IB Mathematics: Applications and Interpretation Quiz: Derivative As Rate Of Change
20 questions · exam conditions
0:00
Derivative As Rate Of ChangeQuestion 1 of 20

A research study models the proportion, P, of a population that has heard a rumour after t days as P(t) = 1 - e⁻⁰·¹ᵗ.

What is the instantaneous rate at which the rumour is spreading at t=5 days?

0.939 per day
0.393 per day
0.607 per day
0.061 per day
← Back to quizzes

IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Derivative As Rate Of Change

Practice Derivative As Rate Of Change in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Derivative As Rate Of Change, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A research study models the proportion, P, of a population that has heard a rumour after t days as P(t) = 1 - e⁻⁰·¹ᵗ.

What is the instantaneous rate at which the rumour is spreading at t=5 days?

  1. 0.939 per day
  2. 0.393 per day
  3. 0.607 per day
  4. 0.061 per day (correct answer)
Explanation: When you see a question asking for an "instantaneous rate" of change, you need to find the derivative of the given function. The instantaneous rate at which the rumor is spreading means you're looking for how fast the proportion P is changing at the specific moment when t = 5 days. Given P(t)=1−e−0.1tP(t) = 1 - e^{-0.1t}, you need to find P′(t)P'(t) first. Using the chain rule, the derivative of −e−0.1t-e^{-0.1t} is −(−0.1)e−0.1t=0.1e−0.1t-(-0.1)e^{-0.1t} = 0.1e^{-0.1t}. So P′(t)=0.1e−0.1tP'(t) = 0.1e^{-0.1t}. Now substitute t = 5: P′(5)=0.1e−0.1(5)=0.1e−0.5=0.1×0.607=0.061P'(5) = 0.1e^{-0.1(5)} = 0.1e^{-0.5} = 0.1 \times 0.607 = 0.061 per day. Option A (0.939) represents the actual proportion of people who have heard the rumor at t = 5, which is P(5), not the rate of change. Option B (0.393) is the value of e−0.5e^{-0.5} without the 0.1 coefficient, showing incomplete derivative calculation. Option C (0.607) is approximately e−0.5e^{-0.5}, which is the exponential term alone without applying the coefficient from the chain rule. The correct answer is D (0.061 per day). Remember: instantaneous rate questions always require derivatives. When working with exponential functions, don't forget to apply the chain rule and include all coefficients from the exponent. Also, distinguish between the function value (proportion at time t) and its derivative (rate of change at time t).

Question 2

Let V(r) be the volume of a sphere with radius r, given by the formula V(r) = (4/3)πr³. Let A(r) be the surface area of a sphere with radius r, given by A(r) = 4πr².

Which of the following statements provides the correct geometric interpretation of the derivative V'(r)?

  1. V'(r) is the rate of change of the surface area with respect to the volume.
  2. V'(r) is the rate of change of the volume with respect to the radius, which is equal to the surface area. (correct answer)
  3. V'(r) is the rate of change of the volume with respect to the radius, which is equal to the circumference of the sphere's great circle.
  4. V'(r) is the ratio of the sphere's volume to its radius.
Explanation: The derivative V'(r) represents the instantaneous rate of change of the volume V with respect to the radius r. To find V'(r), we differentiate the volume formula: V'(r) = d/dr [(4/3)πr³] = (4/3)π * 3r² = 4πr². This resulting expression, 4πr², is the formula for the surface area of the sphere, A(r). Therefore, the rate of change of the volume with respect to the radius is equal to the surface area of the sphere.

Question 3

The displacement, s, in metres of a particle from a fixed point O at time t seconds is given by s(t) = 2t³ - 15t² + 24t + 10.

At which of the following times is the particle momentarily at rest?

