Business Calculus Quiz: Higher Order Derivatives
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Higher Order DerivativesQuestion 1 of 18

Let P(t)P(t) represent the profit of a company, in millions of dollars, tt months after a new product launch. At month t=6t=6, an analyst determines that P(6)>0P'(6) > 0 and P(6)<0P''(6) < 0. Which of the following statements best describes the company's profit at this time?

The company's profit is decreasing, and the rate of decrease is slowing down.
The company's profit is increasing, and the rate of increase is accelerating.
The company's profit has reached a maximum and is about to start decreasing.
The company's profit is increasing, but the rate of increase is slowing down.
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Business Calculus Quiz

Business Calculus Quiz: Higher Order Derivatives

Practice Higher Order Derivatives in Business Calculus 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 Higher Order Derivatives, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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

Let P(t)P(t) represent the profit of a company, in millions of dollars, tt months after a new product launch. At month t=6t=6, an analyst determines that P(6)>0P'(6) > 0 and P(6)<0P''(6) < 0. Which of the following statements best describes the company's profit at this time?

  1. The company's profit is decreasing, and the rate of decrease is slowing down.
  2. The company's profit is increasing, and the rate of increase is accelerating.
  3. The company's profit has reached a maximum and is about to start decreasing.
  4. The company's profit is increasing, but the rate of increase is slowing down. (correct answer)
Explanation: P(t)P'(t) represents the rate of change of profit. Since P(6)>0P'(6) > 0, the profit is increasing at t=6t=6. P(t)P''(t) represents the rate of change of the rate of change of profit (i.e., the acceleration of profit). Since P(6)<0P''(6) < 0, the rate of change is decreasing. Therefore, at t=6t=6, the profit is increasing, but at a decreasing (or slowing) rate.

Question 2

The total cost, in dollars, to produce xx units of a certain electronic component is given by C(x)=0.002x30.9x2+150x+5000C(x) = 0.002x^3 - 0.9x^2 + 150x + 5000. The marginal cost is the rate of change of the total cost. For what range of production levels is the marginal cost increasing?

  1. For all x>0x > 0
  2. For x>150x > 150 (correct answer)
  3. For x<150x < 150
  4. For x=150x = 150 only
Explanation: The marginal cost is given by the first derivative, C(x)C'(x). We want to find where the marginal cost is increasing, which means we need to find where its derivative, C(x)C''(x), is positive. First, find the first derivative (marginal cost): C(x)=0.006x21.8x+150C'(x) = 0.006x^2 - 1.8x + 150. Next, find the second derivative: C(x)=0.012x1.8C''(x) = 0.012x - 1.8. To find where the marginal cost is increasing, set C(x)>0C''(x) > 0: 0.012x1.8>0    0.012x>1.8    x>1.80.012    x>1500.012x - 1.8 > 0 \implies 0.012x > 1.8 \implies x > \frac{1.8}{0.012} \implies x > 150. Therefore, the marginal cost is increasing for production levels greater than 150 units.

Question 3

The number of new users for a mobile app tt weeks after a promotional campaign begins is modeled by N(t)=500te0.2tN(t) = 500t e^{-0.2t} for t0t \ge 0. The rate of growth in new users initially accelerates and then begins to slow down. At what time tt does this rate of growth start to slow down?

  1. t=5t = 5 weeks
  2. t=20t = 20 weeks
  3. t=5e1t = 5e^{-1} weeks
  4. t=10t = 10 weeks (correct answer)
Explanation: When analyzing growth patterns, you need to understand that the "rate of growth" refers to the first derivative, and when this rate "starts to slow down," you're looking for where the second derivative changes from positive to negative - the inflection point. Given N(t)=500te0.2tN(t) = 500te^{-0.2t}, first find the rate of growth by taking the derivative. Using the product rule: N(t)=500e0.2t+500t(0.2)e0.2t=500e0.2t(10.2t)N'(t) = 500e^{-0.2t} + 500t(-0.2)e^{-0.2t} = 500e^{-0.2t}(1 - 0.2t) To find when this rate starts slowing down, take the second derivative: N(t)=500(0.2)e0.2t(10.2t)+500e0.2t(0.2)=100e0.2t(10.2t)100e0.2t=100e0.2t(20.2t)N''(t) = 500(-0.2)e^{-0.2t}(1 - 0.2t) + 500e^{-0.2t}(-0.2) = -100e^{-0.2t}(1 - 0.2t) - 100e^{-0.2t} = -100e^{-0.2t}(2 - 0.2t) The growth rate changes from accelerating to decelerating when N(t)=0N''(t) = 0. Since e0.2te^{-0.2t} is never zero, we need 20.2t=02 - 0.2t = 0, which gives us t=10t = 10 weeks. Answer choice A (t=5t = 5) gives you the time when the growth rate is maximized, not when acceleration stops. Choice B (t=20t = 20) has no mathematical significance in this context. Choice C (t=5e1t = 5e^{-1}) appears to confuse exponential relationships but doesn't solve the correct equation. Remember: when a problem asks about changes in growth patterns, you're typically looking for inflection points where the second derivative equals zero. Always verify which derivative answers the specific question being asked.

