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\institution{SIAST Palliser Campus}
\department{Mathematics}
\course{MAT 226}
\unittitle{Applications of the Derivative}
\unitnumber{4}
\date{January 2011}
\title{STUDY GUIDE}
\author{Alan Dill and Robert G. Petry}
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\publisher{the Department of Mathematics, SIAST Palliser Campus}
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\section{Tangents and Normals}
%The first option says what line to put the picture at
%The second option says whether the picture should be left (l) centre (c)
%or right(r)
\begin{window}[0,r,%
\mbox{\includegraphics[width=3.2in]{image1.png}},]
The slope of the tangent to a curve at a given point is equal to the
value of the derivative at that point.

The normal is perpendicular to the tangent, so its slope is the
negative reciprocal of the slope of the tangent.

To find the equation of the tangent at a given point, we can evaluate
the derivative at the point, and then use the point-slope form of the
equation of a straight line.  The equation of the normal can then be
found in a similar way, using the negative reciprocal of the slope of
the tangent.
\end{window}

\textbf{Example}
\begin{enumerate}
\item Find the equation of the tangent and normal of the curve $\displaystyle y=x^2-4x+6$ at $(1,3)$~.
\item Find the equation of the tangent and normal of the curve $\displaystyle 16x^2+9y^2=144$ at $(2,-2.98)$~.
\item Find the angle of intersection of the curves:
\[y_1=x^2+x-1\]
\[y_2=x^2-5x+5\]
at the point $(1,1)$~.
\end{enumerate}
\textassignment{Sec. 28-1}{Ex. 1 (P. 809) \# 1-13 (odd)}

\section{Maximum, Minimum and Inflection Points}
\subsection{Increasing and decreasing functions}
\begin{window}[0,r,%
\mbox{\includegraphics[width=3in]{image2.png}},]
Consider the function $y = f(x)$ whose graph is shown on the right.

Any tangent line drawn to the left of point $M$ will have a positive
slope, as will any tangent line drawn to the right of point $m$.  Since
the slope of the tangent line at any point is equal to the value of the
first derivative at that point, this means that the first derivative is
positive to the left of $M$ and to the right of $m$.  We also see that, as
$x$ increases, the function increases in the region to the left of point
$M$ and in the region to the right of point $m$.

For points between $M$ and $m$, any tangent line will have a negative slope,
and the function decreases from left to right in this region.
\end{window}

We can summarize these results by stating the following rules:
\begin{itemize}
\item If the first derivative is \textbf{positive}, the function is \textbf{increasing}.
\item If the first derivative is \textbf{negative}, the function is \textbf{decreasing}.
\end{itemize}
(It is assumed here that $x$ is always increasing, i.e. we are observing
the behavior of the function as we move from left to right along the
$x$-axis.)

We can also state the above rules in symbols as follows:
\begin{itemize}
\item If $f'(x)>0$, $f(x)$ is \textbf{increasing}.
\item If $f'(x)<0$, $f(x)$ is \textbf{decreasing}.
\end{itemize}

\textbf{Examples}
\begin{enumerate}
\item Is the function $y=x^3-4x^2+5$ increasing or decreasing at $x=3$?
\item For what values of $x$ is the function $y=x^2-4x+7$ increasing,
  and for what values of $x$ is it decreasing?
\end{enumerate}

\subsection{Concavity}
\begin{window}[0,r,%
\mbox{\includegraphics[width=3in]{image4.png}},]
Concavity refers to the curvature of the graph of a function.  As we
have already seen in the previous unit, when the second derivative is
positive, the slope of the curve is increasing, which means that the
curve is concave upward.  When the second derivative is negative, the
curve is concave downward.  At a point where the curve changes from
being concave upward to concave downward (or vice-versa), the second
derivative is zero.  This is illustrated in the diagram on the right.
\end{window}
\textbf{Examples}
\begin{enumerate}
\item Is the function $y=x^3-4x^2+5$ concave up or down at $x=3$?
\item Find the regions of concavity of $y=x^2-4x+7$.
\end{enumerate}

\subsection{Stationary points}
\begin{window}[0,r,%
\mbox{\includegraphics[width=3in]{image3.png}},]
A \emph{stationary point} is a point at which the first derivative is zero
(i.e. the tangent is horizontal at the point).

Points $M$, $m$  and $X$ in the diagram are all stationary points.

Point $M$ is called a \emph{relative maximum point}.  This means that the value
of $y$ is greater at $M$ than at any other point near it.

Point $m$ is called a \emph{relative minimum point}.  This means that the value
of $y$ is less at $m$ than at any other point near it.

Point $X$ is a stationary point, but it is \emph{neither} a maximum
nor a minimum point.

As can be seen in the diagram, functions change from increasing
to decreasing and vice-versa at stationary points that are relative
maxima and minima.
\end{window}

To find the stationary point(s) of a function, set the first derivative
equal to zero and solve for $x$.  Then substitute in the original
function to find the $y-$coordinates of the stationary points.

\textbf{Example}

Find the stationary point(s) of the function $y=x^3-3x+1$.

\subsection{Testing for maximum or minimum points}

We have seen that the coordinates of the stationary points are found by
setting the first derivative to zero, but this does not tell us whether
each point is a maximum, a minimum or neither.  To make this
determination, we use the ordinate test, the first derivative test, or the second derivative
test which are described below:

\subsubsection{Ordinate test}
\begin{window}[0,r,%
\mbox{\includegraphics[width=1.5in]{image7.png}},]
The ordinate test determines whether a point is a
relative maximum or minimum by evaluating the function $y=f(x)$ at
points neighbouring the stationary point.  It is clear from the
diagrams at right that:

