Contextual Applications of Differentiation Practice Questions
AP (Advanced Placement) · AP Calculus AB · 150 free MCQs with instant results and detailed explanations.
150
Total
41
Easy
80
Medium
29
Hard
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Sample Questions from Contextual Applications of Differentiation
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Q1
Easy
A car's position is given by the function s(t) = 4t^2 + 2t, where s is in meters and t is in seconds. What is the velocity of the car at t = 3 seconds?
A.
26 m/s
B.
30 m/s
C.
20 m/s
D.
18 m/s
Show Answer & Explanation
Correct Answer: A
To find the velocity, we need to differentiate the position function. The derivative, s'(t) = 8t + 2. Substituting t = 3 gives s'(3) = 8(3) + 2 = 26 m/s.
Q2
Easy
A rectangle's length is increasing at a rate of 3 cm/s, and its width is increasing at a rate of 2 cm/s. At what rate is the area of the rectangle increasing when the length is 10 cm and the width is 5 cm?
A.
35 cm²/s
B.
40 cm²/s
C.
25 cm²/s
D.
30 cm²/s
Show Answer & Explanation
Correct Answer: A
The area A of a rectangle is given by A = lw. To find the rate of change of the area, we use dA/dt = l(dw/dt) + w(dl/dt). Substituting l = 10, w = 5, dl/dt = 3, and dw/dt = 2 gives dA/dt = 10(2) + 5(3) = 20 + 15 = 35 cm²/s.
Q3
Easy
A company's profit P (in thousands of dollars) is given by the function P(x) = -2x^2 + 12x - 10, where x is the number of units sold. How many units should the company sell to maximize its profit?
A.
3
B.
6
C.
2
D.
1
Show Answer & Explanation
Correct Answer: A
To maximize profit, we find the vertex of the parabola represented by the quadratic equation. The x-coordinate of the vertex is found using the formula x = -b/(2a). Here, a = -2 and b = 12, so x = -12/(2*-2) = 3. Thus, selling 3 units maximizes profit.
Q4
Medium
A toy rocket is launched vertically from the ground. Its height in meters after t seconds is given by the function h(t) = -5t^2 + 20t. What is the maximum height reached by the rocket?
A.
50 meters
B.
40 meters
C.
20 meters
D.
30 meters
Show Answer & Explanation
Correct Answer: B
The maximum height can be found by determining the vertex of the parabola represented by h(t). The vertex occurs at t = -b/(2a) = 20/(2*5) = 2 seconds. Substituting t = 2 into h(t) gives h(2) = -5(2)^2 + 20(2) = 40 meters.
Q5
Medium
A car's position s(t) in meters at time t in seconds is modeled by the equation s(t) = 4t^2 + 2t. What is the car's velocity at t = 5 seconds?
A.
44 m/s
B.
54 m/s
C.
48 m/s
D.
38 m/s
Show Answer & Explanation
Correct Answer: A
Velocity is found by taking the derivative of the position function s(t). Thus, v(t) = s'(t) = 8t + 2. Evaluating this at t = 5 gives v(5) = 8(5) + 2 = 40 + 2 = 44 m/s.
Q6
Medium
The demand function for a product is given by D(p) = 100 - 2p, where p represents the price. If the price increases by $1, what is the approximate change in demand?
A.
-2
B.
-1
C.
2
D.
1
Show Answer & Explanation
Correct Answer: A
The rate of change of demand with respect to price is given by D'(p) = -2. Therefore, if the price increases by $1, the approximate change in demand would be -2, indicating a decrease in demand.
Q7
Medium
A rectangular garden's area A is given by A(x) = x(10 - x), where x is the length of one side. To maximize the area, what is the optimal length for side x?
A.
5
B.
10
C.
2.5
D.
0
Show Answer & Explanation
Correct Answer: A
To maximize the area, we differentiate A(x) to find A'(x) = 10 - 2x. Setting A'(x) = 0 gives x = 5. This is within the feasible range for x, hence it maximizes the area.
Q8
Hard
A particle moves along a straight line such that its position at time t seconds is given by the function s(t) = 4t^3 - 12t^2 + 9t. What is the acceleration of the particle at t = 2 seconds?
A.
24 m/s²
B.
36 m/s²
C.
12 m/s²
D.
0 m/s²
Show Answer & Explanation
Correct Answer: A
To find acceleration, we need the second derivative of the position function. The first derivative, v(t) = s'(t) = 12t^2 - 24t + 9. The second derivative, a(t) = v'(t) = 24t - 24. Plugging in t = 2 gives a(2) = 24 * 2 - 24 = 24 m/s².
Q9
Hard
A company's revenue R in thousands of dollars from selling x units of a product is given by the function R(x) = -2x^2 + 80x. What is the maximum revenue the company can achieve?
A.
1600
B.
160
C.
3200
D.
320
Show Answer & Explanation
Correct Answer: A
To find the maximum revenue, we need to determine the vertex of the quadratic function R(x). The vertex x-coordinate is given by -b/(2a) = -80/(2 * -2) = 20. Plugging x = 20 back into R(x) gives R(20) = -2(20)^2 + 80(20) = 1600. Therefore, the maximum revenue is 1600.
Q10
Hard
A company wants to maximize its profit, given by the function P(x) = -2x^2 + 40x - 150, where P is the profit in dollars and x is the number of items produced. What is the maximum profit the company can achieve?
A.
$150
B.
$250
C.
$650
D.
$450
Show Answer & Explanation
Correct Answer: C
To find the maximum profit, we need to determine the vertex of the quadratic function P(x) = -2x^2 + 40x - 150. The vertex occurs at x = -b/(2a) = -40/(2*(-2)) = 10. Substituting x = 10 back into P gives P(10) = -2(10)^2 + 40(10) - 150 = 650. Thus, the maximum profit is $650.
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Contextual Applications of Differentiation — AP (Advanced Placement) AP Calculus AB Practice Questions Online
This page contains 150 practice MCQs for the chapter Contextual Applications of Differentiation in AP (Advanced Placement) AP Calculus AB. The questions are organized by difficulty — 41 easy, 80 medium, 29 hard — so you can choose the right level for your preparation.
Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit.