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The Derivative as a Rate of Change Help

By — McGraw-Hill Professional
Updated on Sep 1, 2011

Introduction to The Derivative as a Rate of Change

If f(t) represents the position of a moving body, or the amount of a changing quantity, at time t , then the derivative f′(t) (equivalently, ( d/dt)f(t)) denotes the rate of change of position (also called velocity ) or the rate of change of the quantity. When f′(t) represents velocity, then sometimes we calculate another derivative —( f′)′(t) —and this quantity denotes the rate of change of velocity, or acceleration . In specialized applications, even more derivatives are sometimes used. For example, sometimes the derivative of the acceleration is called jerk and sometimes the derivative of jerk is called surge .

Example 1

The position of a body moving along a linear track is given by p(t) = 3 t 2 − 5 t + 7 feet. Calculate the velocity and the acceleration at time t = 3 seconds.

Solution 1

The velocity is given by

p′(t) = 6 t − 5.

At time t = 3 we therefore find that the velocity is p′ (3) = 18 − 5 = 13 ft/sec.

The acceleration is given by the second derivative:

p″(t) = ( p′ )′( t ) = (6 t − 5)′ = 6.

The acceleration at time t = 3 is therefore 6 ft/sec 2 .

Math Note: As previously noted, velocity is measured in feet per second (or ft/sec). Acceleration is the rate of change of velocity with respect to time; therefore acceleration is measured in “feet per second per second” (or ft/sec 2 ).

Example 2

A massive ball is dropped from a tower. It is known that a falling body descends (near the surface of the earth) with an acceleration of about 32 ft/sec. From this information one can determine that the equation for the position of the ball at time t is

p(t) = −16 t 2 + ν 0 t + h 0 ft.

Here ν 0 is the initial velocity and h 0 is the initial height of the ball in feet. 1 Also t is time measured in seconds. If the ball hits the earth after 5 seconds, then determine the height from which the ball is dropped.

Solution 2

Observe that the velocity is

υ(t) = p′(t) = −32 t + υ 0 .

Obviously the initial velocity of a falling body is 0. Thus

0 = υ (0) = −32·0 + υ 0 .

We shall say more about this equation, and this technique, in Section 3.4.

It follows that υ 0 = 0, thus confirming our intuition that the initial velocity is 0. Thus

p(t) = −16 t 2 + h 0 .

Now we also know that p (5) = 0; that is, at time t = 5 the ball is at height 0. Thus

0 = p (5) = −16·5 2 + h 0 .

We may solve this equation for h 0 to determine that h 0 = 400.

We conclude that

p(t) = −16 t 2 + 400.

Furthermore, p (0) = 400, so the initial height of the ball is 400 feet.

You Try It: Suppose that a massive ball falls from a height of 600 feet. After how many seconds will it strike the ground?

Find practice problems and solutions for these concepts at: Foundations of Calculus Practice Test.

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