  1. t = 1 and t = 4 seconds (correct answer)
  2. t = 2.5 seconds
  3. t = 0 and t = 3.5 seconds
  4. t = 0.84 and t = 6.16 seconds
Explanation: The particle is 'at rest' when its velocity is zero. Velocity is the derivative of displacement, v(t) = s'(t). First, find the derivative: s'(t) = 6t² - 30t + 24. Set the velocity to zero to find when the particle is at rest: 6t² - 30t + 24 = 0. Divide by 6: t² - 5t + 4 = 0. Factor the quadratic: (t - 1)(t - 4) = 0. The solutions are t = 1 and t = 4.

Question 4

A research study models the proportion, P, of a population that has heard a rumour after t days as P(t) = 1 - e⁻⁰·¹ᵗ.

What is the instantaneous rate at which the rumour is spreading at t=5 days?

  1. 0.939 per day
  2. 0.393 per day
  3. 0.607 per day
  4. 0.061 per day (correct answer)
Explanation: When you see a question asking for an "instantaneous rate" of change, you need to find the derivative of the given function. The instantaneous rate at which the rumor is spreading means you're looking for how fast the proportion P is changing at the specific moment when t = 5 days. Given P(t)=1−e−0.1tP(t) = 1 - e^{-0.1t}, you need to find P′(t)P'(t) first. Using the chain rule, the derivative of −e−0.1t-e^{-0.1t} is −(−0.1)e−0.1t=0.1e−0.1t-(-0.1)e^{-0.1t} = 0.1e^{-0.1t}. So P′(t)=0.1e−0.1tP'(t) = 0.1e^{-0.1t}. Now substitute t = 5: P′(5)=0.1e−0.1(5)=0.1e−0.5=0.1×0.607=0.061P'(5) = 0.1e^{-0.1(5)} = 0.1e^{-0.5} = 0.1 \times 0.607 = 0.061 per day. Option A (0.939) represents the actual proportion of people who have heard the rumor at t = 5, which is P(5), not the rate of change. Option B (0.393) is the value of e−0.5e^{-0.5} without the 0.1 coefficient, showing incomplete derivative calculation. Option C (0.607) is approximately e−0.5e^{-0.5}, which is the exponential term alone without applying the coefficient from the chain rule. The correct answer is D (0.061 per day). Remember: instantaneous rate questions always require derivatives. When working with exponential functions, don't forget to apply the chain rule and include all coefficients from the exponent. Also, distinguish between the function value (proportion at time t) and its derivative (rate of change at time t).

Question 5

The cost, C, in euros, of producing x kilograms of a specialty cheese is modelled by the function C(x) = 0.2x² + 5x + 300.

Find the marginal cost when 50 kilograms of cheese are produced.

  1. €25 (correct answer)
  2. €28.50
  3. €1050
  4. €1300
Explanation: Marginal cost is the instantaneous rate of change of the cost function, which is its derivative, C'(x). First, find the derivative: C'(x) = 2(0.2)x + 5 = 0.4x + 5. Then, evaluate at x = 50: C'(50) = 0.4(50) + 5 = 20 + 5 = 25. The marginal cost is €25 per kilogram. This represents the approximate cost of producing the 51st kilogram.

Question 6

The number of active users of a new social media app, U(t), in millions, is modelled by U(t) = 0.5t² + 2t, where t is the number of months since its launch.

Find the average rate of change in the number of users from t = 2 to t = 6, and compare it to the instantaneous rate of change at t = 4.

  1. The average rate is 6 million users/month, which is equal to the instantaneous rate at t=4. (correct answer)
  2. The average rate is 6 million users/month, which is less than the instantaneous rate at t=4.
  3. The average rate is 5 million users/month, which is less than the instantaneous rate at t=4.
  4. The average rate is 5 million users/month, which is greater than the instantaneous rate at t=4.
Explanation: First, calculate the average rate of change: [U(6) - U(2)] / (6 - 2). U(6) = 0.5(6)² + 2(6) = 18 + 12 = 30. U(2) = 0.5(2)² + 2(2) = 2 + 4 = 6. The average rate is (30 - 6) / 4 = 24 / 4 = 6 million users per month. Next, calculate the instantaneous rate of change by finding the derivative, U'(t) = t + 2. Evaluate at t = 4: U'(4) = 4 + 2 = 6 million users per month. The average rate of change over the interval [2, 6] is equal to the instantaneous rate of change at the midpoint t=4. This is a special property of quadratic functions.