Question 4

The revenue R(t)R(t) of a company is changing over time tt (in years). The rate of change of the rate of change of revenue is constant, with R(t)=8R''(t) = -8 (in millions of dollars per year squared). At time t=3t=3 years, the revenue was growing at a rate of 1212 million dollars per year. At what time tt does the company's revenue reach its peak?

  1. t=1.5t = 1.5 years
  2. t=3t = 3 years
  3. t=4.5t = 4.5 years (correct answer)
  4. t=6t = 6 years
Explanation: To find the rate of change of revenue, R(t)R'(t), we need to integrate R(t)R''(t) with respect to tt. R(t)=R(t)dt=8dt=8t+CR'(t) = \int R''(t) dt = \int -8 dt = -8t + C. We are given that at t=3t=3, the rate of growth was 1212, so R(3)=12R'(3) = 12. We use this to solve for the constant CC. 12=8(3)+C    12=24+C    C=3612 = -8(3) + C \implies 12 = -24 + C \implies C = 36. So, the rate of change of revenue is R(t)=8t+36R'(t) = -8t + 36. The revenue reaches its peak when its rate of change is zero, so we set R(t)=0R'(t) = 0. 8t+36=0    8t=36    t=368=4.5-8t + 36 = 0 \implies 8t = 36 \implies t = \frac{36}{8} = 4.5 years.

Question 5

The total cost to produce qq units of a product is given by C(q)=2q3+bq2+192q+4000C(q) = 2q^3 + bq^2 + 192q + 4000, where bb is a constant. The marginal cost is minimized at a production level of q=12q=12 units. What is the value of the constant bb?

  1. b=72b = -72 (correct answer)
  2. b=144b = -144
  3. b=36b = -36
  4. b=12b = 12
Explanation: When you encounter optimization problems involving cost functions, you're looking for points where derivatives equal zero. Here, marginal cost is the first derivative of the total cost function, and you need to minimize this marginal cost. The marginal cost function is C(q)=6q2+2bq+192C'(q) = 6q^2 + 2bq + 192. To find where this is minimized, you take the second derivative: C(q)=12q+2bC''(q) = 12q + 2b. At the minimum, this second derivative equals zero. Since the marginal cost is minimized at q=12q = 12, you can substitute: C(12)=0C''(12) = 0 12(12)+2b=012(12) + 2b = 0 144+2b=0144 + 2b = 0 2b=1442b = -144 b=72b = -72 Let's check why the other answers don't work. Answer B gives b=144b = -144, which would make C(12)=144+2(144)=1440C''(12) = 144 + 2(-144) = -144 \neq 0. Answer C gives b=36b = -36, making C(12)=144+2(36)=720C''(12) = 144 + 2(-36) = 72 \neq 0. Answer D gives b=12b = 12, making C(12)=144+2(12)=1680C''(12) = 144 + 2(12) = 168 \neq 0. None of these satisfy the condition that the second derivative equals zero at q=12q = 12. Therefore, A) b=72b = -72 is correct. Study tip: In optimization problems, remember the sequence: find the first derivative, set its derivative (the second derivative) equal to zero to locate extrema, then solve for unknown parameters. Always verify your answer by substituting back into the condition.

