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) = x2
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)=x2 so
f'(x)=2x2-1=2x1=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 * t2

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 * t1-1 - 9.81 m per s2 * 2 * t2-1
h'(t) = 35.5555 m per s * t0 - 9.81 m per s2 * 2 * t1
h'(t) = 35.5555 m per s * 1 - 9.81 m per s2 * 2 * t
h'(t) = 35.5555 m per s - 9.81 m per s2 * 2 * t

and the minimum occurs at h'(t) = 0, so

0 = h'(t) = 35.5555 m per s - 9.81 m per s2 * 2 * t
0 + 9.81 m per s2 * 2 * t = 35.5555 m per s - 9.81 m per s2 * 2 * t + 9.81 m per s2 * 2 * t
(9.81 m per s2 * 2 * t) / 9.81 m per s2 = 35.5555 m per s / 9.81 m per s2
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 s2 * t2
h(1.8122 s) = 35.5555 m per s * 1.8122 s- 9.81 m per s2 * (1.8122 s)2
h(1.8122 s) = 64.4336 m - 9.81 m per s2 * 3.2840 s2
h(1.8122 s) = 64.4336 m - 9.81 m per s2 * 3.2840 s2
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!

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