Calculus 1 Quiz: Accumulation Functions In Context
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Accumulation Functions In ContextQuestion 1 of 20

A population of insects increases at a rate of R(t)=200+6t2R(t) = 200 + 6t^2 insects per day. If the initial population at t=0t=0 is 1000, what is the population at t=3t=3 days?

254
654
1654
1212
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Calculus 1 Quiz

Calculus 1 Quiz: Accumulation Functions In Context

Practice Accumulation Functions In Context in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Accumulation Functions In Context, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

A population of insects increases at a rate of R(t)=200+6t2R(t) = 200 + 6t^2 insects per day. If the initial population at t=0t=0 is 1000, what is the population at t=3t=3 days?

  1. 254
  2. 654
  3. 1654 (correct answer)
  4. 1212
Explanation: The population at t=3t=3 is the initial population plus the total increase from t=0t=0 to t=3t=3. The total increase is the integral of the rate: ∫03(200+6t2)dt=[200t+2t3]03=(200(3)+2(33))−0=600+2(27)=600+54=654\int_0^3 (200 + 6t^2) dt = [200t + 2t^3]_0^3 = (200(3) + 2(3^3)) - 0 = 600 + 2(27) = 600 + 54 = 654. The total population is 1000+654=16541000 + 654 = 1654.

Question 2

A rocket burns fuel at a rate of r(t)=15tr(t) = 15\sqrt{t} kg per second. The rocket initially contains 20,000 kg of fuel. How much fuel, in kg, remains in the rocket 100 seconds after launch?

  1. 10,000 (correct answer)
  2. 15,000
  3. 18,500
  4. 5,000
Explanation: The amount of fuel remaining is the initial amount minus the total fuel burned. The total fuel burned in the first 100 seconds is ∫0100r(t)dt=∫010015t1/2dt\int_0^{100} r(t) dt = \int_0^{100} 15t^{1/2} dt. The integral evaluates to [15⋅23t3/2]0100=[10t3/2]0100=10(1003/2)−0=10(103)=10,000[15 \cdot \frac{2}{3}t^{3/2}]_0^{100} = [10t^{3/2}]_0^{100} = 10(100^{3/2}) - 0 = 10(10^3) = 10,000 kg. The remaining fuel is the initial amount minus the amount burned: 20,000−10,000=10,00020,000 - 10,000 = 10,000 kg.

Question 3

A car starts at a position 10 miles east of a landmark. The car's velocity is given by v(t)=15t−3t2v(t) = 15t - 3t^2 miles per hour, where tt is in hours and positive velocity indicates eastward travel. What is the car's position relative to the landmark after 2 hours?

  1. 22 miles east
  2. 32 miles east (correct answer)
  3. 18 miles east
  4. 28 miles east
Explanation: The position x(t)x(t) is the initial position plus the displacement. The initial position is x(0)=10x(0)=10. The displacement over the first 2 hours is ∫02v(t)dt=∫02(15t−3t2)dt\int_0^2 v(t) dt = \int_0^2 (15t - 3t^2) dt. This integral evaluates to [152t2−t3]02=(152(4)−23)−0=30−8=22[\frac{15}{2}t^2 - t^3]_0^2 = (\frac{15}{2}(4) - 2^3) - 0 = 30 - 8 = 22 miles. The final position is the initial position plus the displacement: x(2)=x(0)+22=10+22=32x(2) = x(0) + 22 = 10 + 22 = 32 miles east of the landmark.

Question 4

From 9 AM (t=0t=0) to 5 PM (t=8t=8), visitors enter an amusement park at a rate modeled by E(t)E(t) people per hour. During the same period, visitors leave the park at a rate modeled by L(t)L(t) people per hour.

If there were 500 people in the park at 9 AM, which expression represents the total number of people in the park at 1 PM (t=4t=4)?