Question 6

For the function f(x)=x55x4+10x310x2+5xf(x) = x^5 - 5x^4 + 10x^3 - 10x^2 + 5x, if f(4)(x)f^{(4)}(x) represents the fourth derivative, what can be concluded about f(5)(x)f^{(5)}(x) and f(6)(x)f^{(6)}(x)?

  1. f(5)(x)=120f^{(5)}(x) = 120 for all xx, and f(6)(x)=0f^{(6)}(x) = 0 for all xx (correct answer)
  2. f(5)(x)=0f^{(5)}(x) = 0 for all xx, and f(6)(x)f^{(6)}(x) is undefined for all xx
  3. f(5)(x)=5f^{(5)}(x) = 5 for all xx, and f(6)(x)=0f^{(6)}(x) = 0 for all xx
  4. f(5)(x)=5!f^{(5)}(x) = 5! for all xx, and f(6)(x)=6!f^{(6)}(x) = 6! for all xx
Explanation: Since f(x)f(x) is a degree 5 polynomial, its fifth derivative will be a constant (the coefficient of x5x^5 times 5!5!), and its sixth derivative will be zero. The coefficient of x5x^5 is 1, so f(5)(x)=15!=120f^{(5)}(x) = 1 \cdot 5! = 120. Any derivative beyond the fifth will be zero. Choice B incorrectly states the fifth derivative is zero. Choice C gives the wrong constant value. Choice D incorrectly suggests higher derivatives continue to exist as factorials.

Question 7

The temperature of a chemical reaction follows T(t)=25+40e0.2tT(t) = 25 + 40e^{-0.2t} degrees Celsius, where tt is time in minutes. If safety protocols require monitoring when the rate of temperature change is itself changing most rapidly, at what time should this occur?

  1. At t=0t = 0 minutes, when the initial cooling rate is established (correct answer)
  2. At t=5t = 5 minutes, when T(t)T'(t) reaches its minimum value
  3. At t=t = \infty minutes, when the temperature stabilizes completely
  4. This occurs continuously since T(t)T''(t) is always changing at a constant rate
Explanation: The rate of temperature change is T(t)=8e0.2tT'(t) = -8e^{-0.2t}, and its rate of change is T(t)=1.6e0.2tT''(t) = 1.6e^{-0.2t}. The rate at which T(t)T'(t) changes is measured by T(t)|T''(t)|, which is maximized when e0.2te^{-0.2t} is largest, occurring at t=0t = 0. Choice B incorrectly identifies when T(t)T'(t) is minimum rather than when T(t)T''(t) is maximum. Choice C suggests the limit case. Choice D misunderstands that we want maximum rate of change, not constant change.

Question 8

Let C(q)C(q) be a company's cost function for producing a quantity qq of a product. At a production level of q0=500q_0 = 500 units, it is known that C(500)=0C''(500) = 0 and C(500)<0C'''(500) < 0. What does this imply about the marginal cost, C(q)C'(q), at this production level?

  1. The marginal cost is at a local minimum.
  2. The marginal cost is at a local maximum. (correct answer)
  3. The marginal cost is zero.
  4. The marginal cost is increasing at its fastest rate.
Explanation: We can use the second derivative test to analyze the behavior of the marginal cost function, C(q)C'(q). The derivative of C(q)C'(q) is C(q)C''(q), and the second derivative of C(q)C'(q) is C(q)C'''(q). We are given that at q0=500q_0 = 500, the first derivative of C(q)C'(q) is zero (since C(500)=0C''(500) = 0), which means q0=500q_0=500 is a critical point for the marginal cost function. We are also given that the second derivative of C(q)C'(q) is negative (since C(500)<0C'''(500) < 0). By the second derivative test, if the first derivative is zero and the second derivative is negative, the function has a local maximum at that point. Therefore, the marginal cost C(q)C'(q) is at a local maximum at q=500q=500.

Question 9

The productivity P(x)P(x) of a company, in thousands of units per month, is given by the function P(x)=x3+30x2+10x+500P(x) = -x^3 + 30x^2 + 10x + 500, where xx is the amount spent on employee training in thousands of dollars. The company experiences diminishing returns when the rate of change of productivity begins to decrease. At what level of training expenditure xx do diminishing returns set in?

