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\markright{Chapter 2: Answers 1}
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\Bf{Chapter 2: Answers 1 \hfill Jack K. Cohen \hfill Colorado School of Mines}

\begin{enumerate}
\item

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\caption{Ups and downs in a balloon.} 
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	\begin{enumerate}
	\item See Figure 1.
	\item Your textbook has an index.
	\item The slope is positive in the intervals $(a,A), (b, B)$ and $(c, C)$.  These are indeed the same as the intervals where the function is increasing.
	\item Zero.
	\item If the slope is positive in an interval to the left of a point and is negative in an interval to the right of this point, then the point gives a local maximum.
	\item Global maximum is at $B$, global minimum is at $a$.
	\item The velocity is 0 at $A, b, B, c$. The velocity has its local maxima at points where the slope is greatest; velocity has a local minimum where the slope is smallest.  
It looks like the largest slope (roughly 30) occurs at $a$ or $C$, the smallest slope (about -15)
occurs between $B$ and $c$.
	\end{enumerate}

\item (2.1.36)  See Figure 2.  We know that the slope function (derivative) of $y = x^2$ is $2x$, so the slope at $x = x_0$ is $m = 2 x_0$.  Since $(x_0, x_0^2)$ is a point on the tangent, the point-slope formula gives the tangent line as $y  - x_0^2 = 2 x_0 ( x - x_0)$.  This line intersects the $x$-axis when $x = x_1, y = 0$, so $0  - x_0^2 = 2 x_0 ( x_1 - x_0)$.  Solving for $x_1$ gives $x_1 = x_0 / 2$ as was to be shown.	
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\caption{Diagram for problem 2.} 
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\item (2.1.42)  See Figure 3.  Some steps along the way: $m = 4 - 2a$, tangent line is $y - (4a - a^2) = (4 - 2 a) (x - a)$.  Plug in the point (2, 5) for $(x, y)$ to get equation for $a$ and simplify: $(a - 1)(a - 3) = 0$.  Answers: $y = 2x + 1, y = -2x + 9$.

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\caption{Diagram for problem 3.} 
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\item (2.1.43)  See Figure 4.  The slope of the tangent at $x = a$ would be $2a$, so the slope of the normal is $-1/2a$.  Use point-slope to get the equation of the normal: $y - a^2 =( -1/2a) ( x - a)$.  Substitute the given point (3, 0) on the normal to get an equation for $a$: $2a^3 + a - 3 = 0$.  The solution $a = 1$ is ``obvious''.   But move the point (3, 0) to another position and it is almost certain that the corresponding cubic will not have any ``nice'' roots---you could still use \Tt{Plot} and/or \Tt{Table} though.

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\caption{Diagram for problem 4.} 
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\item Following up on previous problems:

	\begin{enumerate}
	\item From a point \Em{exterior} to the parabola there would always be exactly two tangent lines, but from an interior point there would not be any (the corresponding quadratic equation would have complex instead of real roots).
	\item  Again, the answer is yes for exterior points, but for interior points there would be \Em{three} normals!
	\end{enumerate}

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