  1. 500+∫04(E(t)−L(t))dt500 + \int_0^4 (E(t) - L(t)) dt (correct answer)
  2. ∫04(E(t)−L(t))dt\int_0^4 (E(t) - L(t)) dt
  3. 500+E(4)−L(4)500 + E(4) - L(4)
  4. 500+∫04E(t)dt500 + \int_0^4 E(t) dt
Explanation: The total number of people at a given time is the initial number of people plus the net change in the number of people. The net rate of change is E(t)−L(t)E(t) - L(t). The net change from t=0t=0 (9 AM) to t=4t=4 (1 PM) is the integral of the net rate, ∫04(E(t)−L(t))dt\int_0^4 (E(t) - L(t)) dt. Thus, the total number of people at t=4t=4 is the initial amount, 500, plus this net change.

Question 5

A tank initially contains W0W_0 gallons of water. For t≥0t \ge 0, water is pumped into the tank at a constant rate of CC gallons per minute. Simultaneously, water leaks out of the tank at a rate of L(t)L(t) gallons per minute, where L(t)L(t) is a differentiable function. Which of the following expressions represents the amount of water in the tank at time TT?

  1. W0+∫0T(C−L(t))dtW_0 + \int_{0}^{T} (C - L(t)) dt (correct answer)
  2. W0+∫0T(L(t)−C)dtW_0 + \int_{0}^{T} (L(t) - C) dt
  3. ∫0T(C−L(t))dt\int_{0}^{T} (C - L(t)) dt
  4. W0+C⋅T−L(T)W_0 + C \cdot T - L(T)
Explanation: Let W(t)W(t) be the amount of water in the tank at time tt. The rate of change of the amount of water is the rate in minus the rate out. So, W′(t)=C−L(t)W'(t) = C - L(t). To find the total amount of water at time TT, we must add the initial amount, W0W_0, to the net change in water from t=0t=0 to t=Tt=T. The net change is the integral of the net rate of change. Therefore, the amount of water at time TT is given by W(T)=W(0)+∫0TW′(t)dt=W0+∫0T(C−L(t))dtW(T) = W(0) + \int_{0}^{T} W'(t) dt = W_0 + \int_{0}^{T} (C - L(t)) dt.

Question 6

The population of a certain species of fish in a lake is changing at a rate of r(t)r(t) fish per month. At time t=3t=3 months, the population is measured to be 1,500 fish. Which expression represents the fish population at time t=12t=12 months?

  1. 1500+∫312r(t)dt1500 + \int_{3}^{12} r(t) dt (correct answer)
  2. ∫312r(t)dt\int_{3}^{12} r(t) dt
  3. 1500+∫012r(t)dt1500 + \int_{0}^{12} r(t) dt
  4. 1500−∫312r(t)dt1500 - \int_{3}^{12} r(t) dt
Explanation: Let P(t)P(t) be the population of fish at time tt. We are given that P′(t)=r(t)P'(t) = r(t) and P(3)=1500P(3) = 1500. According to the Fundamental Theorem of Calculus, the net change in population from t=3t=3 to t=12t=12 is ∫312r(t)dt=P(12)−P(3)\int_{3}^{12} r(t) dt = P(12) - P(3). To find the population at t=12t=12, we can rearrange this equation: P(12)=P(3)+∫312r(t)dtP(12) = P(3) + \int_{3}^{12} r(t) dt. Substituting the known value P(3)=1500P(3) = 1500, we get P(12)=1500+∫312r(t)dtP(12) = 1500 + \int_{3}^{12} r(t) dt.

Question 7

The rate at which water flows into a reservoir is modeled by the function r(t)=300−20tr(t) = 300 - 20t cubic meters per hour, where tt is the number of hours since midnight. At midnight (t=0t=0), the reservoir contains 10,000 cubic meters of water. The model is valid for 0≤t≤200 \le t \le 20. At what time does the reservoir hold the maximum amount of water?