  1. x=5x = 5
  2. x=10x = 10 (correct answer)
  3. x=20x = 20
  4. x=30x = 30
Explanation: The point of diminishing returns occurs at the inflection point of the productivity function P(x)P(x), which is where the second derivative P(x)P''(x) is zero and changes sign. First, find the first and second derivatives of P(x)P(x). P(x)=3x2+60x+10P'(x) = -3x^2 + 60x + 10 P(x)=6x+60P''(x) = -6x + 60 Set the second derivative equal to zero to find potential inflection points: 6x+60=0    6x=60    x=10-6x + 60 = 0 \implies 6x = 60 \implies x = 10. For x<10x < 10, P(x)>0P''(x) > 0 (concave up), and for x>10x > 10, P(x)<0P''(x) < 0 (concave down). This confirms that x=10x=10 is the inflection point where the rate of change of productivity begins to decrease. The expenditure is 1010 thousand dollars.

Question 10

A marketing director states, "Our market share is still increasing, but the gains we're making each month are smaller than the gains from the month before." If M(t)M(t) is the market share at time tt in months, which pair of conditions must be true for the current time t0t_0?

  1. M(t0)<0M'(t_0) < 0 and M(t0)>0M''(t_0) > 0
  2. M(t0)<0M'(t_0) < 0 and M(t0)<0M''(t_0) < 0
  3. M(t0)>0M'(t_0) > 0 and M(t0)<0M''(t_0) < 0 (correct answer)
  4. M(t0)>0M'(t_0) > 0 and M(t0)>0M''(t_0) > 0
Explanation: The statement "Our market share is still increasing" means that the rate of change of market share, M(t)M'(t), is positive. So, M(t0)>0M'(t_0) > 0. The statement "the gains we're making each month are smaller than the gains from the month before" means that the rate of increase is slowing down. The rate of change of the rate of increase is the second derivative, M(t)M''(t). If the rate is slowing down, its derivative must be negative. So, M(t0)<0M''(t_0) < 0. The correct pair of conditions is M(t0)>0M'(t_0) > 0 and M(t0)<0M''(t_0) < 0.

Question 11

The efficiency E(h)E(h) of an assembly line worker, measured in units assembled per hour, is modeled by E(h)=0.02h3+0.9h2+12hE(h) = -0.02h^3 + 0.9h^2 + 12h after hh hours of training, for 0h300 \le h \le 30. The worker's rate of improvement is the rate of change of their efficiency. After an initial period, this rate of improvement begins to decline. At how many hours of training, hh, does this decline begin?

  1. h=15h = 15 hours (correct answer)
  2. h=10h = 10 hours
  3. h=20h = 20 hours
  4. h=30h = 30 hours
Explanation: When you encounter questions about rates of change and when those rates begin to decline, you're dealing with optimization using derivatives. The "rate of improvement" is the first derivative of efficiency, and when this rate "begins to decline," you need to find where the second derivative equals zero. The efficiency function is E(h)=0.02h3+0.9h2+12hE(h) = -0.02h^3 + 0.9h^2 + 12h. The rate of improvement is the first derivative: E(h)=0.06h2+1.8h+12E'(h) = -0.06h^2 + 1.8h + 12. To find when this rate begins to decline, you need the second derivative: E(h)=0.12h+1.8E''(h) = -0.12h + 1.8. The rate of improvement begins to decline when E(h)=0E''(h) = 0: 0.12h+1.8=0-0.12h + 1.8 = 0 h=1.80.12=15h = \frac{1.8}{0.12} = 15 This confirms answer A) h=15h = 15 hours is correct. Let's examine why the other answers are wrong. Answer B) h=10h = 10 gives E(10)=1.2+1.8=0.6>0E''(10) = -1.2 + 1.8 = 0.6 > 0, meaning the rate is still increasing. Answer C) h=20h = 20 gives E(20)=2.4+1.8=0.6<0E''(20) = -2.4 + 1.8 = -0.6 < 0, so the decline has already begun. Answer D) h=30h = 30 is the endpoint of the domain and gives E(30)=3.6+1.8=1.8<0E''(30) = -3.6 + 1.8 = -1.8 < 0, far past when the decline began. Study tip: Remember that "rate begins to decline" means finding where the second derivative equals zero, not where it's negative. The second derivative tells you about the concavity and turning points of the first derivative.

