We have already talked about the average rate of change back in supplemental notes1, but let us review the definition of this topic.

DEFINITION:

The average rate of change of a function f (x) with respect to x over the interval from x ₀ to x ₀ + h is

Now here is the definition of the instantaneous rate of change.

DEFINITION:

The instantaneous rate of change of f with respect to x at x ₀ is the derivative

provided the limit exists.

Sometimes this derivative is called the rate of change. Let us work an example to illustrate a use of this concept.

EXAMPLE 1:

The volume V = (4/3)p r ³ of a spherical balloon changes with the radius. At what rate does the volume change with respect to the radius when r = 2 ft.?

SOLUTION: Here we are talking about the instantaneous rate of change, so let us find the derivative dV/dr.

When r = 2 ft., the volume is changing at a rate of 16p ft ³/ft. This means that a small change in D r ft in the radius would result in a change of 16p ft ³ in the volume of the sphere.

Suppose that an object is moving along a coordinate line so that we know its position s on that line as a function of t, s = f (t).

The displacement of the object over the time interval from t to t + D t is

D s = f (t + D t) - f (t).

The average velocity of the object over that time interval is

DEFINITION:

The (instantaneous) velocity is the derivative of the position function s = f (t) with respect to time. At time t, the velocity is

DEFINITION:

Speed is the absolute value of velocity.

DEFINITION:

Acceleration is the derivative of velocity with respect to time. If a body's position at time t is s = f (t), then the body's acceleration at time t is

What the velocity tells us is how fast and what direction the object is moving. If it is moving forward (s increasing), the velocity is positive. If it is moving backward (s decreasing), the velocity is negative. Here is a practical example of the difference between speed and velocity. Suppose you drive 65 mph to NorthPark mall from Clinton and back. Your speedometer will show 65 mph going, but not - 65 mph when you are coming back from the mall. Speed is the absolute value of the velocity. 65 mph is the speed, but the plus is when you are going and the negative means you are returning.

Another way of illustrating these concepts is with free-fall of an object. When air resistance is absent or insignificant, and the only force acting on a falling body is the force of gravity, the generic formula for free-fall is s = 0.5g t ². There are two forms of this equation. One for the English measurement of gravity, and the other for the metric measurement.

ENGLISH

g = 32 ft/sec ²

s = 16t ²

METRIC

g = 9.8 m/sec ²

s = 4.9t ²

EXAMPLE 2:

Jar Jar Binks falls off a cliff at time, t = 0. (A) How many feet does Jar Jar Binks fall in 5 seconds? (B) What are his velocity, speed, and acceleration then?

SOLUTION: Since we want the answer in feet, we will use the equation s = 16t ².

1. How many feet does Jar Jar Binks fall in 5 seconds?

s (5) = 16(5) ² = 400 feet

2. What are his velocity, speed, and acceleration then?

EXAMPLE 3:

Given s = (t ⁴/4) - t ³ + t ² of a body moving on a coordinate line for 0 £ t £ 2, with s in meters and t in seconds. (A) Find the body's displacement and average velocity for the given time interval. (B) Find the body's speed and acceleration at the endpoints of the interval. (C) When during the interval does the body change direction (if ever)?

SOLUTION: (A) Remember that the displacement is D s = f (t + D t) - f (t) or D s = f (b) - f (a) and the average velocity is D s/D t.

(B) To find the body's speed and acceleration at the endpoints of the interval, we must find the first and second derivatives of s.

(C) When during the interval does the body change direction (if ever)?

If the particle is moving forward, then v > 0, and if the particle is moving backward, then v < 0. So if the particle changes direction, then v has to go from positive to negative or vice versa. Therefore, the velocity would have to equal zero at some point.

So there is a direction change at 1 second.

EXAMPLE 4:

A 45-caliber bullet fired straight up from the surface of the moon would reach a height of s = 832t - 2.6t ² feet after t seconds. On earth, in the absence of air, its height would be s = 832t - 16t ² feet after t seconds. How long will the bullet be aloft in each case? How high would the bullet go?

SOLUTION: To find out how long the bullet is aloft, set s = 0 and solve for t. The reason why we would set the position, s, equal to zero, is when the bullet is fired, the position is 0 and when it hits the ground, again the position will be zero.

So on the earth; the bullet will be 52 seconds aloft.

So on the moon; the bullet will be 320 seconds aloft.

To find how high the bullet will go, we will find the velocity. Then we will set it equal to zero and determine the time when the velocity changes direction. This will be the time when the bullet is at the highest point on its flight. So to find the maximum height, I will plug t into s and evaluate.