# Instantaneous Rate Of Change

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I-heng_mccomb
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Can you interpret and/or estimate instantaneous rate of change in multiple representations?

• 1.

### The graph shows y = f(x) in purple. Which line has a slope equal to the rate expression shown?

• A.

Line a

• B.

Line b

• C.

Line c

• D.

Line d

B. Line b
Explanation
Line b has a slope equal to the rate expression shown in the graph. The slope of a line represents the rate of change between two points on the line. In this case, the slope of Line b matches the rate expression shown in the graph, indicating that the function f(x) has a constant rate of change along Line b. Therefore, Line b is the correct answer.

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• 2.

### Estimate the instantaneous rate using the graph of y = f(x).

• A.

0.02

• B.

–0.02

• C.

7.25

• D.

–7.25

D. –7.25
Explanation
The correct answer is -7.25. This is because the graph of y = f(x) has a steep negative slope at the given point, indicating a high rate of change in the negative direction. Therefore, the instantaneous rate at this point is -7.25.

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• 3.

### Acceleration is the rate of change of velocity. If the velocity of a car driving on a straight road is given by v(t), which of the following expressions represent the car's instantaneous expression at t = 5 seconds? (Check all that apply.)

• A.

Expression 1:

• B.

Expression 2:

• C.

Expression 3:

• D.

Expression 4:

• E.

Expression 5:

• F.

Expression 6:

C. Expression 3:
D. Expression 4:
F. Expression 6:
Explanation
The expressions that represent the car's instantaneous acceleration at t = 5 seconds are Expression 3, Expression 4, and Expression 6.

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• 4.

### Estimate the instantaneous rate of change of H(x) = sin(ln x2) at x = –3

• A.

0.1

• B.

0.4

• C.

0.8

• D.

3.5

• E.

Undefined

B. 0.4
Explanation
The instantaneous rate of change of a function at a specific point can be found by taking the derivative of the function and evaluating it at that point. In this case, the function H(x) = sin(ln x^2) needs to be differentiated. The derivative of sin(x) is cos(x), and the derivative of ln(x^2) is 2/x. Therefore, the derivative of H(x) is cos(ln x^2) * 2/x. Evaluating this derivative at x = -3 gives cos(ln 9) * 2/-3. Since cos(ln 9) is a constant value, the instantaneous rate of change at x = -3 is 2/(-3) = -2/3. None of the given answer choices match this value, so the correct answer is undefined.

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• 5.

### Which statement(s) correctly contrast average rate of change and instantaneous rate of change?

• A.

Average rate of change requires specifying two points on a graph while instantaneous rate of change requires only one.

• B.

Average rate of change can be interpreted as a slope while instantaneous rate of change can not.

• C.

Average rate of change can be for any function while instantaneous rate of change is only for nonlinear functions.

• D.

Average rate of change can be calculated as a difference quotient while instantaneous rate of change is a limit of a difference quotient.

A. Average rate of change requires specifying two points on a graph while instantaneous rate of change requires only one.
D. Average rate of change can be calculated as a difference quotient while instantaneous rate of change is a limit of a difference quotient.
Explanation
The correct answer states that average rate of change requires specifying two points on a graph while instantaneous rate of change requires only one. This is because average rate of change is calculated by finding the slope between two points on a graph, while instantaneous rate of change is the rate of change at a specific point on a graph. Additionally, the correct answer states that average rate of change can be calculated as a difference quotient, which is the change in the dependent variable divided by the change in the independent variable, while instantaneous rate of change is the limit of a difference quotient as the change in the independent variable approaches zero.

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