%We need a minipage to but an item list in the window environment
\begin{minipage}{4in}\raggedright
\begin{itemize}
\item The height of the curve on either side of a \textbf{minimum} point is
  \textbf{greater} than it is at the minimum point
\item The height of the curve on either side of a \textbf{maximum} point is
  \textbf{less} than it is at the maximum point
\end{itemize}
\end{minipage}
\end{window}
To use the ordinate test, find the $y$-value at a point just to the
left of the stationary point, and at a point just to the right of the
stationary point.  If both calculated $y$-values are greater than the
$y$-value at the stationary point, it is a relative minimum.  If both
calculated $y$-values are less than the $y$-value at the stationary
point, it is a relative maximum.  If one of the $y$-values is greater
and one is less than the $y$-value at the stationary point, it is
neither a minimum nor a maximum.
\clearpage
\subsubsection{First derivative test}
\begin{window}[0,r,%
\mbox{\includegraphics[width=2in]{image5.png}},]
Referring to the diagrams on the right, and based on what we have
already said about increasing and decreasing functions, we can state
the following rules:

\begin{minipage}{4in}\raggedright
\begin{itemize}
\item The first derivative is \textbf{negative} to the \textbf{left}
  and \textbf{positive} to the \textbf{right} of a relative
  \textbf{minimum} point.
\item The first derivative is \textbf{positive} to the \textbf{left}
  and \textbf{negative} to the \textbf{right} of a relative
  \textbf{maximum} point.
\end{itemize}
\end{minipage}
\end{window}

\vspace{2em}

\subsubsection{Second derivative test}
\begin{window}[0,r,%
\mbox{\includegraphics[width=1.5in]{image6.png}},]
We have already seen that the second derivative is a measure of the
rate of change of the slope of a curve.  Thus, if the second
derivative is positive in an interval, the curve is concave upward in
that interval.  If the second derivative is negative in an interval,
the curve is concave downward in that interval.

Since a curve is concave upward at a minimum point and concave
downward at a maximum point (see diagram), we have the following rules
for determining whether a given stationary point is a maximum or
minimum:
\end{window}
\begin{itemize}
\item If the second derivative is \textbf{positive}, the stationary point is a relative \textbf{minimum}.
\item If the second derivative is \textbf{negative}, the stationary point is a relative \textbf{maximum}.
\item If the second derivative is \textbf{zero}, the test is \textbf{inconclusive}.
\end{itemize}

Note that, if the second derivative is zero, the first derivative test
or the ordinate test must be applied to determine whether the point is
a maximum, a minimum, or neither.

\textbf{Examples}
\begin{enumerate}
\item Apply all three of the above tests to the stationary points of the
function \mbox{$y = x^3-3x+1$} found in the previous example.
\item Find the maximum and minimum points for the implicit relation
\mbox{$x^2+y^2-2x+4y=4$}.
\end{enumerate}

\subsection{Inflection Points}
\begin{window}[0,r,%
\mbox{\includegraphics[width=2.5in]{image8.png}},]
An \emph{inflection point} is a point where the curvature changes from
concave upward to concave downward, or vice-versa.  This means that
the second derivative changes sign at an inflection point.

Points $A$ and $B$ in the graph on the right are inflection points.

To find inflection points, set the second derivative equal to zero and
solve for $x$.  Then check to see whether the second derivative changes
sign on either side of the point(s).
\end{window}

\textbf{Examples}

Find the inflection point(s) of the following functions:
\begin{enumerate}
\item $\displaystyle y=3x^4-4x^3$
\item $\displaystyle y=x^4$
\end{enumerate}

\subsection{Systematic Procedure for Finding Maximum, Minimum and Inflection Points of a Function}
The formula sheets at the end of the unit summarize the steps for
finding \emph{critical points} (stationary or inflection points).  Use
that procedure to find any maximum, minimum and inflection points for
the following function:
\[y=x^4-18x^2\]

\textassignment{Sec. 28-2}{Ex. 2 (P. 819) \# 1-23 (odd), 27, 29}

\section{Curve Sketching}

While it is usually possible to graph a function simply by plotting
pairs of points, this is not always the best way to get an overall
picture of how the function behaves.  By looking at certain
characteristics of the function, we can often get a good idea of what
the graph of the function will look like, and make a sketch without
using a table of x{}- and y{}-values.  Going through these steps will
enable us to analyze and interpret the characteristics of the function
as well as allowing us to sketch the function.

Depending on the particular function, some or all of the following
characteristics may be used in sketching the graph of a function.

\subsection{Type of function}
Certain types of functions have characteristic shapes, a few of which
are illustrated below:
\begin{center}
\begin{tabular}{c|c|c}\hline
Function Type & Example & Graph \\ \hline
Quadratic & $\displaystyle y=x^2-3x+5$ & \raisebox{-.5in}{\includegraphics[width=1.5in]{image9.png}} \\
Reciprocal & $\displaystyle y=\frac{5}{x}$ & \raisebox{-.5in}{\includegraphics[width=1.5in]{image10.png}} \\
Square Root & $\displaystyle y=2\sqrt{x}$ & \raisebox{-.5in}{\includegraphics[width=1.5in]{image11.png}} \\
Exponential & $\displaystyle y=e^x$ & \raisebox{-.5in}{\includegraphics[width=1.5in]{image12.png}} \\ \hline
\end{tabular}
\end{center}

\subsection{Intercepts}
Intercepts are points where a curve crosses a coordinate axis.  To
find them do the following:
\begin{description}
\item[$\bm y$-intercepts]: Set $x$ equal to zero and solve for $y$.
\item[$\bm x$-intercepts]: Set $y$ equal to zero and solve for $x$.
\end{description}
\textbf{Example:}
Find the $x$ and $y$-intercepts of the function $y = x^2 - 5x + 6$.
\begin{description}
\item[$\mathbf y$-intercepts:] ($x = 0$)
\[y = 0^2-5(0)+6=6\]
\begin{center}
$y$-intercept: $(0,6)$
\end{center}
\item[$\mathbf x$-intercepts:] ($y = 0$)
\[x^2-5x+6=0\]
\[(x-2)(x-3)=0\]
\[x-2=0\ \text{or}\ x-3=0\]
\[x=2\ \text{or}\ x=3\] 
\begin{center}
$x$-intercepts: $(2,0)$, $(3,0)$
\end{center}
\end{description}

\subsection{Symmetry}
\raggedright
If the equation remains unchanged when $-y$ is substituted for $y$, the
curve is \textbf{symmetric about the $\bm x$-axis}.