  1. t=0t=0 hours
  2. t=10t=10 hours
  3. t=15t=15 hours (correct answer)
  4. t=20t=20 hours
Explanation: Let V(t)V(t) be the volume of water in the reservoir at time tt. The rate of change of the volume is V′(t)=r(t)=300−20tV'(t) = r(t) = 300 - 20t. The amount of water at time tt is given by the accumulation function V(t)=V(0)+∫0tr(x)dx=10000+∫0t(300−20x)dxV(t) = V(0) + \int_{0}^{t} r(x) dx = 10000 + \int_{0}^{t} (300 - 20x) dx. To find the maximum volume, we apply the Extreme Value Theorem on the interval [0,20][0, 20]. First, find the critical points by setting the derivative V′(t)=r(t)V'(t) = r(t) to zero: 300−20t=0300 - 20t = 0, which gives t=15t=15. Now, we evaluate the volume at the critical point and the endpoints: V(0)=10000V(0) = 10000. V(15)=10000+∫015(300−20x)dx=10000+[300x−10x2]015=10000+4500−2250=12250V(15) = 10000 + \int_{0}^{15} (300 - 20x) dx = 10000 + [300x - 10x^2]_0^{15} = 10000 + 4500 - 2250 = 12250. V(20)=10000+∫020(300−20x)dx=10000+[300x−10x2]020=10000+6000−4000=12000V(20) = 10000 + \int_{0}^{20} (300 - 20x) dx = 10000 + [300x - 10x^2]_0^{20} = 10000 + 6000 - 4000 = 12000. Comparing these values, the maximum volume of 12,250 cubic meters occurs at t=15t=15 hours.

Question 8

The temperature of a chemical substance, in degrees Celsius, is changing at a rate given by R(t)=10cos⁡(πt12)R(t) = 10 \cos(\frac{\pi t}{12}) for 0≤t≤240 \le t \le 24 hours. At t=0t=0, the temperature is 50∘C50^{\circ}C. On which open interval is the temperature of the substance decreasing but at an increasing rate?

  1. The interval (6,12)(6, 12)
  2. The interval (12,18)(12, 18) (correct answer)
  3. The interval (6,18)(6, 18)
  4. The interval (18,24)(18, 24)
Explanation: This problem tests your understanding of the relationship between a function's first and second derivatives. When you see "decreasing but at an increasing rate," you need to analyze both the sign of the derivative (for direction) and the sign of the second derivative (for concavity). Given R(t)=10cos⁡(πt12)R(t) = 10\cos(\frac{\pi t}{12}), this represents the rate of temperature change, so it's the first derivative of temperature: T′(t)=R(t)T'(t) = R(t). For temperature to be decreasing, we need T′(t)<0T'(t) < 0, which means 10cos⁡(πt12)<010\cos(\frac{\pi t}{12}) < 0. This occurs when cos⁡(πt12)<0\cos(\frac{\pi t}{12}) < 0, happening on intervals (6,18)(6, 18). For "increasing rate" of decrease, we need the second derivative T′′(t)>0T''(t) > 0. Taking the derivative: T′′(t)=R′(t)=−10⋅π12sin⁡(πt12)T''(t) = R'(t) = -10 \cdot \frac{\pi}{12} \sin(\frac{\pi t}{12}). For this to be positive, we need sin⁡(πt12)<0\sin(\frac{\pi t}{12}) < 0, which occurs on (12,24)(12, 24). Both conditions must be satisfied simultaneously: decreasing (T′(t)<0T'(t) < 0 on (6,18)(6, 18)) AND increasing rate (T′′(t)>0T''(t) > 0 on (12,24)(12, 24)). The intersection is (12,18)(12, 18), making B correct. Choice A (6,12)(6, 12) has decreasing temperature but decreasing rate of change. Choice C (6,18)(6, 18) includes the portion where the rate is still decreasing. Choice D (18,24)(18, 24) has increasing temperature. Study tip: For rate problems, always identify what the given function represents in the derivative hierarchy, then work systematically through first and second derivative conditions.

Question 9

The marginal cost of producing xx widgets is given by the function M(x)=10+500x+25M(x) = 10 + \frac{500}{x+25} dollars per widget. The total cost to produce the first 100 widgets is $5000. Which expression represents the total cost to produce 300 widgets?

