Applications of the derivative :: Finding
the
minima and maxima
Geometrically speaking, the derivative can
be thought of as a measurement of how quickly
a function is changing. The derivative can
only be found where a function is continuous
which has a special mathematic definition,
but essentially means that the derivative
for a function exists everywhere where the
graph of that function has no breaks in the line
or sharp corners.
Consider the function f(x) = x^{2
}
The graph for this bad boy of a function will
look something like this 
.  .


.  .

.  .
.  .
. < (x=0)




^ (f(x)=y=0)
(imagine those '.'s are connected)
Now you'll notice that as you trace your
finger down your screen on the left hand side,
the line is going down. Hence the function
has a negative rate of change. On the right,
the function goes up  positive rate of
change.
Right in the middle, at x=0 and y=0, you find
the minimum of the function  the lowest point
on the graph. In paragraph one, I mentioned
that the derivative can be thought of as how
quickly a function is changing, the rate of
change. Using the equations given in
Enzondio's node on derivative, we can figure
out what the equation for the derivative is
f(x)=x^{2} so
f'(x)=2x^{21}=2x^{1}=2x
where f'(x) is the derivative
We said before that the lowest point occurs
where the derivative equals zero. To find
where this occurs, we have to solve the equation
f'(x)=0. This is an easy one...
f'(x)=2x=0
2x / 2 = 0 / 2
x = 0
so f'(x)=0 where x=0. This is not surprising.
We can see it from the graph. Duh.
Example
Note: I play it fast and loose with the numerical accuracy on this example, keeping all the numbers to 4 decimal places. If you do the math yourself, you'll prolly notice differences in the less significant digits. If you see any glaring errors, pls. msg me!
Where it gets interesting is when you don't have
a graph. Let's say you have a buddy who throws
a 128 kilometers per hour fastball. Let's say
he throws it
straight up in the air. How long would it take
to hit the top of the arc and how far up would
it go? Let's find out!
K, first, remember that gravity moves things
downwards at 9.81 meters per second per second,
and that
km means kilometers
m means meters
h means hours
s means seconds
128 km per h * 1000 m per km =
128,000 m per h
128,000 m per h / 3600 s per h =
35.555555 m per s
With this info, we can make an equation for how
high that baseball is at a certain time approximately.
There are other factors that come into play,
the air slows the ball down a bit and your
theoretical friend probably doesn't practice
throwing his fastball straight up, for example,
but for the purposes of this example, let's not
worry about it. The equation we'd come up with
is as follows:
Let's say the height of the fastball at a time 't'
is h(t), where
h(t) = 35.5555 m per s * t  9.81 m per s * t^{2}
Using what we learned above, and the information on
derivation in Enzondio's node on derivative, we know
h'(t) = 35.5555 m per s * t^{11}  9.81 m per s^{2} * 2 * t^{21}
h'(t) = 35.5555 m per s * t^{0}  9.81 m per s^{2} * 2 * t^{1}
h'(t) = 35.5555 m per s * 1  9.81 m per s^{2} * 2 * t
h'(t) = 35.5555 m per s  9.81 m per s^{2} * 2 * t
and the minimum occurs at h'(t) = 0, so
0 = h'(t) = 35.5555 m per s  9.81 m per s^{2} * 2 * t
0 + 9.81 m per s^{2} * 2 * t = 35.5555 m per s
 9.81 m per s^{2} * 2 * t + 9.81 m per s^{2} * 2 * t
(9.81 m per s^{2} * 2 * t) / 9.81 m per s^{2} = 35.5555 m per s / 9.81 m per s^{2}
2 * t = 3.6244 s
t = 1.8122 s
So the ball would reach the top height at just about
2 seconds. To find the height the ball gets to, use
our first equation with t=1.8122...
h(t) = 35.5555 m per s * t  9.81 m per s^{2} * t^{2}
h(1.8122 s) = 35.5555 m per s * 1.8122 s 9.81 m per s^{2} * (1.8122 s)^{2}
h(1.8122 s) = 64.4336 m  9.81 m per s^{2} * 3.2840 s^{2}
h(1.8122 s) = 64.4336 m  9.81 m per s^{2} * 3.2840 s^{2}
h(1.8122 s) = 64.4336 m  32.2167 m
h(1.8122 s) = 64.4336 m  32.2167 m
h(1.8122 s) = 32.2169 m
So, your buddy's fastball would go up 32.2169 m, and take 1.8122 s
to get there!