40} \doteq 6.32$.  Hence
$$12\cdot 80 \doteq 12.64 \frac{dz}{dt} \Rightarrow \frac{dz}{dt} \doteq \frac{960}{12.64} \doteq 75.95$$
(units are km/hr).



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1^+} f(x) = 1$.
This implies $\lim_{x\to 1} f(x)$ \emph{does not exist} (the one-sided limits are not
equal).
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