If the equation remains unchanged when $-x$ is substituted for $x$, the
curve is \textbf{symmetric about the $\bm y$-axis}.

If the equation does not change when $-x$ is substituted for $x$ and $-y$
is substituted for $y$, the curve is \textbf{symmetric about the origin}.
\begin{center}
\includegraphics[width=\textwidth]{image13.png}
\end{center}

\textbf{Examples}
\begin{enumerate}
\item $y^2-3x=0$

If $-y$ is substituted for $y$ we get
\begin{eqnarray*}
(-y)^2-3x & = & 0 \\
y^2-3x & = & 0
\end{eqnarray*}
Thus the curve is symmetric about the $x$-axis.

\item $y=x^4+2x^2$

If $-x$ is substituted for $x$ we get
\begin{eqnarray*}
y & = & (-x)^4+2(-x)^2 \\
  & = & x^4+2x^2
\end{eqnarray*}
Thus the curve is symmetric about the $y$-axis.

\item $y=x^3-x$

If $-x$ is substituted for $x$ and $-y$ is substituted for $y$ we get
\[-y =(-x)^3-(-x)=-x^3+x\]
Dividing both sides by $-1$ gives
\[y=x^3-x\]
Thus the curve is symmetric about the origin.
\end{enumerate}
\subsection{Extent (Domain and Range)}

For some functions, there may be certain $x$ or $y$ values that are
not allowed.  For example, some $x$ values may lead to division by
zero, or may result in a negative number under a square root sign. The
set of allowed $x$ values is the \emph{domain} of the function.  The
set of $y$ values that correspond to the allowed $x$ values is the
\emph{range} of the function.

\textbf{Example}
\[y=\sqrt{5-x}\]
For this function, only $x$ values less than or equal to $5$ are
allowed, so the domain of the function is $x\le 5$.  Since the square
root function by definition gives only non-negative values, and since
the $x$ values in the domain make possible any such value, only $y$
values greater than or equal to zero are possible. (The range of the
function is $y\ge 0$.)

\subsection{Asymptotes}

If $y$ approaches infinity (positive or negative) as x approaches some
value (from below or from above), there will be a \emph{vertical asymptote} at
that $x$ value.

If $y$ approaches some value as $x$ approaches positive or negative
infinity, there will be a \emph{horizontal asymptote} at that $y$ value.

\textbf{Example}
\[y=\frac{3}{x-1}\]

As $x$ approaches $1$ from below, $y$ becomes infinite in the negative
direction.  As $x$ approaches $1$ from above, $y$ becomes infinite in
the positive direction.  In symbols:
\[\lim_{x\rightarrow 1^{-}}\frac{3}{x-1}=-\infty\ \ \ \ \ \ \ \lim_{x\rightarrow 1^{+}}\frac{3}{x-1}=+\infty\]
Thus $x=1$ is a vertical asymptote.

As $x$ approaches positive infinity, $y$ approaches $0$ from above.
As $x$ approaches negative infinity, $y$ approaches $0$ from below.
In symbols:
\[\lim_{x\rightarrow \infty}\frac{3}{x-1}=0\ \ \ \ \ \ \ \lim_{x\rightarrow -\infty}\frac{3}{x-1}=0\]
 Thus $y=0$ is a horizontal asymptote.

\subsection{Large x values}
What happens to y as x gets very large in the positive or negative
direction?  In addition to horizontal asymptotes one may find the
function is approximately equal to a simpler function for large $x$
which will determine its large $x$ behaviour.

\textbf{Example}
\[y = x^4 - 3x^2 + 5\]
Since
\[\lim_{x\rightarrow\pm\infty}\left(x^4 - 3x^2 + 5\right)=\lim_{x\rightarrow\pm\infty}x^4\left(1-\frac{3}{x^2}+\frac{5}{x^4}\right)\approx x^4\ ,\]
it follows that as $x\rightarrow\pm\infty$, $y\rightarrow\infty$
  since the $x^4$ term predominates.

\subsection{Stationary Points (Maxima and Minima)}

Find stationary points by setting $y'$ equal to zero and solve for
$x$.  Apply the second derivative test (or the first derivative or
ordinate tests if the second derivative test fails) to determine if a
stationary point is a maximum, a minimum or neither.

\textbf{Example}
\[y=x^2+3x\]
Take the derivatives:
\begin{eqnarray*}
y'&=&2x+3 \\
y''&=& 2
\end{eqnarray*}
Solving $0=2x+3$ gives $x=-3/2$.  The corresponding $y$ value is
\[y=(-3/2)^2+3(-3/2)=-9/4\ .\] Since $y''(-3/2)=2$ is greater than
zero we have a relative minimum at $(-3/2,-9/4)$.

\subsection{Increasing and Decreasing Functions} 
\begin{itemize}
\item When $y'$ is positive the function is increasing.
\item When $y'$ is negative the function is decreasing.
\end{itemize}

\textbf{Example}

Continuing with the last example, we now solve $y'<0$:
\begin{eqnarray*}
2x+3 & <  & 0 \\
2x & <  & -3 \\
x & < & -\frac{3}{2}
\end{eqnarray*}
Hence when $x<-3/2$, $y'<0$, so the function is decreasing when $x<-3/2$.
Similarly solving $2x+3>0$ shows $y'>0$ when $x>-3/2$ so the function
is increasing when $x>-3/2$.

Note that regions of increasing or decreasing are often broken up by
maxima and minima since $y'$ often changes sign at such points. In
this example there was a minimum with $x$-coordinate of $-3/2$.

\subsection{Inflection Points}

\begin{itemize}
\item Set $y''$ equal to zero and solve for $x$.  Test to see whether
the sign of $y'$ changes on either side of the $x$ value. 
If it does, it is an inflection point.
\end{itemize}
\clearpage
\textbf{Example}
\[y=-x^3+2x^2\]
Take the derivatives:
\begin{eqnarray*}
y' & = & -3x^2+4x \\
y'' & = & -6x+4
\end{eqnarray*}
Solving $y''=0$ results in $x=-4/-6=2/3$ with corresponding $y$ coordinate
of $y=-(2/3)^3+2(2/3)^2=16/27$.  To the left and right of $x=2/3=.667$
we have $y''(.6)=.4>0$ and $y''(.7)=-.2<0$ so there is an inflection
point at $(2/3,16/27)$.

