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2.1 INSTANTANEOUS RATE OF CHANGE 1
2.1 Instantaneous Rate of Change
In this section, we discuss the concept of the instantaneous rate of change of a given function. As an application, we use the velocity of a moving object. The motion of an object along a line at a particular instant is very difficult to define precisely. The modern approach consists of computing the average velocity over smaller and smaller time intervals. To be more precise, let s(t) be the position function or displacement of a moving object at time t. We would like to compute the velocity of the object at the instant t = t 0.
Average Velocity We start by finding the average velocity of the object over the time interval t 0 โค t โค t 0 + โt given by the expression
v =
Distance T raveled Elapsed T ime
s(t 0 + โt) โ s(t 0 ) โt
Geometrically, the average velocity over the time interval [t 0 , t 0 + โt] is just the slope of the line joining the points (t 0 , s(t 0 )) and (t 0 + โt, s(t 0 + โt)) on the graph of s(t).(See Figure 2.1.1)
Figure 2.1.
Example 2.1. A freely falling body experiencing no air resistance falls s(t) = 16t^2 feet in t seconds. Complete the following table
time interval [1.8,2] [1.9,2] [1.99,2] [1.999,2] [2,2.0001] [2,2.001] [2,2.01] Average velocity
Solution. From time t = 1.8 to time t = 2, formula (2.1.1) gives
s(2) โ s(1.8) 2 โ 1. 8
โ16(2)^2 โ [โ16(1.8)^2 ]
= โ 60 .8 ft/sec.
Using formula (2.1.1) on each of the remaining intervals, we find
time interval [1.8,2] [1.9,2] [1.99,2] [1.999,2] [2,2.0001] [2,2.001] [2,2.01] Average velocity 60.8 62.4 63.84 63.98 64.0016 64.016 64.
Instantaneous Velocity and Speed The next step is to calculate the average velocity on smaller and smaller time intervals ( that is, make โt close to zero). The average velocity in this case approaches what we would intuitively call the instantaneous velocity at time t = t 0 which is defined using the limit notation by
v(t 0 ) = lim โtโ 0
s(t 0 + โt) โ s(t 0 ) โt
Geometrically, the instantaneous velocity at t 0 is the slope of the tangent line to the graph of s(t) at the point (t 0 , s(t 0 )).(See Figure 2.1.2)
Figure 2.1.
Example 2.1. For the distance function in Example 2.1.1, find the instantaneous velocity at t = 2.
Example 2.1. (a) Find f โฒ(1) for f (x) = x^2. (b) Find the equation of the tangent line to the graph of f (x) at the point (1, f (1)).
Solution. Completing the following chart
x [0.9,1] [0.99,1] [0.999,1] [1,1.0001] [1,1.001] [1,1.01] [1,1.1] f (b)โf (a) bโa 1.9^ 1.99^ 1.999^ 2.0001^ 2.001^ 2.01^ 2.
we see that f โฒ(1) = 2. (b) The equation of the tangent line is
y โ f (1) = f โฒ(1)(x โ 1)
or y โ 1 = 2(x โ 1).
In point-intercept form, we have y = 2x โ 1
Example 2.1.4 (Numerical Estimation of the Derivative) Find approximate values for f โฒ(x) at each of the xโvalues given in the fol- lowing table
x 0 5 10 15 20 f (x) 100 70 55 46 40
Solution. The derivative can be estimated by using the average rate of change or the difference quotient
f โฒ(a) โ
f (a + h) โ f (a) h
If a is a left-endpoint then f โฒ(a) is estimated by
f โฒ(a) โ
f (b) โ f (a) b โ a
where b > a. If a is a right-endpoint then f โฒ(a) is estimated by
f โฒ(a) โ
f (a) โ f (b) a โ b
2.1 INSTANTANEOUS RATE OF CHANGE 5
where b < a. If a is an interior point then f โฒ(a) is estimated by
f โฒ(a) โ
f (a) โ f (b) a โ b
f (c) โ f (a) c โ a
where b < a < c. For example,
f โฒ(0) โ
f (5) โ f (0) 5
f โฒ(5) โ
f (10) โ f (5) 5
f (5) โ f (0) 5
f โฒ(10) โ
f (15) โ f (10) 5
f (10) โ f (5) 5
f โฒ(15) โ
f (20) โ f (15) 5
f (15) โ f (10) 5
f โฒ(20) โ
f (20) โ f (15) 5
The quantity f^ (10) 10 โโf 5 (5) is known as the right slope estimation of f โฒ(5). Similarly, we can estimate f โฒ(5) by using a left slope estimation,i.e.,
f โฒ(5) โ
f (5) โ f (0) 5 โ 0
An improved estimation consists of taking the average of the left slope and the right slope, that is,
f โฒ(5) โ