Question 12

The profit P(x)P(x) from producing xx items is such that the marginal profit at a production level of x=1000x=1000 is zero. The second derivative of the profit function is given by P(x)=0.04x30P''(x) = 0.04x - 30. Based on this information, what can be concluded about the profit at the production level of x=1000x=1000?

  1. The profit is at a local minimum. (correct answer)
  2. The profit is at a local maximum.
  3. The profit function has a point of inflection.
  4. The profit is neither a maximum nor a minimum.
Explanation: When you encounter a problem about profit optimization, you're working with the second derivative test to classify critical points. Since marginal profit is the first derivative P(x)P'(x), and you're told that P(1000)=0P'(1000) = 0, you have a critical point at x=1000x = 1000. To determine whether this critical point is a maximum or minimum, you need to examine the second derivative at that point. Calculate P(1000)P''(1000) using the given formula: P(1000)=0.04(1000)30=4030=10P''(1000) = 0.04(1000) - 30 = 40 - 30 = 10. Since P(1000)=10>0P''(1000) = 10 > 0, the second derivative test tells us the profit function is concave up at x=1000x = 1000, confirming a local minimum. Let's examine why the other answers are incorrect. Answer B suggests a local maximum, but this would require P(1000)<0P''(1000) < 0, which contradicts our calculation. Answer C claims a point of inflection, but inflection points occur where P(x)=0P''(x) = 0, and we found P(1000)=100P''(1000) = 10 \neq 0. Answer D suggests neither maximum nor minimum, but since we have P(1000)=0P'(1000) = 0 and P(1000)>0P''(1000) > 0, the second derivative test definitively classifies this as a local minimum. Study tip: Remember the second derivative test pattern: when f(a)=0f'(a) = 0 and f(a)>0f''(a) > 0, you have a local minimum; when f(a)=0f'(a) = 0 and f(a)<0f''(a) < 0, you have a local maximum. Always calculate the second derivative value at the critical point to classify it properly.

Question 13

The total cost C(q)C(q) to produce qq items has a marginal cost C(q)C'(q) that is observed to be decreasing when production is less than 200 units (q<200q < 200) and increasing when production is greater than 200 units (q>200q > 200). Assuming marginal cost is always positive, which statement accurately describes the total cost function at a production level of q=250q=250?

  1. The total cost is decreasing at an increasing rate.
  2. The total cost is increasing at a decreasing rate.
  3. The total cost is increasing at an increasing rate. (correct answer)
  4. The total cost is decreasing at a decreasing rate.
Explanation: The behavior of the marginal cost function, C(q)C'(q), gives us information about the second derivative of the total cost function, C(q)C''(q). If C(q)C'(q) is increasing, its derivative, C(q)C''(q), must be positive. We are told that for q>200q > 200, the marginal cost is increasing. Therefore, at q=250q=250, we have C(250)>0C''(250) > 0. This means the total cost function C(q)C(q) is concave up at q=250q=250. We are also told that marginal cost is always positive, so C(q)>0C'(q) > 0 for all qq. This means the total cost function is always increasing. Combining these facts, at q=250q=250, the total cost is increasing (C>0C' > 0) and it is concave up (C>0C'' > 0). This means the total cost is increasing at an increasing rate.

Question 14

A company's revenue function is R(t)=100ln(t+1)+50tR(t) = 100\ln(t+1) + 50t thousand dollars, where tt is time in years since startup. What does R(2)R''(2) represent, and what is its value?