  1. ∫100300M(x)dx\int_{100}^{300} M(x) dx
  2. ∫0300M(x)dx\int_{0}^{300} M(x) dx
  3. 5000+∫0100M(x)dx5000 + \int_{0}^{100} M(x) dx
  4. 5000+∫100300M(x)dx5000 + \int_{100}^{300} M(x) dx (correct answer)
Explanation: Let C(x)C(x) be the total cost to produce xx widgets. The marginal cost is the derivative of the total cost, so C′(x)=M(x)C'(x) = M(x). We are given that C(100)=5000C(100) = 5000. We want to find C(300)C(300). The net change in cost from producing 100 widgets to 300 widgets is given by the definite integral ∫100300M(x)dx\int_{100}^{300} M(x) dx. This represents the additional cost to produce the next 200 widgets after the first 100. By the Fundamental Theorem of Calculus, ∫100300M(x)dx=C(300)−C(100)\int_{100}^{300} M(x) dx = C(300) - C(100). To find the total cost for 300 widgets, we rearrange the formula: C(300)=C(100)+∫100300M(x)dxC(300) = C(100) + \int_{100}^{300} M(x) dx. Substituting the given cost, we get C(300)=5000+∫100300M(x)dxC(300) = 5000 + \int_{100}^{300} M(x) dx.

Question 10

The rate at which items are produced at a factory is given by p(t)p(t) items per hour, where tt is the number of hours since the beginning of the workday at 8:00 AM. Let A(x)=∫2xp(t)dtA(x) = \int_{2}^{x} p(t) dt. If A(5)=150A(5)=150, what is the correct interpretation of this statement?

  1. At 1:00 PM, the rate of production is 150 items per hour.
  2. The total number of items produced during the first five hours of the workday is 150.
  3. The net number of items produced between 10:00 AM and 1:00 PM is 150. (correct answer)
  4. The average rate of production between 10:00 AM and 1:00 PM is 150 items per hour.
Explanation: The function A(x)=∫2xp(t)dtA(x) = \int_{2}^{x} p(t) dt represents the net accumulation of items produced from time t=2t=2 to time t=xt=x. The variable tt represents hours since 8:00 AM. Therefore, t=2t=2 corresponds to 10:00 AM, and t=5t=5 corresponds to 1:00 PM. The statement A(5)=150A(5)=150 means that the value of the integral from t=2t=2 to t=5t=5 is 150. In context, this is the net number of items produced during the time interval from 10:00 AM to 1:00 PM.

Question 11

A car's rate of fuel consumption is given by r(s)r(s) gallons per mile, where ss is the speed of the car in miles per hour. The car's speed over time is given by the function s(t)s(t) for t≥0t \ge 0. Which integral represents the total fuel, in gallons, consumed by the car during the first two hours of a trip?

  1. ∫02r(s(t))⋅dsdtdt\int_{0}^{2} r(s(t)) \cdot \frac{ds}{dt} dt
  2. ∫02r(s(t))s(t)dt\int_{0}^{2} r(s(t)) s(t) dt (correct answer)
  3. ∫02s(t)⋅r′(s(t))dt\int_{0}^{2} s(t) \cdot r'(s(t)) dt
  4. ∫02r(s(t))s(t)dt\int_{0}^{2} \frac{r(s(t))}{s(t)} dt
Explanation: When you encounter problems involving rates and accumulation over time, think about how different quantities relate through the chain rule and units analysis. The key is tracking what each function represents and how they combine. To find total fuel consumed, you need to integrate the rate of fuel consumption over the trip. Since r(s)r(s) gives gallons per mile and you want total gallons over time, you need to account for how distance changes with time. The correct approach is option B: ∫02r(s(t))s(t)dt\int_{0}^{2} r(s(t)) s(t) dt. Here's why: r(s(t))r(s(t)) gives the fuel consumption rate in gallons per mile at the current speed, while s(t)s(t) gives the speed in miles per hour. Multiplying these gives gallons per hour (since gallonsmile×mileshour=gallonshour\frac{\text{gallons}}{\text{mile}} \times \frac{\text{miles}}{\text{hour}} = \frac{\text{gallons}}{\text{hour}}). Integrating this rate over 2 hours yields total gallons consumed. Option A uses dsdt\frac{ds}{dt} (acceleration), which would give you something involving gallons per unit time squared - not what we want. Option C involves r′(s(t))r'(s(t)), the derivative of the consumption rate, which doesn't directly relate to total fuel consumed. Option D divides by speed, giving you gallons/milemiles/hour=gallons⋅hoursmile2\frac{\text{gallons/mile}}{\text{miles/hour}} = \frac{\text{gallons} \cdot \text{hours}}{\text{mile}^2}, which has incorrect units. Strategy tip: In rate problems, always check units! The integrand should have units that, when multiplied by the differential's units, give you the units you want for the final answer.