Question 7

The temperature, T, in degrees Celsius, of a chemical reaction t minutes after it begins is given by the function T(t) = 150 - 80e⁻⁰·²ᵗ. A value is calculated using a GDC: T'(5) ≈ 5.89.

Which of the following is the best interpretation of the value T'(5) ≈ 5.89?

  1. After 5 minutes, the temperature is 5.89°C.
  2. In the first 5 minutes, the temperature increased by an average of 5.89°C per minute.
  3. Exactly 5 minutes after the reaction begins, the temperature is increasing at a rate of approximately 5.89°C per minute. (correct answer)
  4. The temperature will increase by 5.89°C during the fifth minute of the reaction.
Explanation: The derivative T'(t) represents the instantaneous rate of change of temperature with respect to time. Therefore, T'(5) is the rate of change at the specific moment t = 5 minutes. A positive value indicates the temperature is increasing. The correct interpretation is that at t = 5, the temperature is increasing at a rate of 5.89°C per minute. The other options confuse the instantaneous rate with the function's value, the average rate, or the total change over an interval.

Question 8

The concentration of a drug in a patient's bloodstream, C, in mg/L, is modelled by the function C(t) = 20te⁻⁰·⁵ᵗ, where t is the number of hours after the drug was administered.

At what rate is the concentration of the drug changing 2 hours after it was administered?

  1. The concentration is decreasing at 7.36 mg/L per hour.
  2. The concentration is increasing at 7.36 mg/L per hour.
  3. The concentration is 0 mg/L per hour. (correct answer)
  4. The concentration is decreasing at 14.72 mg/L per hour.
Explanation: The rate of change is the derivative, C'(t). Using the product rule, C'(t) = (20)(e⁻⁰·⁵ᵗ) + (20t)(-0.5e⁻⁰·⁵ᵗ) = 20e⁻⁰·⁵ᵗ(1 - 0.5t). To find the rate at t = 2, substitute into the derivative: C'(2) = 20e⁻⁰·⁵²(1 - 0.52) = 20e⁻¹(1 - 1) = 20e⁻¹(0) = 0. A rate of 0 mg/L per hour means the concentration has reached a maximum or minimum at this point; in this context, it is the maximum concentration.

Question 9

The displacement, s, in metres of a particle from a fixed point O at time t seconds is given by s(t) = 2t³ - 15t² + 24t + 10.

At which of the following times is the particle momentarily at rest?

  1. t = 1 and t = 4 seconds (correct answer)
  2. t = 2.5 seconds
  3. t = 0 and t = 3.5 seconds
  4. t = 0.84 and t = 6.16 seconds
Explanation: The particle is 'at rest' when its velocity is zero. Velocity is the derivative of displacement, v(t) = s'(t). First, find the derivative: s'(t) = 6t² - 30t + 24. Set the velocity to zero to find when the particle is at rest: 6t² - 30t + 24 = 0. Divide by 6: t² - 5t + 4 = 0. Factor the quadratic: (t - 1)(t - 4) = 0. The solutions are t = 1 and t = 4.

Question 10

The curve of a function is given by y = x⁴ - 3x.

Find the equation of the tangent line to the curve at the point where x = 1.

  1. y = x - 2
  2. y = x - 3 (correct answer)
  3. y = 4x - 6
  4. y = -2x
Explanation: First, find the y-coordinate of the point of tangency by substituting x = 1 into the function: y = (1)⁴ - 3(1) = 1 - 3 = -2. The point is (1, -2). Next, find the gradient of the tangent by finding the derivative, dy/dx. The derivative is dy/dx = 4x³ - 3. Evaluate the gradient at x = 1: m = 4(1)³ - 3 = 1. Using the point-slope form y - y₁ = m(x - x₁), we get y - (-2) = 1(x - 1). This simplifies to y + 2 = x - 1, or y = x - 3.