  1. The concavity indicator of revenue at t=2t=2; value is 1009-\frac{100}{9} thousand dollars per year²
  2. The second derivative of revenue at t=2t=2; value is 1009\frac{100}{9} thousand dollars per year²
  3. The acceleration of revenue growth at t=2t=2; value is 1009-\frac{100}{9} thousand dollars per year² (correct answer)
  4. The instantaneous rate of revenue change at t=2t=2; value is 1003+50\frac{100}{3} + 50 thousand dollars per year
Explanation: When you encounter questions about second derivatives in business contexts, remember that the second derivative measures how the rate of change is itself changing - essentially the "acceleration" of the function. To find R(2)R''(2), you need to take two derivatives of the revenue function. Starting with R(t)=100ln(t+1)+50tR(t) = 100\ln(t+1) + 50t, the first derivative is R(t)=100t+1+50R'(t) = \frac{100}{t+1} + 50. Taking the derivative again gives R(t)=100(t+1)2R''(t) = -\frac{100}{(t+1)^2}. Evaluating at t=2t=2: R(2)=100(2+1)2=1009R''(2) = -\frac{100}{(2+1)^2} = -\frac{100}{9} thousand dollars per year². The negative value indicates that revenue growth is decelerating - the company is still gaining revenue, but at a slower and slower rate each year. Option A correctly calculates the value but uses vague terminology. While "concavity indicator" isn't wrong, it's less precise than "acceleration of revenue growth." Option B has the right general idea about second derivatives but gets the sign wrong - the value is negative, not positive. Option D confuses the second derivative with the first derivative, giving you R(2)=1003+50R'(2) = \frac{100}{3} + 50, which measures the instantaneous rate of change, not the acceleration. Study tip: Remember that in business calculus, the second derivative always represents acceleration or deceleration. A negative second derivative means the growth rate is slowing down, while a positive one means growth is speeding up. This interpretation is crucial for understanding business trends.

Question 15

An economist models consumer demand as D(p)=1000p2+1D(p) = \frac{1000}{p^2 + 1}, where DD is quantity demanded and pp is price. The second derivative D(p)D''(p) indicates how the rate of change of demand responds to price changes. For what price does D(p)=0D''(p) = 0?

  1. p=0p = 0, where demand is at its maximum level
  2. p=3p = \sqrt{3}, where the second derivative reaches its minimum value
  3. p=1p = 1, where the denominator effect balances the numerator effect
  4. p=13p = \frac{1}{\sqrt{3}}, where the demand curve changes concavity (correct answer)
Explanation: When you see a question asking where the second derivative equals zero, you're looking for inflection points—places where the curve changes from concave up to concave down (or vice versa). This is crucial in economics because it tells you where the rate of change in demand switches behavior. To find D(p)=0D''(p) = 0, you need to differentiate twice. Starting with D(p)=1000p2+1D(p) = \frac{1000}{p^2 + 1}, use the quotient rule or rewrite as D(p)=1000(p2+1)1D(p) = 1000(p^2 + 1)^{-1}. First derivative: D(p)=2000p(p2+1)2D'(p) = -2000p(p^2 + 1)^{-2} Second derivative: D(p)=2000(p2+1)2+8000p2(p2+1)3D''(p) = -2000(p^2 + 1)^{-2} + 8000p^2(p^2 + 1)^{-3} Factoring out common terms: D(p)=2000(p2+1)+8000p2(p2+1)3=6000p22000(p2+1)3D''(p) = \frac{-2000(p^2 + 1) + 8000p^2}{(p^2 + 1)^3} = \frac{6000p^2 - 2000}{(p^2 + 1)^3} Setting D(p)=0D''(p) = 0: 6000p22000=06000p^2 - 2000 = 0, so p2=13p^2 = \frac{1}{3}, giving p=13p = \frac{1}{\sqrt{3}}. Answer A is wrong because p=0p = 0 makes D(0)=20000D''(0) = -2000 \neq 0. Answer B confuses the critical point with an extremum of the second derivative itself. Answer C incorrectly assumes some mysterious "balance" occurs at p=1p = 1, but substituting shows D(1)0D''(1) \neq 0. Remember: when finding inflection points, always set the second derivative equal to zero and solve. The economic interpretation—where concavity changes—comes after the math.

Question 16

A population model is given by N(t)=1000e0.05tN(t) = 1000e^{0.05t}, where N(t)N(t) is the population after tt years. If N(t)N'(t) represents the growth rate and N(t)N''(t) represents the acceleration of growth, what is the relationship between these quantities?