\subsection{Curvature}
\begin{itemize}
\item When $y''$ is positive the function is concave upward.
\item When $y''$ is negative the function is concave downward.
\end{itemize}

\textbf{Example}

Contining the last example we solve the inequality $y''<0$:
\begin{eqnarray*}
 -6x+4 & <  & 0 \\
-6x & < & -4 \\
x & > & \frac{-4}{-6} \\
x & > & \frac{2}{3}
\end{eqnarray*}
Hence when $x>2/3$, $y''<0$, so the function is concave downward for
$x>2/3$.  Similarly solving $-6x+4>0$ shows $y''>0$ when $x<2/3$ so
the function is concave upward when $x<2/3$. (Note in solving the
above inequality we had to change the direction of the inequality
because we divided (or multiplied) by a negative quantity.)

Note that regions of curvature may be separated by inflection points
as can be seen here with an inflection point with $x$-coordinate of
$x=2/3$.

\subsection{Summary of Steps in Curve Sketching}
The above steps for curve sketching are summarized on the formula sheets
at the end of the unit.  Follow through them as you do systematic curve sketches
of the following functions.
\clearpage
\textbf{Examples}

Sketch the graphs of the following functions:
\begin{enumerate}
\item $\displaystyle y=\frac{2}{x^2+1}$
\item $\displaystyle y=\frac{x^2}{x-2}$
\end{enumerate}

\textassignment{Sec. 28-3}{Ex. 3 (P. 823) \# 1-21 (odd)}

\section{Newton's Method}

Newton's method is a numerical method for solving equations.  Newton's
method involves making an initial guess as to the solution of an
equation of the form $f(x) = 0$, then refining this guess to get a
better approximation of the solution.  The process is repeated until
the desired degree of accuracy is obtained.

The root, or solution, of the equation, is the value of x at which the
graph of the function crosses the $x$-axis.
\begin{window}[0,r,%
\mbox{\includegraphics[width=2.5in]{image14.png}},]
Let $x_1$ be the first guess of the root.

The slope of the tangent at $x_1$ is given by

$\displaystyle f'\left(x_1\right)=\frac{f\left(x_1\right)}{h}$

Solving for $h$, we get

$\displaystyle h =\frac{f\left(x_1\right)}{f'\left(x_1\right)}$

\end{window}
Let $x_2$ be the point where the tangent crosses the $x$-axis.
\begin{eqnarray*}
x_2 & = & x_1-h \\
    & = & x_1-\frac{f\left(x_1\right)}{f'\left(x_1\right)}
\end{eqnarray*}

In general, if $x_n$ is the $n^{\text th}$ approximation of the
solution, the $(n+1)^{\text{st}}$ approximation is given by the
following formula:

\begin{boxedequation}
x_{n+1}=x_n-\frac{f\left(x_n\right)}{f'\left(x_n\right)}
\end{boxedequation}


\textbf{Examples}
\begin{enumerate}
\item Use Newton's method to find the positive root of the equation
  $x^4+8x-12=0$ to two decimal places.
\clearpage
\item Use Newton's method to find the positive root of
the equation $x^2=\sqrt{2x+1}$  to three decimal places.
\end{enumerate}
\textassignment{Sec. 28-4}{Ex. 4 (P. 825) \# 1-11 (odd)}

\section{Rates of Change}
\subsection{Introduction}
Recall that the value of the first derivative gives the rate of change
of a function with respect to the independent variable.

\textbf{Example}

Illumination ($I$) from a light source is inversely proportional to
the distance squared~($d^2$) from the source.  If the illumination
from a light source is 702 lux at a distance of 3.25 m, what is the
rate of change of illumination with respect to distance at 3.25 m?

\subsection{Equations of Motion}
An important case of a rate of change is that of position with time.
Position can be measured in one spatial dimension or higher.  As well we can
speak of angular position changing in time.
\subsubsection{Straight-line motion}
If an object is moving along one dimension (perhaps straight up and
down or along a straight horizontal line) we can give its position by
stating its displacement $s$ along that axis.  If it is moving along the
axis then $s$ is a function of time, $s(t)$.

The \emph{velocity} of a moving object is defined as the instantaneous rate of
change of displacement with respect to time, i.e. the derivative of
displacement with respect to time.
\begin{boxedequation}
v=\frac{ds}{dt}
\end{boxedequation}

The \emph{acceleration} of a moving object is defined as the instantaneous rate
of change of the velocity with respect to time.  This is the derivative
of the velocity, or the second derivative of the displacement.
\begin{boxedequation}
a = \frac{dv}{dt}=\frac{d^2 s}{dt^2}
\end{boxedequation}
Note that the velocity and acceleration themselves will, in general,
depend on time.

\textbf{Example}

A ball thrown straight up has a vertical displacement given by
\[s=30t-4.9t^2\]
where $s$ is the vertical displacement in metres and $t$ is the time in
seconds.  Find
\begin{enumerate}[a)]
\item Its velocity as a function of time
\item Its acceleration as a function of time
\item The greatest height reached by the ball
\item The velocity of the ball when it reaches the ground. (Answer to 2
significant digits.)
\end{enumerate}

\subsubsection{Curvilinear Motion}
Curvilinear motion is motion which is not in a straight line.  It may
occur in two or three spatial dimensions.  We will consider two
spatial dimensions but our results are easily generalized to three.
In two spatial dimensions the position is determined by the displacent
vector $\bm s=(x,y)$ from some fixed origin.  Since the position is
changing in time it is given by the vector function $\bm
s(t)=\left(x(t),y(t)\right)$, where $x(t)$ and $y(t)$ are just real
functions giving the $x$- and $y$-coordinates at some time $t$.

With two spatial dimensions the velocity $\bm v=(v_x,v_y)$ and
acceleration $\bm a=(a_x,a_y)$ are also vectors.  Their components are
determined by differentiating $\bm s=(x,y)$ component-wise with
respect to time as shown on the following table:
\begin{center}
\begin{tabular}{|c|c|c|}\hline
Vector & x-component & y-component \\ \hline
\rule[-1.1em]{0em}{2.8em}Velocity $\bm v=(v_x,v_y)$ & $\displaystyle v_x=\frac{dx}{dt}$ & $\displaystyle v_y=\frac{dy}{dt}$ \\ \hline
\rule[-1.1em]{0em}{2.8em}Acceleration $\bm a=(a_x,a_y)$ & $\displaystyle a_x=\frac{dv_x}{dt}=\frac{d^2x}{dt^2}$ & $\displaystyle a_y=\frac{dv_y}{dt}=\frac{d^2y}{dt^2}$ \\ \hline
\end{tabular}
\end{center}
To find the magnitude and direction of the velocity or acceleration,
find the $x$ and $y$ components of the velocity or acceleration then
find the magnitude and direction from the components as done when
converting from rectangular to polar vectors in Math~120.

\textbf{Example}

A point moves so that its horizontal and vertical displacements
are given by
\begin{eqnarray*}
x & = & 4t^2-5\\
y & = & 2-3t^3
\end{eqnarray*}
where $x$ and $y$ are in metres, $t$ is in seconds.
\begin{enumerate}
\item Find the magnitude and direction of its velocity when $t=1.25\ \mbox{s}$. 
(Give the answers to 3 significant digits)
\item Find the acceleration ($x$ and $y$ components) at $t=1.25\ \mbox{s}$.
\end{enumerate}

%This problem is unnecessarily tricky, so I omitted it and added acceleration
%to the last one.
%A point moves along the curve $y=2x^3-3x^2-3$ at a speed of
%14.0 m/s.  Find the direction of its velocity when x = 1.50 m, and the
%$x$- and $y$-components of its velocity at that point.

\paragraph{Projectiles}

Projectile motion is the special case of curvilinear motion which
occurs when an object is shot or thrown with an initial velocity
$\bm{v_0}=(v_{0x},v_{0y})$ and subsequently only feels a downward
acceleration (negative $y$ direction) due to gravity.

The motion of a projectile moving in the earth's
gravitational field is then described by the following equations for $x$ and
$y$:
\begin{eqnarray*}
x & = & v_{0x} t \\
y & = & v_{0y}t-\frac{gt^2}{2}
\end{eqnarray*}
Here $v_{0x}$ is the initial velocity $x$ component, $v_{0y}$ is the
initial velocity $y$ component, $g$ is the acceleration of gravity
($9.8\ \mbox{m}/\mbox{s}^2$ or $32\ \mbox{ft}/\mbox{s}^2$).  These are
all constants while $t$ is time.  One can solve for the velocity and
acceleration of a projectile once and for all by differentiating
components.  The veolocity $\bm v=(v_x,v_y)=(dx/dt,dy/dt)$ has
components:
\begin{eqnarray*}
v_x & = & v_{0x} \\
v_y & = & v_{0y}-gt\ ,
\end{eqnarray*}
while the acceleration $\bm a=(a_x,a_y)=(dv_x/dt,dv_y/dt)$ has components:
\begin{eqnarray*}
a_x & = & 0 \\
a_y & = & -g 
\end{eqnarray*}

\textbf{Example}

An arrow is shot with initial speed of 220 ft/s at an angle of
$33.0^\circ$.  Find the position and velocity (components) at time
$t=4.00\ \mbox{s}$
\clearpage

\subsubsection{Rotational motion}
An object that is rotating, such as a horse on a merry-go-round, can
have its position specified by the angle $\theta$ it makes with a
fixed set of coordinates with origin at the centre of rotation.  If
the merry-go-round is moving in time then one has the angle
$\theta(t)$ at time $t$.

The angular velocity, $\omega$, of a rotating body is defined as the
instantaneous rate of change of the angular displacement, $\theta$.
\begin{boxedequation}
\omega=\frac{d\theta}{dt}
\end{boxedequation}

The angular acceleration, $\alpha$, of a rotating body is defined as
the instantaneous rate of change of the angular velocity, $\omega$.
\begin{boxedequation}
\alpha=\frac{d\omega}{dt}=\frac{d^2 \theta}{dt^2}
\end{boxedequation}

The SI units of angular velocity are rad/s, and the SI units of angular
acceleration are $\mbox{rad}/\mbox{s}^2$.

\textbf{Example}

The angular displacement of a rotating body is given by
\[\theta=184+271 t^3\]
Find the angular velocity and angular acceleration at t = 1.25 s.

\textassignment{Sec. 29-1; 29-2}{Exercise 1, p. 830 \# 1-7 (odd); Exercise 2, p. 837 \# 1-19, 22, 24}

\section{Related Rate Problems}
If a problem involves two variables, both of which vary with respect
to time, and the relationship between the two variables is known, it
is possible to express the time rate of change of one in terms of the
time rate of change of the other.  This is done by differentiating the
equation relating the variables with respect to time.  The steps to be
followed are outlined on page 838 of the text and are summarized on
page~\pageref{sec:relatedratesteps} at the end of this unit.

\textbf{Examples}
\begin{enumerate}
\item A leaking pipe drips water on a floor creating a circular pool
  of water whose area is increasing a constant rate of $2.00\
  \mbox{cm}^2/\mbox{s}$. Find the rate at which the radius of the pool is
  changing when the radius is $5.00~\mbox{cm}$.

\item A spherical balloon is being blown up at a constant rate of
  $2.00\ \mbox{m}^3/\mbox{min}$.  Find the rate at which the radius is
  increasing when it is 3.00 m.
\item A tank in the shape of an inverted cone has a radius of 8.00~m
  at the top and a height of 15.0~m.  At the instant when the water in
  the tank is 5.00~m deep, the surface level is rising at a rate of
  0.500~m/min.  Find the rate at which water is being added.
\end{enumerate}

\textassignment{Sec. 29-3}{Exercise 3, p. 841 \# 1-29 (odd) -- except \# 13}

\section{Optimization Problems}

Optimization problems require us to find the value of the independent
variable that results in either a maximum or minimum value of the
dependent variable.  The steps for solving optimization problems are
given on page 845 of the text and summarized at the end of these notes
on page~\pageref{sec:optimizationsteps}.  Effectively one is just
finding the maxima and minima of functions derived from a word
problem.