Question 11

Let V(r) be the volume of a sphere with radius r, given by the formula V(r) = (4/3)πr³. Let A(r) be the surface area of a sphere with radius r, given by A(r) = 4πr².

Which of the following statements provides the correct geometric interpretation of the derivative V'(r)?

  1. V'(r) is the rate of change of the surface area with respect to the volume.
  2. V'(r) is the rate of change of the volume with respect to the radius, which is equal to the surface area. (correct answer)
  3. V'(r) is the rate of change of the volume with respect to the radius, which is equal to the circumference of the sphere's great circle.
  4. V'(r) is the ratio of the sphere's volume to its radius.
Explanation: The derivative V'(r) represents the instantaneous rate of change of the volume V with respect to the radius r. To find V'(r), we differentiate the volume formula: V'(r) = d/dr [(4/3)πr³] = (4/3)π * 3r² = 4πr². This resulting expression, 4πr², is the formula for the surface area of the sphere, A(r). Therefore, the rate of change of the volume with respect to the radius is equal to the surface area of the sphere.

Question 12

The value of an investment, V, in US dollars, after t years is modelled by V(t) = 5000(1.04)ᵗ. The value of a second investment is modelled by W(t) = 4000e⁰·⁰⁵ᵗ.

Which investment is growing at a faster rate after 6 years?

  1. Investment V, because V(6) > W(6).
  2. Investment W, because W(6) > V(6).
  3. Investment V, because V'(6) > W'(6).
  4. Investment W, because W'(6) > V'(6). (correct answer)
Explanation: The rate of growth is found by comparing the derivatives of the two functions at t = 6. For V(t), the derivative is V'(t) = 5000 * ln(1.04) * (1.04)ᵗ. At t = 6, V'(6) = 5000 * ln(1.04) * (1.04)⁶ ≈ 249.46. For W(t), the derivative is W'(t) = 4000 * 0.05 * e⁰·⁰⁵ᵗ = 200e⁰·⁰⁵ᵗ. At t = 6, W'(6) = 200e⁰·⁰⁵*⁶ = 200e⁰·³ ≈ 269.97. Since 269.97 > 249.46, W'(6) > V'(6), meaning investment W is growing at a faster rate after 6 years. Comparing the values V(6) and W(6) tells us which investment is worth more, not which is growing faster.

Question 13

The elevation, E(x) in metres, of a trail at a horizontal distance x kilometres from the start is modelled by the function E(x) = 0.1x³ - 1.2x² + 3.6x + 50, for 0 ≤ x ≤ 10.

At which two horizontal distances is the trail momentarily flat?

  1. At x = 2 km and x = 6 km (correct answer)
  2. At x = 0 km and x = 8 km
  3. At x = 4 km only
  4. At x = 3 km and x = 5 km
Explanation: The trail is 'flat' when its gradient (slope) is zero. The gradient is given by the derivative of the elevation function, E'(x). First, find the derivative: E'(x) = 0.3x² - 2.4x + 3.6. Set the derivative to zero to find where the trail is flat: 0.3x² - 2.4x + 3.6 = 0. This is a quadratic equation. We can simplify by dividing by 0.3: x² - 8x + 12 = 0. Factoring the quadratic gives (x - 2)(x - 6) = 0. The solutions are x = 2 km and x = 6 km.

Question 14

The sales of a product, S (in thousands of units), are modelled as a function of advertising spending, a (in thousands of dollars), by S(a) = -0.01a³ + 0.9a² + 5a + 100.

What is the rate of change of sales with respect to advertising spending when $20,000 is spent on advertising?

