**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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