Question 12

Let f(t)f(t) be a continuous function and define G(x)=∫3xf(t)dtG(x) = \int_3^x f(t) dt. Which of the following statements is always true?

  1. G(x)G(x) is the antiderivative of f(x)f(x).
  2. G(3)=f(3)G(3) = f(3)
  3. G′(x)=f(x)G'(x) = f(x) (correct answer)
  4. G(0)=3G(0) = 3
Explanation: This question directly tests the Fundamental Theorem of Calculus, Part 1. The theorem states that if G(x)=∫axf(t)dtG(x) = \int_a^x f(t) dt, then G′(x)=f(x)G'(x) = f(x). Choice A is incorrect because G(x)G(x) is an antiderivative, but not necessarily the antiderivative; antiderivatives can differ by a constant. Choice B is incorrect; by definition, G(3)=∫33f(t)dt=0G(3) = \int_3^3 f(t) dt = 0. Choice D makes an unsupported assumption.

Question 13

An oil tanker is leaking oil at a rate of R(t)=100−t2R(t) = 100 - t^2 barrels per hour for 0≤t≤100 \le t \le 10. A clean-up crew removes the oil at a constant rate of 64 barrels per hour. Assuming there was no oil in the water at t=0t=0, what is the maximum amount of oil in the water, in barrels?

  1. 6
  2. 36
  3. 144 (correct answer)
  4. 528
Explanation: The net rate of change of the amount of oil is A′(t)=R(t)−64=(100−t2)−64=36−t2A'(t) = R(t) - 64 = (100 - t^2) - 64 = 36 - t^2. To find the maximum amount, we find critical points by setting A′(t)=0A'(t) = 0, which gives 36−t2=036 - t^2 = 0, so t=6t=6. Since A′(t)>0A'(t) > 0 for t<6t<6 and A′(t)<0A'(t) < 0 for t>6t>6, the maximum amount occurs at t=6t=6. The maximum amount is the total accumulation up to that time: A(6)=∫06(36−t2)dt=[36t−t33]06=36(6)−633=216−2163=216−72=144A(6) = \int_0^6 (36 - t^2) dt = [36t - \frac{t^3}{3}]_0^6 = 36(6) - \frac{6^3}{3} = 216 - \frac{216}{3} = 216 - 72 = 144 barrels.

Question 14

If F′(x)=e−x2F'(x) = e^{-x^2} and F(2)=7F(2) = 7, which of the following is an expression for F(5)F(5)?

  1. 7+∫52e−t2dt7 + \int_5^2 e^{-t^2} dt
  2. ∫25e−t2dt\int_2^5 e^{-t^2} dt
  3. 7+∫25e−t2dt7 + \int_2^5 e^{-t^2} dt (correct answer)
  4. 7−∫25e−x2dx7 - \int_2^5 e^{-x^2} dx
Explanation: By the Net Change Theorem, ∫25F′(t)dt=F(5)−F(2)\int_2^5 F'(t) dt = F(5) - F(2). Substituting the given information, we have ∫25e−t2dt=F(5)−7\int_2^5 e^{-t^2} dt = F(5) - 7. To solve for F(5)F(5), we add 7 to both sides: F(5)=7+∫25e−t2dtF(5) = 7 + \int_2^5 e^{-t^2} dt. The variable of integration can be any letter, so using xx or tt doesn't change the value.