  1. N(t)=0.05N(t)+N(t)N''(t) = 0.05 \cdot N(t) + N'(t) for all values of tt
  2. N(t)=(0.05)2N(t)N''(t) = (0.05)^2 \cdot N(t) for all values of tt
  3. N(t)=0.05N(t)N''(t) = 0.05 \cdot N'(t) for all values of tt (correct answer)
  4. N(t)=2.5N(t)N''(t) = 2.5 \cdot N(t) for all values of tt
Explanation: When you encounter exponential growth models like N(t)=1000e0.05tN(t) = 1000e^{0.05t}, you're dealing with functions where the derivatives have special relationships due to the properties of exponential functions. To find the relationship between N(t)N'(t) and N(t)N''(t), let's calculate each derivative. Using the chain rule, N(t)=1000e0.05t0.05=50e0.05tN'(t) = 1000 \cdot e^{0.05t} \cdot 0.05 = 50e^{0.05t}. Taking the derivative again: N(t)=50e0.05t0.05=2.5e0.05tN''(t) = 50 \cdot e^{0.05t} \cdot 0.05 = 2.5e^{0.05t}. Now examine the relationship: N(t)=2.5e0.05tN''(t) = 2.5e^{0.05t} and N(t)=50e0.05tN'(t) = 50e^{0.05t}. Notice that N(t)=0.0550e0.05t=0.05N(t)N''(t) = 0.05 \cdot 50e^{0.05t} = 0.05 \cdot N'(t). This confirms answer choice C is correct. Looking at the wrong answers: Choice A incorrectly adds N(t)N'(t) to a multiple of N(t)N(t), which doesn't match our calculated relationship. Choice B uses (0.05)2=0.0025(0.05)^2 = 0.0025, but our calculation shows the coefficient should be 0.050.05, not its square. Choice D states N(t)=2.5N(t)N''(t) = 2.5 \cdot N(t), which would mean 2.5e0.05t=2.51000e0.05t2.5e^{0.05t} = 2.5 \cdot 1000e^{0.05t}, clearly false since 110001 ≠ 1000. Study tip: For exponential growth models N(t)=aertN(t) = ae^{rt}, remember that N(t)=rN(t)N'(t) = r \cdot N(t) and N(t)=rN(t)N''(t) = r \cdot N'(t). The growth constant rr always appears as the proportionality factor between consecutive derivatives.

Question 17

For a production function Q(L)=20L2/3Q(L) = 20L^{2/3}, where QQ is output and LL is labor input, economists define the marginal product as Q(L)Q'(L) and the rate of change of marginal product as Q(L)Q''(L). If Q(L)=809L5/3Q''(L) = -\frac{80}{9}L^{-5/3}, what can be concluded about the production process?

  1. Marginal product is always increasing because Q(L)>0Q'(L) > 0 for all L>0L > 0
  2. Marginal product is always decreasing because Q(L)<0Q''(L) < 0 for all L>0L > 0 (correct answer)
  3. The production function exhibits increasing returns to scale throughout
  4. Output decreases as more labor is added because Q(L)<0Q''(L) < 0
Explanation: Since Q(L)=809L5/3<0Q''(L) = -\frac{80}{9}L^{-5/3} < 0 for all L>0L > 0, the marginal product Q(L)Q'(L) is always decreasing. This represents diminishing marginal returns. Choice A confuses positive marginal product with increasing marginal product. Choice C discusses returns to scale, which is different from marginal productivity. Choice D incorrectly suggests output decreases, when actually Q(L)=403L1/3>0Q'(L) = \frac{40}{3}L^{-1/3} > 0, so output still increases but at a decreasing rate.

Question 18

The velocity of a particle is given by v(t)=6t224t+18v(t) = 6t^2 - 24t + 18 meters per second. At t=1t = 1 second, what is the rate of change of the acceleration?

  1. 00 m/s³, indicating constant acceleration at that instant
  2. 1212 m/s³, indicating acceleration is increasing at that rate (correct answer)
  3. 12-12 m/s³, indicating acceleration is decreasing at that rate
  4. 66 m/s³, indicating acceleration is increasing at that rate
Explanation: The rate of change of acceleration is the third derivative of position, or the second derivative of velocity. Given v(t)=6t224t+18v(t) = 6t^2 - 24t + 18, we have a(t)=v(t)=12t24a(t) = v'(t) = 12t - 24 and j(t)=a(t)=12j(t) = a'(t) = 12. At t=1t = 1, the jerk is 1212 m/s³, meaning acceleration is increasing at 1212 m/s³. Choice A incorrectly uses acceleration value instead of its rate of change. Choice C has the wrong sign. Choice D uses half the correct value.