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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Applications of the derivative.}

For most applications, we think of \( f'(x) \) as measuring the \textit{rate
of change} of the function \( f \) with respect to change of the variable \( x \).
That is one reason why we denote the derivative sometimes by \( df/dx \), thinking
of it as the ratio. For functions of time (usually written as \( t \)) the
idea of rate makes the most sense, but we can also talk about the rate of change
of a function with respect to, say, distance as well. 

Here are some standard examples of how the derivative is interpreted.


\subsubsection{Position and velocity:}

If \( f(t) \) measures the position of a particle (or whatever) as it moves
along a line, then the derivative of the position, \( f'(t)=df/dx\, :=v(t) \)
is its \textit{velocity}, and the derivative of the velocity, \( v'(t)=dv/dt\, :=a(t) \)
is the \textit{acceleration} of the particle.

\begin{example}
If the position of a particle (in meters, along a line) is given by \( s=t^{3}-4.5t^{2}-7t \),
for \( t\geq 0 \), find the velocity \( v(t) \), and find when the velocity
reaches \( 5\, m/sec \). What motion did the particle have at \( t=0 \)?
\end{example}

\subsubsection*{Solution:}

\begin{example}
A ball is thrown vertically upward with a velocity of \( 80\, ft/sec \), and
its position \( t \) seconds later is \( s(t)\, :=80t-16t^{2} \) feet up.
How high does it get, and how fast is it moving when it hits the ground again?

{[}Do the ball-toss animation again, but this time hard-coded with s(t)=80{*}t-16{*}t\textasciicircum{}2{]}
Only show the ball going up and down, no need for the velocity and acceleration
plots this time.{]}
\end{example}

\subsubsection*{Solution:}

\begin{xca}
I shot an arrow, into the air. It fell to earth, I know not where. However,
I shot the arrow from a height of 5 feet off the ground, at an angle of \( 45^{\circ } \)
to the horizontal, pointing due North. 
\end{xca}
\textbf{a)} Find a formula for the height \( h(t) \) of the arrow at time \( t \)
(in seconds after I shot the arrow). \( h(t)= \)

\emph{Hint:} Only that part of the initial velocity which is upward pointing
counts. Use trigonometry to decide what the magnitude of the vertical part of
the velocity is.

\textbf{b)} Find when the arrow will hit the ground. \( t= \)

\textbf{c)} Find a formula for the horizontal position \( s(t) \) of the arrow
at time \( t \), how far North of my position it is at time \( t \). \( s(t)= \)

\textbf{d)} Find the position when the arrow hits the ground. \( s= \) This
will answer the poet's question.


\subsubsection{Density}

The \textit{density} \( \rho  \) of a bar or rod is the \textit{mass per unit
length}. But if the bar is not uniform (perhaps not uniform thickness), the
density will not be constant, so to really have a good idea of the density of
the rod at a given point \( x \) along its length, we have to take the infinitesimal
ratio of mass per length at \( x \). If you think of the mass of the rod up
to the point \( x \) as \( m(x) \), then the density is the rate of change
of \( m(x) \), \( \rho =dm/dx \). 

Later on, in other courses, you'll get more interpretations of this notion of
density. Clearly, really, density should be mass per unit \emph{volume}, but
that requires, in its full generality, techniques of third-semester calculus.
So, for the time being, we'll stick to a one-dimensional model. 

\begin{example}
Assume that the thickness of a wire increases slightly along the length, so
that the mass up to \( x \) meters is \( m(x)\, :=3x+x^{2}/10 \) grams. Find
the density three inches from the beginning of the wire.
\end{example}

\subsubsection*{Solution:}

\( \rho (x)=dm/dx=3+x/5 \), so \( \rho (3)=3.6\, g/m \).


\subsubsection{Cost}

The total cost of producing \( x \) items is called (surprisingly) the \textit{cost
function}, \( C(x) \). The derivative, \( C'(x) \), is the \textit{marginal
cost}, and represents (approximately) the cost of producing the next item. The
word {}``marginal{}'' in economics refers to a rate of change, so that your
{}``marginal tax rate{}'' is the tax percentage on the next dollar you earn. 

Typically, the costs involved in producing \( x \) items include a certain
amount of \textit{fixed costs}, which would include the cost of the factory,
machinery, and other things you need no matter how many of the widgets you produce.
In addition, materials and labor costs typically are proportional to the number
of items produced, but there might be some costs that grow more quickly as the
number of items increase, such as warehousing costs. So, a typical cost function
might look like \( C(x)=300+20x+x^{2}/10 \).

\begin{example}
Presume that the cost function to produce \( x \) widgets is \( C(x)=300+20x+x^{2}/10 \).
Then, find the marginal cost at the production level of \( 100 \) widgets.
\end{example}

\subsubsection*{Solution:}

The marginal cost is \( C'(x)=20+x/5 \), so the marginal cost at a production
level of \( 100 \) widgets is \( C'(100)=40 \).