  1. 37 thousand units per thousand dollars
  2. 31 thousand units per thousand dollars
  3. 25 thousand units per thousand dollars
  4. 29 thousand units per thousand dollars (correct answer)
Explanation: When you see a question asking for the "rate of change" of one quantity with respect to another, you're being asked to find the derivative. The rate of change of sales with respect to advertising spending is S′(a)S'(a), which tells you how many additional units are sold per additional thousand dollars spent on advertising. To find S′(a)S'(a), differentiate the given function S(a)=−0.01a3+0.9a2+5a+100S(a) = -0.01a³ + 0.9a² + 5a + 100: S′(a)=−0.03a2+1.8a+5S'(a) = -0.03a² + 1.8a + 5 Since the question asks for the rate when $20,000 is spent on advertising, and $aa ismeasuredinthousandsofdollars,substituteis measured in thousands of dollars, substitute a=20a = 20 $: S'(20) = -0.03(20)² + 1.8(20) + 5 S'(20) = -0.03(400) + 36 + 5 S'(20) = -12 + 36 + 5 = 29 This confirms answer D: 29 thousand units per thousand dollars. Answer A (37) likely comes from incorrectly substituting a = 20 into the original function S(a) instead of its derivative. Answer B (31) might result from a sign error when calculating -0.03(400) , treating it as -10 instead of -12 . Answer C (25) could come from forgetting to add the constant term 5 when differentiating, giving S'(20) = -12 + 36 = 24 , then rounding. Remember: "rate of change" always means derivative. When working with real-world functions, pay careful attention to the units given in the problem to ensure you're substituting the correct value.

Question 15

The population of a species of fish in a lake, N, t years after a conservation program begins is modelled by the function N(t) = 500e⁰.¹⁵ᵗ.

Calculate the rate at which the fish population is growing at the end of the 8th year.

  1. Approximately 75 fish per year
  2. Approximately 126 fish per year (correct answer)
  3. Approximately 1660 fish per year
  4. Approximately 2490 fish per year
Explanation: The rate of growth is the derivative of the population function, N'(t). Using the chain rule, N'(t) = 500 * 0.15 * e⁰.¹⁵ᵗ = 75e⁰.¹⁵ᵗ. To find the rate at t = 8, we calculate N'(8) = 75e⁰.¹⁵*⁸ = 75e¹·² ≈ 125.86. This rounds to 126 fish per year. Distractor A is N'(0), the initial rate. Distractor C is N(8), the total population at 8 years. Distractor D is a miscalculation.

Question 16

The height of a model rocket, h metres, t seconds after launch is given by h(t) = -4.9t² + 80t.

At what time, t > 0, is the instantaneous velocity of the rocket equal to 31 m/s?

  1. 1.0 second
  2. 5.0 seconds (correct answer)
  3. 8.16 seconds
  4. 11.22 seconds
Explanation: Velocity, v(t), is the derivative of the height function, h'(t). First, find the derivative: v(t) = h'(t) = -9.8t + 80. We need to find the time t when the velocity is 31 m/s. Set h'(t) = 31: -9.8t + 80 = 31. Solve for t: -9.8t = 31 - 80 = -49. Thus, t = -49 / -9.8 = 5. The velocity is 31 m/s at t = 5 seconds.

Question 17

The curve of a function is given by y = x⁴ - 3x.

Find the equation of the tangent line to the curve at the point where x = 1.

  1. y = x - 2
  2. y = x - 3 (correct answer)
  3. y = 4x - 6
  4. y = -2x
Explanation: First, find the y-coordinate of the point of tangency by substituting x = 1 into the function: y = (1)⁴ - 3(1) = 1 - 3 = -2. The point is (1, -2). Next, find the gradient of the tangent by finding the derivative, dy/dx. The derivative is dy/dx = 4x³ - 3. Evaluate the gradient at x = 1: m = 4(1)³ - 3 = 1. Using the point-slope form y - y₁ = m(x - x₁), we get y - (-2) = 1(x - 1). This simplifies to y + 2 = x - 1, or y = x - 3.

Question 18

The value of an investment, V, in US dollars, after t years is modelled by V(t) = 5000(1.04)ᵗ. The value of a second investment is modelled by W(t) = 4000e⁰·⁰⁵ᵗ.

