College

Finding Rates of Change

In Exercises 91-94, find the average rate of change of the function over the given interval. Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval.

91. [tex]f(t) = 3t + 5, \quad [1, 2][/tex]

92. [tex]f(t) = t^2 - 7, \quad [3, 3.1][/tex]

93. [tex]f(x) = \frac{-1}{x}, \quad [1, 2][/tex]

94. [tex]f(x) = \sin x, \quad \left[0, \frac{\pi}{6}\right][/tex]

Answer :

Sure! Let's go through each of the exercises one by one to find and compare the average rate of change with the instantaneous rates of change at the endpoints of the given intervals.

### Exercise 91: [tex]\( f(t) = 3t + 5 \)[/tex], Interval: [tex]\([1, 2]\)[/tex]
1. Average Rate of Change:
- Calculate [tex]\( f(2) \)[/tex] and [tex]\( f(1) \)[/tex]:
[tex]\[
f(2) = 3 \cdot 2 + 5 = 11
\][/tex]
[tex]\[
f(1) = 3 \cdot 1 + 5 = 8
\][/tex]
- The average rate of change is:
[tex]\[
\frac{f(2) - f(1)}{2 - 1} = \frac{11 - 8}{1} = 3
\][/tex]

2. Instantaneous Rates of Change:
- The derivative of [tex]\( f(t) = 3t + 5 \)[/tex] is [tex]\( f'(t) = 3 \)[/tex].
- At [tex]\( t = 1 \)[/tex] and [tex]\( t = 2 \)[/tex], the rate is 3.

Comparison: Both the average rate of change and the instantaneous rates at the endpoints are 3.

### Exercise 92: [tex]\( f(t) = t^2 - 7 \)[/tex], Interval: [tex]\([3, 3.1]\)[/tex]
1. Average Rate of Change:
- Calculate [tex]\( f(3.1) \)[/tex] and [tex]\( f(3) \)[/tex]:
[tex]\[
f(3.1) = (3.1)^2 - 7 = 2.61
\][/tex]
[tex]\[
f(3) = 3^2 - 7 = 2
\][/tex]
- The average rate of change is:
[tex]\[
\frac{f(3.1) - f(3)}{3.1 - 3} = \frac{2.61 - 2}{0.1} = 6.1
\][/tex]

2. Instantaneous Rates of Change:
- The derivative of [tex]\( f(t) = t^2 - 7 \)[/tex] is [tex]\( f'(t) = 2t \)[/tex].
- At [tex]\( t = 3 \)[/tex], the rate is [tex]\( 2 \cdot 3 = 6 \)[/tex].
- At [tex]\( t = 3.1 \)[/tex], the rate is [tex]\( 2 \cdot 3.1 = 6.2 \)[/tex].

Comparison: The average rate of change is 6.1, while the instantaneous rates at the endpoints are 6 and 6.2.

### Exercise 93: [tex]\( f(x) = \frac{-1}{x} \)[/tex], Interval: [tex]\([1, 2]\)[/tex]
1. Average Rate of Change:
- Calculate [tex]\( f(2) \)[/tex] and [tex]\( f(1) \)[/tex]:
[tex]\[
f(2) = \frac{-1}{2} = -0.5
\][/tex]
[tex]\[
f(1) = \frac{-1}{1} = -1
\][/tex]
- The average rate of change is:
[tex]\[
\frac{f(2) - f(1)}{2 - 1} = \frac{-0.5 + 1}{1} = 0.5
\][/tex]

2. Instantaneous Rates of Change:
- The derivative of [tex]\( f(x) = \frac{-1}{x} \)[/tex] is [tex]\( f'(x) = \frac{1}{x^2} \)[/tex].
- At [tex]\( x = 1 \)[/tex], the rate is [tex]\( 1 \)[/tex].
- At [tex]\( x = 2 \)[/tex], the rate is [tex]\( \frac{1}{4} \)[/tex].

Comparison: The average rate of change is 0.5, while the instantaneous rates at the endpoints are 1 and 0.25.

### Exercise 94: [tex]\( f(x) = \sin x \)[/tex], Interval: [tex]\([0, \frac{\pi}{6}]\)[/tex]
1. Average Rate of Change:
- Calculate [tex]\( f(\frac{\pi}{6}) \)[/tex] and [tex]\( f(0) \)[/tex]:
[tex]\[
f\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = 0.5
\][/tex]
[tex]\[
f(0) = \sin(0) = 0
\][/tex]
- The average rate of change is:
[tex]\[
\frac{f\left(\frac{\pi}{6}\right) - f(0)}{\frac{\pi}{6} - 0} = \frac{0.5 - 0}{\frac{\pi}{6}} = \frac{3}{\pi} \approx 0.9549
\][/tex]

2. Instantaneous Rates of Change:
- The derivative of [tex]\( f(x) = \sin x \)[/tex] is [tex]\( f'(x) = \cos x \)[/tex].
- At [tex]\( x = 0 \)[/tex], the rate is [tex]\( \cos(0) = 1 \)[/tex].
- At [tex]\( x = \frac{\pi}{6} \)[/tex], the rate is [tex]\( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \approx 0.866 \)[/tex].

Comparison: The average rate of change is approximately 0.9549, with instantaneous rates at the endpoints being 1 and approximately 0.866.