\subsubsection{Linear approximation}

The Leibniz notation for derivatives, \( df/dx \), isn't supposed to mean a
fraction, really. But, in reality it does have some meaning like that. Since
the  derivative is the \textit{limit} of fractions, if \( h \) is \textit{near}
\( a \), then  \( (f(a+h)-f(a))/h \) will be \textit{near} \( f'(a) \). That
is, replacing \( a+h \) by the more generic \( x \) and \( h \) by \( x-a \):
\[
\frac{f(x)-f(a)}{x-a}\approx f'(a),\]


or, solving for \( f(x) \), \[
f(x)\approx f(a)+f'(a)(x-a).\]


Another way to look at this is that the equation of the tangent line \( L \)
to the curve \( y=f(x) \) at \( (a,f(a)) \) is \( y=f(a)+f'(a)(x-a) \). Since
the tangent line is near the graph, the \( y \) will be \textit{almost} the
same as \( f(x) \). 

The point here is to approximate \( f(x) \), which might be a very complicated
function, by a simpler function -- a linear one, which is why it is called \textit{linear
approximation}.

We sometimes write \( \Delta x \) for \( x-a \) (or \( h \)), to indicate
the \textit{change} in \( x \). In that notation: \[
f(a+\Delta x)\approx f(a)+f'(a)\Delta x,\]
 or writing \( \Delta f:=f(a+\Delta x)-f(a) \) to denote the \emph{change in}
\( f \) (the \( \Delta  \) usually refers to change in the quantity),  \( \Delta f\approx f'(a)\Delta x. \)
We sometimes call the right-hand side of that equation the \textit{differential}
of \( f \), but when we do we usually write \( dx \) instead of \( \Delta x \).
That is, \[
df=f'(a)dx,\]
 getting us back to the Leibniz notation for \( f'(x) \), \( df/dx \). The
point of that is to approximate the function by the \textit{linearization},
the \textit{differential}. 

\begin{example}
Find an approximate value of \[
\sqrt{4.2}.\]
 
\end{example}

\subsubsection*{Solution:}

\begin{example}
Use differentials to find the relative error (the percentage of error) in the
computation of the volume of a cube, if the measurements of the sides might
be off by as much as 1\%.
\end{example}

\subsubsection*{Solution}

\begin{xca}
Approximate \[
\sqrt{8.7}\]
 using differentials  (equivalently, using linear approximation). 
\end{xca}

\subsubsection*{Answer}

\begin{xca}
Find the differential of \( x^{3}+2x \). 
\end{xca}

\subsubsection*{Answer}

\begin{xca}
Find the linear approximation of \[
\sqrt{1+x}\]
 for \( x \) near 0. 
\end{xca}

\subsubsection*{Answer}

\begin{xca}
Find the linear approximation of \[
\sqrt{1+x^{2}}\]
 for \( x \) near 0. 
\end{xca}

\subsubsection*{Answer}


\subsubsection{Related Rates Problems}

Related Rates problems are one of the classic types of ``word problems'' in
calculus classes. They provide a first introduction to mathematical \textit{modeling}:
taking a problem, distilling from it the mathematical essentials, then, using
standard mathematical techniques which can be applied to \textit{any} problem
of the same general type, solve the resulting mathematical problem. 

For the problems we'll be looking at now, the underlying mathematics is quite
simple. Let's say we have two functions, \( f \) and \( g \). We can make
them both functions of \( t \) to  keep in the spirit of many of these applications.
Then, if we know a  relationship between \( f \) and \( g \) (some equation
involving them), we can  differentiate both sides of the equation to get a relationship
between their derivatives. In applications, we usually know something about
one of the  derivatives, and we want to figure out the other. 

For example, if  \[
(f(t))^{2}+(g(t))^{2}=1,\]
 then differentiate both sides. The derivative of 1 is 0, and the derivative
of \( (f(t))^{2} \) (with respect to \( t \)) is \( 2f(t)f'(t) \). Similarly,
\( ((g(t))^{2})'=2g(t)g'(t) \), so that  \[
2f(t)f'(t)+2g(t)g'(t)=0,\]


an equation involving the derivatives of \( f \) and \( g \) (the \textit{rates}
of the  functions). 


\subsubsection*{Here is how the idea works in practice }

\begin{example}
A boater has his boat tied to the pier by a  long rope. He feeds the pier-end
of the rope through a winch and pulls his boat in towards the pier. The tide
is out, so the bow of the boat is 6 feet below the level of the pier. If the
winch is pulling in the line at a constant rate of 2 feet per second, how fast
is the boat moving towards the pier when there is 10 feet of rope between the
boat and the winch?  

