Page 107 - Calculus Demystified

CHAPTER 3 Applications of the Derivative
94
Substituting the values for r, dr/dt, h, and dh/dt into the right-hand side yields
dV 1 2
1 24π
= π 2 · 4 · (0.2) · 6 + 4 · (0.3) = π[9.6 + 4.8]= .
dt 3 3 5
You Try It: Intheheatofthesun,asheetofaluminumintheshapeofanequilateral
triangle is expanding so that its side length increases by 1 millimeter per hour. When
the side length is 100 millimeters, how is the area increasing?

3.4 Falling Bodies

It is known that, near the surface of the earth, a body falls with acceleration (due
2
to gravity) of about 32 ft/sec .Ifwelet h(t) be the height of the object at time t
(measured in seconds), then our information is that

h (t) =−32.
Observe the minus sign to indicate that height is decreasing.
Now we will do some organized guessing. What could h be? It is some function

whose derivative is the constant −32. Our experience indicates that polynomials
decrease in degree when we differentiate them. That is, the degree goes from 5 to 4,

or from 3 to 2. Since, h is a polynomial of degree 0, we therefore determine

that h will be a polynomial of degree 1. A moment’s thought then suggests that

h (t) =−32t. This works! If h (t) =−32t then h (t) =[h (t)] =−32. In fact

we can do a bit better. Since constants differentiate to zero, our best guess of what

the velocity should be is h (t) =−32t +v 0 , where v 0 is an undetermined constant.
Now let us guess what form h(t) should have. We can learn from our experience
in the last paragraph. The “antiderivative” of −32t (a polynomial of degree 1)
2
should be a polynomial of degree 2.After a little fiddling, we guess −16t .And this
works. The antiderivative of v 0 (a polynomial of degree 0) should be a polynomial
of degree 1. After a little fiddling, we guess v 0 t. And this works. Taking all this

information together, we find that the “antiderivative” of h (t) =−32t + v 0 is
2
h(t) =−16t + v 0 t + h 0 . (†)
Notice that we have once again thrown in an additive constant h 0 . This does no
harm:

2
h (t) =[−16t ] +[v 0 t] +[h 0 ] =−32t + v 0 ,
just as we wish. And, to repeat what we have already confirmed,

h (t) =[h (t)] =[−32t] +[v 0 ] =−32.