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v ave = s ( t ) s ( a ) t a .

As the values of t approach a , the values of v ave approach the value we call the instantaneous velocity    at a . That is, instantaneous velocity at a , denoted v ( a ) , is given by

v ( a ) = s ( a ) = lim t a s ( t ) s ( a ) t a .

To better understand the relationship between average velocity and instantaneous velocity, see [link] . In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time t = a whose position at time t is given by the function s ( t ) . The slope of the secant line (shown in green) is the average velocity of the object over the time interval [ a , t ] .

This figure consists of the Cartesian coordinate plane with 0, a, and t1 marked on the t-axis. The function y = s(t) is graphed in the first quadrant along with two lines marked tangent and secant. The tangent line touches y = s(t) at only one point, (a, s(a)). The secant line touches y = s(t) at two points: (a, s(a)) and (t1, s(t1)).
The slope of the secant line is the average velocity over the interval [ a , t ] . The slope of the tangent line is the instantaneous velocity.

We can use [link] to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using [link] .

Estimating velocity

A lead weight on a spring is oscillating up and down. Its position at time t with respect to a fixed horizontal line is given by s ( t ) = sin t ( [link] ). Use a table of values to estimate v ( 0 ) . Check the estimate by using [link] .

A picture of a spring hanging down with a weight at the end. There is a horizontal dashed line marked 0 a little bit above the weight.
A lead weight suspended from a spring in vertical oscillatory motion.

We can estimate the instantaneous velocity at t = 0 by computing a table of average velocities using values of t approaching 0 , as shown in [link] .

Average velocities using values of t Approaching 0
t sin t sin 0 t 0 = sin t t
−0.1 0.998334166
−0.01 0.9999833333
−0.001 0.999999833
0.001 0.999999833
0.01 0.9999833333
0.1 0.998334166

From the table we see that the average velocity over the time interval [ −0.1 , 0 ] is 0.998334166 , the average velocity over the time interval [ −0.01 , 0 ] is 0.9999833333 , and so forth. Using this table of values, it appears that a good estimate is v ( 0 ) = 1 .

By using [link] , we can see that

v ( 0 ) = s ( 0 ) = lim t 0 sin t sin 0 t 0 = lim t 0 sin t t = 1 .

Thus, in fact, v ( 0 ) = 1 .

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A rock is dropped from a height of 64 feet. Its height above ground at time t seconds later is given by s ( t ) = −16 t 2 + 64 , 0 t 2 . Find its instantaneous velocity 1 second after it is dropped, using [link] .

−32 ft/s

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As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.


The instantaneous rate of change    of a function f ( x ) at a value a is its derivative f ( a ) .

Chapter opener: estimating rate of change of velocity

The same sports car speeding along a winding road from the beginning of the chapter.
(credit: modification of work by Codex41, Flickr)

Reaching a top speed of 270.49 mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from 0 to 60 mph in 3.05 seconds, from 0 to 100 mph in 5.88 seconds, from 0 to 200 mph in 14.51 seconds, and from 0 to 229.9 mph in 19.96 seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration ) as it approaches 229.9 mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?

First observe that 60 mph = 88 ft/s, 100 mph 146.67 ft/s, 200 mph 293.33 ft/s, and 229.9 mph 337.19 ft/s. We can summarize the information in a table.

v ( t ) At different values of t
t v ( t )
0 0
3.05 88
5.88 147.67
14.51 293.33
19.96 337.19

Now compute the average acceleration of the car in feet per second on intervals of the form [ t , 19.96 ] as t approaches 19.96 , as shown in the following table.

Average acceleration
t v ( t ) v ( 19.96 ) t 19.96 = v ( t ) 337.19 t 19.96
0.0 16.89
3.05 14.74
5.88 13.46
14.51 8.05

The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph ( 337.19 ft/s).

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Questions & Answers

What is derivative of antilog x dx ?
Tanmay Reply
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an area under a curve is continuous because you are looking at an area that covers a range of numbers, it is over an interval, such as 0 to 4
using product rule x^3,x^5
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Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3. ∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx ∫x2lnxdx=13x3lnx−∫13x2dx ∫x2lnxdx=13x3lnx−19x3+C
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you just have to recognize the problem. there can be multiple ways to solve 1 problem. that's the hardest part about integration
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Practice Key Terms 4

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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