Which investment is growing at a faster rate after 6 years?

  1. Investment V, because V(6) > W(6).
  2. Investment W, because W(6) > V(6).
  3. Investment V, because V'(6) > W'(6).
  4. Investment W, because W'(6) > V'(6). (correct answer)
Explanation: The rate of growth is found by comparing the derivatives of the two functions at t = 6. For V(t), the derivative is V'(t) = 5000 * ln(1.04) * (1.04)ᵗ. At t = 6, V'(6) = 5000 * ln(1.04) * (1.04)⁶ ≈ 249.46. For W(t), the derivative is W'(t) = 4000 * 0.05 * e⁰·⁰⁵ᵗ = 200e⁰·⁰⁵ᵗ. At t = 6, W'(6) = 200e⁰·⁰⁵*⁶ = 200e⁰·³ ≈ 269.97. Since 269.97 > 249.46, W'(6) > V'(6), meaning investment W is growing at a faster rate after 6 years. Comparing the values V(6) and W(6) tells us which investment is worth more, not which is growing faster.

Question 19

The elevation, E(x) in metres, of a trail at a horizontal distance x kilometres from the start is modelled by the function E(x) = 0.1x³ - 1.2x² + 3.6x + 50, for 0 ≤ x ≤ 10.

At which two horizontal distances is the trail momentarily flat?

  1. At x = 2 km and x = 6 km (correct answer)
  2. At x = 0 km and x = 8 km
  3. At x = 4 km only
  4. At x = 3 km and x = 5 km
Explanation: The trail is 'flat' when its gradient (slope) is zero. The gradient is given by the derivative of the elevation function, E'(x). First, find the derivative: E'(x) = 0.3x² - 2.4x + 3.6. Set the derivative to zero to find where the trail is flat: 0.3x² - 2.4x + 3.6 = 0. This is a quadratic equation. We can simplify by dividing by 0.3: x² - 8x + 12 = 0. Factoring the quadratic gives (x - 2)(x - 6) = 0. The solutions are x = 2 km and x = 6 km.

Question 20

The sales of a product, S (in thousands of units), are modelled as a function of advertising spending, a (in thousands of dollars), by S(a) = -0.01a³ + 0.9a² + 5a + 100.

What is the rate of change of sales with respect to advertising spending when $20,000 is spent on advertising?

  1. 37 thousand units per thousand dollars
  2. 31 thousand units per thousand dollars
  3. 25 thousand units per thousand dollars
  4. 29 thousand units per thousand dollars (correct answer)
Explanation: When you see a question asking for the "rate of change" of one quantity with respect to another, you're being asked to find the derivative. The rate of change of sales with respect to advertising spending is S′(a)S'(a), which tells you how many additional units are sold per additional thousand dollars spent on advertising. To find S′(a)S'(a), differentiate the given function S(a)=−0.01a3+0.9a2+5a+100S(a) = -0.01a³ + 0.9a² + 5a + 100: S′(a)=−0.03a2+1.8a+5S'(a) = -0.03a² + 1.8a + 5 Since the question asks for the rate when $20,000 is spent on advertising, and $aa ismeasuredinthousandsofdollars,substituteis measured in thousands of dollars, substitute a=20a = 20 $: S'(20) = -0.03(20)² + 1.8(20) + 5 S'(20) = -0.03(400) + 36 + 5 S'(20) = -12 + 36 + 5 = 29 This confirms answer D: 29 thousand units per thousand dollars. Answer A (37) likely comes from incorrectly substituting a = 20 into the original function S(a) instead of its derivative. Answer B (31) might result from a sign error when calculating -0.03(400) , treating it as -10 instead of -12 . Answer C (25) could come from forgetting to add the constant term 5 when differentiating, giving S'(20) = -12 + 36 = 24 , then rounding. Remember: "rate of change" always means derivative. When working with real-world functions, pay careful attention to the units given in the problem to ensure you're substituting the correct value.