\textbf{Examples}
\begin{enumerate}
\item What are the dimensions of the rectangle with the largest area
that can be enclosed with a perimeter of 16.0 m?
\item An architect is designing a rectangular building in which the
  front wall costs twice as much per linear metre as the other 3
  walls.  The building is to cover $1350\ \mbox{m}^2$.  What
  dimensions must it have to minimize the cost of the walls?
\item \begin{minipage}[t]{4in} A construction company desires to build
    an apartment building in the shape of a rectangular parallelpiped
    (shown) with fixed volume of $32\bar{0}00\; \mbox{m}^3$. The
    building is to have a square base.  In order to minimize heat
    loss, the total above ground surface area (the area of the four
    sides and the roof) is to be minimimized.  Find the optimal
    dimensions (base length $x$ and height $h$) of the building.
\end{minipage}
\raisebox{-1.5in}{\resizebox{1in}{!}{\input optimization.pdf_t}} % sets the width to 1in
\item A can is to be made in the shape of a right circular cylinder.
  Find the optimal dimensions (radius $r$ and height $h$) which
  maximize its volume if it is to have a fixed total surface area $a$.
  (Your expressions for $r$ and $h$ should be in terms of the constant
  $a$.)  Find the ratio $\frac{h}{r}$ of this optimal can. Why would a
  manufacturer want to have a maximal container volume for a fixed
  surface area?
\end{enumerate}
\textassignment{Sec. 29-4}{Ex. 4 (p. 850) \# 7, 9, 11, 17, 27, 29}

\section{The Differential of a Function}

\subsection{Definition}
The \emph{differentials} $dx$ and $dy$ for a function $f(x)$ are
defined as follows:
\begin{description}
\item[dx] is an arbitrary change in x (analagous to an increment
$\Delta x$).
\item[dy] is defined by the equation
\begin{boxedequation}
dy=f'(x)dx
\end{boxedequation}
\end{description}

Note that the above equation can be rearranged to write the ratio of dy
to dx as follows:
\[\frac{dy}{dx}=f'(x)\]

Thus the derivative, $\frac{dy}{dx}$, can now be viewed as the
quotient of the two differentials, $dy$ and $dx$.

\begin{window}[0,r,%
\mbox{\includegraphics[width=2.8in]{image17.png}},]
\begin{minipage}{3in}
Geometric meaning of the increment, $\Delta y$ and the differential,
$dy$:
\begin{itemize}
\item $\Delta y$ is the change in $y$ on the curve for a given change in $x$,
$\Delta x$.
\item $dy$ is the change in $y$ along the tangent for a given change in $x$, $dx$ (or $\Delta x$).
\end{itemize}
\end{minipage}
\end{window}

\textbf{Examples}

Find the differential dy for each of the following:
\begin{enumerate}
\item $\displaystyle y=4x^3-x^2$
\item $\displaystyle y=\frac{x}{2x+1}$
\item $\displaystyle x^2-2xy+y^3=7$
\end{enumerate}
\clearpage
\subsection{Estimating Small Changes in a Function Using Differentials}

If $\Delta x$ is small, then the increment $\Delta y$ is approximately
equal to the differential $dy$, i.e., setting $dx=\Delta x$ in our differential formula
\[dy=\frac{dy}{dx} dx\]
gives
\[\Delta y\approx \frac{dy}{dx} \Delta x\ .\]
The calculation of the differential is often simpler than calculation of
the increment, as illustrated in the first example below.

\textbf{Examples}
\begin{enumerate}
\item Estimate the change in the function $y = 3x^2$ when $x$ changes from
3.00 to 3.01.  Compare the estimate with the exact value of
$\Delta y$.
\item A metal ball when heated increases in radius from 1.250 cm to
1.252 cm.  Estimate the resulting change in volume.
\end{enumerate}

\subsection{Approximate Volumes of Shells and Rings}
The volume of a thin shell or ring is estimated as the change in volume
when the radius (or other linear dimension) increases by an amount
equal to the wall thickness.  In symbols,
\[\Delta V\approx \frac{dV}{dr}\Delta r\]

\textbf{Examples}
\begin{enumerate}
\item
\begin{enumerate}
\item Derive a formula for the volume of material in a closed
thin-walled cubical box of edge length $L$ and wall thickness $t$.
\item Use the formula to estimate the volume of material in a
cubical box 75.0 cm on a side with walls 0.240 cm thick.
\end{enumerate}
\item
\begin{enumerate}
\item Derive a formula for the volume of material in a metal ring
of radius $r$, height $h$ and thickness $t$.
\item Use the formula to find the volume of material in a ring 25.0
cm in diameter, 3.00 cm high and 0.120 cm thick.
\end{enumerate}
\end{enumerate}

\subsection{Errors in Calculated Quantities}

If a measured quantity has an \emph{absolute error} $\Delta x$, a quantity
$y$ calculated from it will have an absolute error given, to a good
approximation, by
\begin{boxedequation}
\Delta y\approx \frac{dy}{dx}\Delta x
\end{boxedequation}
The \emph{relative error}, $\Delta y/y$, can be approximated by dividing the
right hand side of the last equation by $y$.