{[}Animation: Have a pair of arrows to run the winch; one arrow pulls it in,
the other out. Also put in a stop button. The winch always runs at a fixed speed
of 2 feet/sec, and the idea is to see the boat moving faster or slower as the
distance to the dock changes. It's be nice if (1) we came up with a rationale
for why the boat moves outward when the rope is played out (offshore wind?),
(2) if the boat is allowed to run up to the dock, it should crash into the dock,
damaging the dock some, and the boat should sink...{]}
\end{example}
\vspace{0.3cm}
{\par\centering \includegraphics{boat.ps} \par}
\vspace{0.3cm}


\subsubsection*{Solution}

\begin{xca}
Egbert is flying his kite. It is 75 feet off the ground, moving horizontally
away from Egbert at a rate of 5 feet per second. How fast is Egbert letting
out the string when the kite is 100 feet  (horizontally) downwind of him? 
\end{xca}
\vspace{0.3cm}
{\par\centering \includegraphics{kite.ps} \par}
\vspace{0.3cm}


\subsubsection*{Answer}

\begin{xca}
Harpo and Chico are walking away from Groucho, in paths perpendicular to each
other. Harpo is 4 feet from Groucho, walking at 3 feet per second, and Chico
is 3 feet from Groucho, walking at 5 feet per  second. How fast is the distance
between Harpo and Chico changing? 
\end{xca}
\vspace{0.3cm}
{\par\centering \includegraphics{marx.ps} \par}
\vspace{0.3cm}

\begin{note}
If you're curious about who these characters are, take a look at this link http://www.whyaduck.com/contents.htm
to one of their fan sites. They were called the Marx Brothers, and starred in
several hilarious movies in the thirties and forties.
\end{note}
\vspace{0.3cm}
{\par\centering \includegraphics{marxbros.ps} \par}


\subsubsection*{Answer}

\begin{example}
A man, walking at night, is walking directly towards a streetlight. The light
is 10 feet off the ground, and the man is 6 feet tall and walking at 4 feet
per second.  When the man is 8 feet from the streetlight, how fast is the  length
of his shadow changing? 

{[}Animate this as well. Just have the man walk towards, or away, from the streetlight
(again, use buttons like for the boat). The shadow should correspondingly stretch
or shrink. In this one, it is the speed of the man walking that is constant{]}
\end{example}
\vspace{0.3cm}
{\par\centering \includegraphics{streetlight.ps} \par}
\vspace{0.3cm}


\subsubsection*{Solution}

\begin{xca}
A 20 foot ladder is leaning against a wall. A painter stands on the  top of
the ladder, minding his own business. Some fool comes by and ties his dog to
the base of the ladder, a cat comes along, and the dog chases after the cat,
dragging the base of the ladder with him at a rate of 2 feet per second directly
away from the wall. How fast is the painter falling when he is 12 feet from
the ground? 
\end{xca}
\vspace{0.3cm}
{\par\centering \includegraphics{painter.ps} \par}
\vspace{0.3cm}


\subsubsection*{Answer}

\begin{xca}
Egbert is drinking a daquiri (non-alcoholic) out of a glass that is a cone with
a radius at the top of 3 inches and a height of 5 inches. He drinks his daquiri
at a constant rate of 2 cubic inches per second (through a straw). How fast
is the top of the daquiri falling when it  is 4 inches above the bottom of the
glass? 
\end{xca}
\vspace{0.3cm}
{\par\centering \includegraphics{daquiri.ps} \par}
\vspace{0.3cm}


\subsubsection*{Answer}

\begin{xca}
A baseball diamond is in the shape of a square, with sides of length 90 feet.
Who's on first base, What's on second, and Eyedonno's on third. When the ball
is hit, What runs towards Eyedonno on third, and Who runs towards second. When
What is 60 feet from Eyedonno (still on third base), running towards third at
20 feet/sec, and Who is 40 feet from second, running towards second at 25 feet/sec,
how fast is the distance between Who and What changing?
\end{xca}
\vspace{0.3cm}
{\par\centering \includegraphics{who.ps} \par}
\vspace{0.3cm}

\begin{note}
If you sort of recognize what this is all about, then you win a prize. The plot
of this problem is shamelessly stolen from one of the funniest stand-up routines
of all time, the \emph{Who's on First?} routine of Abbott and Costello. Read
about it here. http://netdetours.com/archive/whosonfirst.html The real routine
is a lot funnier than this problem.
\end{note}

\subsubsection*{Answer}
\end{document}
