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Finding instantaneous rates of change

Many applications of the derivative involve determining the rate of change at a given instant of a function with the independent variable time—which is why the term instantaneous is used. Consider the height of a ball tossed upward with an initial velocity of 64 feet per second, given by s ( t ) = −16 t 2 + 64 t + 6 , where t is measured in seconds and s ( t ) is measured in feet. We know the path is that of a parabola. The derivative will tell us how the height is changing at any given point in time. The height of the ball is shown in [link] as a function of time. In physics, we call this the “ s - t graph.”

Graph of a negative parabola with a vertex at (2, 70) and two points at (1, 55) and (3, 55).

Finding the instantaneous rate of change

Using the function above, s ( t ) = −16 t 2 + 64 t + 6 , what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?

The velocity at t = 1 and t = 3 is the instantaneous rate of change of distance per time, or velocity. Notice that the initial height is 6 feet. To find the instantaneous velocity, we find the derivative    and evaluate it at t = 1 and t = 3 :

f ( a ) = lim h 0 f ( a + h ) f ( a ) h          = lim h 0 16 ( t + h ) 2 + 64 ( t + h ) + 6 ( 16 t 2 + 64 t + 6 ) h Substitute  s ( t + h )  and  s ( t ) .          = lim h 0 16 t 2 32 h t h 2 + 64 t + 64 h + 6 + 16 t 2 64 t 6 h Distribute .          = lim h 0 32 h t h 2 + 64 h h Simplify .          = lim h 0 h ( 32 t h + 64 ) h Factor the numerator .          = lim h 0 32 t h + 64 Cancel out the common factor  h . s ( t ) = 32 t + 64 Evaluate the limit by letting  h = 0.

For any value of t , s ( t ) tells us the velocity at that value of t .

Evaluate t = 1 and t = 3.

s ( 1 ) = −32 ( 1 ) + 64 = 32 s ( 3 ) = −32 ( 3 ) + 64 = −32

The velocity of the ball after 1 second is 32 feet per second, as it is on the way up.

The velocity of the ball after 3 seconds is −32 feet per second, as it is on the way down.

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The position of the ball is given by s ( t ) = −16 t 2 + 64 t + 6. What is its velocity 2 seconds into flight?

0

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Using graphs to find instantaneous rates of change

We can estimate an instantaneous rate of change at x = a by observing the slope of the curve of the function f ( x ) at x = a . We do this by drawing a line tangent to the function at x = a and finding its slope.

Given a graph of a function f ( x ) , find the instantaneous rate of change of the function at x = a .

  1. Locate x = a on the graph of the function f ( x ) .
  2. Draw a tangent line, a line that goes through x = a at a and at no other point in that section of the curve. Extend the line far enough to calculate its slope as
    change in  y change in  x .

Estimating the derivative at a point on the graph of a function

From the graph of the function y = f ( x ) presented in [link] , estimate each of the following:

  1. f ( 0 )
  2. f ( 2 )
  3. f ' ( 0 )
  4. f ' ( 2 )

Graph of an odd function with multiplicity of two and with two points at (0, 1) and (2, 1).

To find the functional value, f ( a ) , find the y -coordinate at x = a .

To find the derivative    at x = a , f ( a ) , draw a tangent line at x = a , and estimate the slope of that tangent line. See [link] .

Graph of the previous function with tangent lines at the two points (0, 1) and (2, 1). The graph demonstrates the slopes of the tangent lines. The slope of the tangent line at x = 0 is 0, and the slope of the tangent line at x = 2 is 4.
  1. f ( 0 ) is the y -coordinate at x = 0. The point has coordinates ( 0 , 1 ) , thus f ( 0 ) = 1.
  2. f ( 2 ) is the y -coordinate at x = 2. The point has coordinates ( 2 , 1 ) , thus f ( 2 ) = 1.
  3. f ( 0 ) is found by estimating the slope of the tangent line to the curve at x = 0. The tangent line to the curve at x = 0 appears horizontal. Horizontal lines have a slope of 0, thus f ( 0 ) = 0.
  4. f ( 2 ) is found by estimating the slope of the tangent line to the curve at x = 2. Observe the path of the tangent line to the curve at x = 2. As the x value moves one unit to the right, the y value moves up four units to another point on the line. Thus, the slope is 4, so f ( 2 ) = 4.
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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