\textbf{Examples}
\begin{enumerate}
\item What error in the area of a square room results from an error of
  0.02 m in the length, which is measured as 15.25 m?
\item To what percent accuracy must the radius of a ball bearing be
  measured so the calculated volume will be correct to 0.3\%?
\end{enumerate}

\textbf{Exercise}
\begin{enumerate}
\item Find the differential of each of the following functions:
\begin{enumerate}
\item $y=3x^2+6$
\item $y=(x^2-1)^4$
\end{enumerate}
\item Find $\Delta y$ and $dy$ for each of the following:
\begin{enumerate}
\item $y=7x^2 + 4x$; $x=4$, $\Delta x=0.2$
\item $y=(1-3x)^5$; $x=1$, $\Delta x=0.01$
\end{enumerate}
\item Estimate the volume of rubber in a rubber-covered ball 25.0 cm
  in diameter if the rubber is 0.125 cm thick. (The volume of a sphere
  is given by $V=\frac{4}{3}\pi r^3$\ .)
\item If the measurement of the radius of a circular patio has an error
of $0.50\ \mbox{cm}$, what is the error in the calculated area if the
radius is measured as 1.35 m? (Area of a circle is given by $A=\pi r^2$\ .)
\end{enumerate}
\textassignment{Sec. 27-6}{Ex. 6 (p. 801) \# 29-39 (odd)}
\textbf{Answers}
\begin{enumerate}
\item
\begin{enumerate}
\item $dy = 6x\,dx$
\item $dy = 8x(x^2-1)^3\,dx$
\end{enumerate}
\item
\begin{enumerate}
\item $\Delta y = 12.28$,  $dy = 12$
\item $\Delta y = -2.473$, $dy = -2.4$
\end{enumerate}
\item $245\ \mbox{cm}^3$
\item $0.042\ \mbox{m}^2$
\end{enumerate}
\clearpage

\section*{Review Problems: Applications of the Derivative}
\begin{enumerate}
\item Write the equations, in general form, of the tangent and normal
  to the curve $y^2=4x$ at the point $(1,2)$.

\item Use Newton's method to find the positive root of
  the equation
\[x^3+3x-8=0\]
to $2$ decimal places.

\item Find the maximum, minimum and inflection points of the curve
\[y=x^3-3x+4\]

\item Using curve sketching techniques demonstrated in class, sketch
  the graph of the function $\displaystyle y=\frac{3}{x}+x^2$ (give
  critical points in decimal form to 3 significant digits).

\item The height $h$ (in kilometres) to which a balloon rises in $t$
minutes is given by the formula
\[h=\frac{10t}{\sqrt{4000+t^2}}\]
At what rate is the balloon rising at the end of $30$ minutes?

\item The horizontal and vertical displacements of a projectile moving
in the earth's gravitational field are given by
\[x=v_{0x}t\ \ y=v_{0y}t-\frac{gt^2}{2}\] where $v_{0x}$ and $v_{0y}$
  are the initial horizontal and vertical velocities respectively, and
  $g=9.80\ \mbox{m}/\mbox{s}^2$.  A cannonball is fired with an
  initial velocity of 175 m/s at an angle of $45.0^\circ$
  to the horizontal.  Find its horizontal and vertical velocities
  after 10.0 s.

\item A ladder 10.0 m long is leaning against a vertical wall.  If the
foot of the ladder is being pulled away from the wall at 0.500 m/s, how
fast is the top of the ladder moving down the wall when the foot of the
ladder is 2.00 m from the wall?

\item A rope attached to a boat is being pulled in by a winch at a rate
of 2.50 m/s.  If the boat is 5.00 m below the level of the winch, how
fast is the boat approaching the wharf when 13.0 m of rope are out?

\item Sand is emptied down a chute at the rate of $3.00\ \mbox{m}^3/\mbox{s}$,
  forming a conical pile whose height is always twice the radius.  At
  what rate is the radius changing when the height is 2.00 m?  (Volume
  of a cone is given by $V=\frac{1}{3}\pi r^2h$.)

\item \begin{minipage}[t]{4in}
    A window at the front of a house with a cathedral ceiling is to be
    designed in the shape of a rectangle with a semicircle on top, as
    shown.  If the perimeter of the window is to be 8.00 m, find the
    dimensions, w and h, of the window that will maximize the area of
    the window.
\end{minipage}\raisebox{-1.2in}{\includegraphics[width=1.3in]{image18.png}}
\item An open rectangular box is to be formed by cutting a square from
  each corner of a square piece of cardboard 24.0 cm on each side, as
  shown.  After the squares are removed the sides are folded up to
  form the box.
\begin{window}[0,r,%
    \mbox{\includegraphics[width=1.5in]{image19.png}},]
\begin{minipage}{4in}
\begin{enumerate}
\item How large a square must be removed from each corner in order
for the box to have the maximum volume?
\item What will the maximum volume of the box be?
\end{enumerate}
\vspace{.2in}
\end{minipage}
\end{window}

\item A coat of paint 0.050 mm thick is to be applied to the outside
surface of a spherical storage tank with a radius of 20.0 m.  Use
differentials to estimate the number of litres of paint that will be
needed. 

(Volume of a sphere is given by  $V=\frac{4}{3}\pi r^3$.)

\item A disk-shaped solar cell has a measured radius of 120.0 mm.  The
error in measurement of the radius of the cell is 0.050 mm.  Use
differentials to estimate:
\begin{enumerate}
\item The absolute error in the area of the top surface of the solar
cell
\item The relative (percent) error in the area of the top surface of the
solar cell
\end{enumerate}
\end{enumerate}
\clearpage
\section*{Answers:}
\begin{enumerate}
\item Tangent: $x-y+1 = 0$; Normal:$x+y-3=0$
\item $x = 1.51$
\item Maximum: $(-1, 6)$; Minimum: $(1, 2)$; Inflection point: $(0, 4)$
\item Minimum: $(1.14,3.93)$; Inflection Point: $(-1.44,0)$

Plot:

\includegraphics[width=2in]{image15.png}

\item 0.117 km/min
\item $v_x=124$ m/s, $v_y=25.7$ m/s
\item 0.102 m/s
\item 2.71 m/s
\item 0.477 m/s
\item $w=2.24$ m, $h=1.12$ m
\item \begin{enumerate}
\item 4.00 cm
\item $1020\ \mbox{cm}^3$
\end{enumerate}
\item 251 L
\item \begin{enumerate}
\item $38\ \mbox{mm}^2$
\item $0.083\%$
\end{enumerate}
\end{enumerate}
\clearpage
\section*{Additional Review Problems}
\begin{itemize}
\item Chapter 28 Review (P. 826) \# 1-16 (all except \# 12)
\item Chapter 29 Review (p. 854) \# 3, 5, 6, 7, 11, 13, 19, 21
\end{itemize}

