Calculus - what is it good for ?

So now you know some of the basics, what can we do with the derivative ?

One big use is in physics or the real world when dealing with cars and rockets and baseballs and something we call speed or velocity.

It turns out that the change in distance over the change in time is SPEED or Velocity and that is the SLOPE we have been talking about called the derivative (neat huh ?).

Now when an object like a rocket or cannon ball or baseball hits its PEAK or maximum height, it has ZERO velocity.

(That is why those towers in the medieval times were so effective. By the time the spear reach the top of the tower it had no velocity...just pluck it out of the sky and use it as a weapon against the enemy below. The same was true with most arrows of that time. In today's world it would be a water balloon fight at the beach motel. If you are on the 3rd or 4th floor, by the time the balloon makes it up to our balcony, you can just pluck it our of the air and throw it back at your sons below you...lol)


Maximum Height

So now if we look at the typical parabola of a flight object the equation might be:
y = -5 x^2 + 10 X + 2

To find the velocity, all we need to do is take the derivative. ( if y = x^n, then y' = n * x^(n-1)....if y = x^2 then y' = 2x...if y = x^3, then y' = 3x^2 ).

So the derivative of the distant formula about would be
Vel. = -10 x + 10.

And if we want to find the maximum height of the ball or object all we need to do is set the velocity = 0.

so 0 = -10x + 10....now some simple algebra gives us that x = 1.

the ball reaches its maximum height at x = 1 and we plug that into the distance equation given and we get y = -5 (1)^2 + 10 (1) + 2 = 7. The object reaches the max height of 7 !!

We can check this with our graphing calculator by graphing y = -5x^2 +10x + 2 and then hitting the buttons 2nd function/Calc max...we also get x=1, y=7.


Minimum Cost

so we can also use the derivative to fine the minimum cost which is very useful in the business world.

Lets assume we have to make a box with a fixed volume of 8 cubic feet. The box has a square base and an open top. What dimensions will minimize or cost of material ?

Take my word for it the formula is SA (Surface Area) = x^2 + 32/x

the derivative SA' = 2x - 32/ x^2

so now we set the derivative equal to zero and we get x = 2.52 and SA = 19.

lets check: if the box was 2 X 2 X 2 (volume of 8) the SA would be SA = 4 + 16 = 20.

if we made it 1 X 1 X 8 (volume of 8) the SA = 1 + 32 = 33

if we made it 2.5 X 2.5 X 1.28 (vol. of 8) the SA = 6.25 + 12.8 = 19.05 (very close)

But the best is what the Calculus tells us using the derivative, x = 2.52