Now that you can calculate the derivative of oodles of different
functions, it probably occurs to you to ask why you would wish to do
so. In this unit we introduce three of the most common applications
of the derivative: curve sketching, max-min problems, and related
rates.
Curve sketching uses the first and second derivative to locate points
of interest on the curve. To sketch y=f(x), we first find where it
reaches maximums and minimums. Other than places where the curve is
not "smooth" (i.e. |x| is not smooth at zero.), a maximum or
minimum occurs only when the first derivative is zero. (One way to
see this: for a maximum, the curve must stop going up and start going
down. Thus either it passes through a point where it is level, and
thus y'=0, or the derivative is not continuous, which is another
way of saying the curve isn't smooth.) To aid the accuracy of our
sketch, we also obtain information form the second derivative. This
informs us of the rate at which the slope of the curve is changing.
If this rate is positive, the slope is becoming steadily larger.
Geometrically, the curve is concave up or "holding water."
Similarly for y''<0. Solving for the points where the second
derivative is zero gives us the places where concavity might change.
A few cautions about curve sketching: Just because a derivative
goes to zero doesn't mean is changes sign; it could go from positive
to zero and back to positive. (An example is y=x3
around the
origin.) Also, remember which function y, y',or y'' you
want to use at the moment. If, for example, y''=0 at x=5,
graph the point by finding the value of y at x=5. Finally,
be careful of discontinuities in the function or its derivatives.
Position, slope and concavity can all change drastically at a
discontinuity.
The next application of the derivative is related rates. The usual
problem here is to have a function, say
and a
derivative, say
, and wish to
find another
derivative, in this case dA/dt. In other words, to
relate the unknown rate to the known rate. There are two basic methods,
which are roughly equivalent:
Use implicit differentiation. In this case, differentiate by
time:
Use the chain rule:
Note that in both cases we need to know the current value of the
radius to specify the change in the area. When r=.5 m,
dA/dt =
m/s. Note that these problems can get
tricky, either because of a large number of derivatives floating
about, or because of difficulty in setting up the geometry of the
situation.
Finally, we have max-min problems. The basic situation here is we
have several quantities related to each other by various formulae.
We wish to either minimize or maximize one of the quantities by
appropriate choices for the values of the others. Recall that, in a
smooth region, the maxima and minima occur when the first derivative
is zero. So our method is to set up a derivative, solve it for zero,
then solve backwards to get the desired values. The usual example is
the case of a farmer who lives next to a straight stream. He has
bought 100 m of fence and wishes to enclose as much pasture as
possible, using the stream as one side of the pasture. He also
insists that the pasture be rectangular. To set this up, we call the
length of the side opposite the stream y, the length of one of the
sides adjacent to the river x and obtain
100 = x + y
A = xy
We can eliminate y now or later. Eliminating it now is obvious, so
let's leave it alone for a while. Differentiate to obtain
Differentiate the other equation to get
0 = 2 + dy/dx
Plugging in dy/dx = -2 and setting dA/dx = 0 gives
y = 2x
Solving backwards gives x=25 m, y=50 m, and A=1250
m2.
How do I know this is the maximum, and not a minimum? Well, it's
fairly obvious in a simple example like this, but in general:
Check near the supposed maximum. x=24, y=52 gives
A=1248.x=26, y=48 gives A=1248. Both are smaller than the
solution, so we're okay.
Check borderline cases and discontinuities. Here, x=0,
y=100, and x=50, y=0 both give A=0.
Use common sense. Don't let lengths, areas or volumes go
negative, etc.
The trick to max-min problems is setting things up so that when you
solve for the derivative equaling zero, you actually have an
equation you are capable of solving.
Objectives:
Upon completion of this unit you should be able to
Sketch curves and indicate regions of positive slope, negative
slope, positive and negative concavity.
Solve related rate problems.
Solve max-min problems.
Suggested Procedure:
Read Simmons 4.1-4.5
Sorry, there are not yet any World Web Math entries on this topic :-(
Again, you need to work lots of problems, especially to get the
hang of related rates and max-min.
Simmons has problems on pp. 91,
94, 101, 108, and 112. The problems vary widely in difficulty. Work
your way up to the harder ones by solving many of the easier
ones.
Unit 5 : More Applications of the Derivative