\subsubsection*{Even-Numbered Answers}
\begin{itemize}
\item Chapter 28:
\begin{description}
\item[2.] (0,0)
\item[4.] Maximum: (-1, 5.33); Minimum: (3, -5.33); Inflection Point: (1,0)
\item[6.] (2,3)
\item[8.] Graph
\item[10.] Decreasing
\item[14.] $x$-intercept of tangent: (1.29, 0), $x$-intercept of normal: (818, 0)
\item[16.] Graph
\end{description}
\item Chapter 29:
\begin{description}
\item[6.] $v=10\ \mbox{units}/\mbox{s}$, $a = 8\ \mbox{units}/\mbox{s}^2$
\end{description}
\end{itemize}

\clearpage
\section*{Formulas}
\setcounter{equation}{0}
\subsection*{Newton's Method}
\begin{boxedequation}
x_{n+1}=x_n-\frac{f\left(x_n\right)}{f'\left(x_n\right)}
\end{boxedequation}
\subsection*{Equations of Motion}
\begin{boxedequation}
v=\frac{ds}{dt}
\end{boxedequation}
\begin{boxedequation}
a = \frac{dv}{dt}=\frac{d^2 s}{dt^2}
\end{boxedequation}
\begin{boxedequation}
\omega=\frac{d\theta}{dt}
\end{boxedequation}
\begin{boxedequation}
\alpha=\frac{d\omega}{dt}=\frac{d^2 \theta}{dt^2}
\end{boxedequation}
\subsection*{Differentials}
\begin{boxedequation}
dy=f'(x)dx
\end{boxedequation}
\begin{boxedequation}
\Delta y\approx \frac{dy}{dx}\Delta x
\end{boxedequation}

\clearpage
\section*{Systematic Procedure for Critical Points}
\begin{description}
\item[Stationary Points (Maxima, Minima):]~
\begin{enumerate}
\item Write down the original function $y$, the first derivative $y'$,
  and the second derivative $y''$.
\item Set the first derivative, $y'$, to zero and solve for $x$.  This
  gives the \mbox{$x$-coordinates} of any stationary points.
\item Substitute these $x$-values into the original function, $y$, to
  find the $y$-coordinates of the stationary points.
\item Perform the \textbf{second derivative test} on each stationary
  point by substituting its x-value into the second derivative,~$y''$.
\begin{itemize}  
\item If $y''>0$, the point is a minimum point.
\item If $y''<0$, the point is a maximum point.
\item If $y''=0$, the test is inconclusive.
\end{itemize}
\item If $y''$ turns out to be zero, use one of the following tests to
  determine if the stationary point is a maximum, a minimum or
  neither:
\begin{description}
\item[First derivative test:] The first derivative $y'$ is:
\begin{itemize} 
\item negative to the left and positive to the right of a minimum
  point.
\item positive to the left and negative to the right of a maximum
  point.
\item the same sign on both sides of a stationary point that is
  neither a maximum nor a minimum.
\end{itemize}
\item[Ordinate test:] the $y$-value is:
\begin{itemize}
\item greater on both sides of a minimum point.
\item less on both sides of a maximum point.
\item greater on one side and less on the other side of a stationary
  point that is neither a maximum nor a minimum.
\end{itemize}
\end{description}
\item Write down the coordinates of each stationary point, and state
  whether each one is a maximum, a minimum or neither.
\end{enumerate}
\item[Inflection Points:]~
\begin{enumerate}
\item To find inflection points, set the second derivative equal to
  zero and solve for x.
\item Check to see whether the second derivative changes sign on
  either side of the point(s).
\item Substitute the $x$-coordinates of any inflection points into the
  original function to find the $y$-coordinates of the inflection
  points.
\item Write down the $x$ and $y$-coordinates of the inflection
  points.
\end{enumerate}
\end{description}
\clearpage
\section*{Steps for Curve Sketching}
\begin{enumerate}
\item Check to see if the equation represents a standard type of curve
(e.g. power function, exponential function, etc.).
\item Determine the $x$ and $y$-intercepts (if possible).
\item Determine whether the curve is symmetric about the $x$-axis,
$y$-axis or origin.
\item Determine the domain and (if possible) the range of the function.
\item Determine whether there are any vertical or horizontal asymptotes.
\item Determine what happens to the function for large (positive and
negative) $x$-values.
\item Locate any stationary points, and determine if they are maximum or
minimum points, or neither.
\item Determine for what $x$-values the function is increasing or
decreasing.
\item Locate any inflection points.
\item Determine for what $x$ values the curve is concave upward or
concave downward.
\end{enumerate}

\clearpage
\section*{Steps for Related Rate Problems}
\label{sec:relatedratesteps}
\begin{enumerate}
\item Locate the given rate and express it as a derivative with
  respect to time.
\item Determine the unknown rate and express it as a derivative with
  respect to time.
\item Find an equation linking the variables in the known and unknown
  rates.  If there are other variables in the equation, eliminate them
  using other known relationships.
\item Take the derivative of the equation with respect to time.
\item Subsitute known values and solve for the unknown rate.
\end{enumerate}

\vspace{1em}
\textbf{\Large Steps for Optimization Problems}
\label{sec:optimizationsteps}
\begin{enumerate}
\item Locate the quantity (for purpose of these steps call it $y$) to
  be maximized (or minimized).
\item Locate the quantity (call it $x$) which is to be varied to
  maximize (or minimize) $y$.
\item Write an equation linking $y$ and $x$.  If the equation contains
  any other variables, eliminate them using a second equation.  A
  graph of $y$ versus $x$ would show maximum (or minimum) points.
\item Take the derivative $\frac{dy}{dx}$.
\item Set the derivative equal to zero and solve for $x$.  Find the
  corresponding $y$ values for these stationary points.
\item Check the stationary points found to see if they are relative
  maxima (or minima) using either the second derivative test, the
  first derivative test or the ordinate test.
\item Find the absolute maximum (or absolute minimum) by comparing $y$
  values of the relative maxima (or relative minima) found.  Also
  check the $y$ values at the endpoints of the physically allowed
  values of $x$ since the largest (or smallest) $y$ value may also lie
  there.
\end{enumerate}
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