Question 15

A factory produces widgets at a rate of P(t)=100+5sin⁡(2πt)P(t) = 100 + 5\sin(2\pi t) widgets per day, where tt is in days. Which expression represents the average daily production of widgets during the first 10 days?

  1. ∫010(100+5sin⁡(2πt))dt\int_0^{10} (100 + 5\sin(2\pi t)) dt
  2. 100+5sin⁡(20π)100 + 5\sin(20\pi)
  3. 110∫010(100+5sin⁡(2πt))dt\frac{1}{10} \int_0^{10} (100 + 5\sin(2\pi t)) dt (correct answer)
  4. 12(P(10)+P(0))\frac{1}{2} (P(10) + P(0))
Explanation: The average value of a function f(x)f(x) on an interval [a,b][a, b] is given by 1b−a∫abf(x)dx\frac{1}{b-a} \int_a^b f(x) dx. In this context, the function is the production rate P(t)P(t) and the interval is [0,10][0, 10]. Therefore, the average daily production is 110−0∫010P(t)dt=110∫010(100+5sin⁡(2πt))dt\frac{1}{10-0} \int_0^{10} P(t) dt = \frac{1}{10} \int_0^{10} (100 + 5\sin(2\pi t)) dt.

Question 16

Sand is added to a beach at a rate of S(t)=60−2tS(t) = 60 - 2t tons per hour. The tide removes sand at a rate of R(t)=15+sin⁡(πt6)R(t) = 15 + \sin(\frac{\pi t}{6}) tons per hour, for 0≤t≤60 \le t \le 6. Let A(t)A(t) be the total amount of sand on the beach. At t=3t=3, which of the following is true?

  1. The amount of sand is increasing, and the rate of accumulation is decreasing. (correct answer)
  2. The amount of sand is increasing, and the rate of accumulation is increasing.
  3. The amount of sand is decreasing, and the rate of accumulation is decreasing.
  4. The amount of sand is decreasing, and the rate of accumulation is increasing.
Explanation: The net rate of change of the amount of sand is A′(t)=S(t)−R(t)=(60−2t)−(15+sin⁡(πt6))=45−2t−sin⁡(πt6)A'(t) = S(t) - R(t) = (60 - 2t) - (15 + \sin(\frac{\pi t}{6})) = 45 - 2t - \sin(\frac{\pi t}{6}). At t=3t=3, A′(3)=45−6−sin⁡(π2)=39−1=38A'(3) = 45 - 6 - \sin(\frac{\pi}{2}) = 39 - 1 = 38. Since A′(3)>0A'(3) > 0, the amount of sand is increasing. To determine if the rate of accumulation is increasing or decreasing, we check the sign of A′′(t)A''(t). A′′(t)=−2−π6cos⁡(πt6)A''(t) = -2 - \frac{\pi}{6}\cos(\frac{\pi t}{6}). At t=3t=3, A′′(3)=−2−π6cos⁡(π2)=−2−0=−2A''(3) = -2 - \frac{\pi}{6}\cos(\frac{\pi}{2}) = -2 - 0 = -2. Since A′′(3)<0A''(3) < 0, the rate of accumulation is decreasing.

Question 17

A company's revenue is generated at a rate of R(t)R(t) dollars per month and its costs accrue at a rate of C(t)C(t) dollars per month, where tt is the number of months from the start of the year. What does the quantity ∫36(R(t)−C(t))dt\int_3^6 (R(t) - C(t)) dt represent?

  1. The average rate of profit, in dollars per month, from the end of month 3 to the end of month 6.
  2. The total change in profit from the end of month 3 to the end of month 6. (correct answer)
  3. The total profit earned by the company at the end of month 6.
  4. The rate of change of profit, in dollars per month, at the end of month 6.
Explanation: The expression R(t)−C(t)R(t) - C(t) represents the rate of change of profit, P′(t)P'(t). By the Net Change Theorem, the integral of a rate of change over an interval gives the net change in the quantity over that interval. Therefore, ∫36(R(t)−C(t))dt=∫36P′(t)dt=P(6)−P(3)\int_3^6 (R(t) - C(t)) dt = \int_3^6 P'(t) dt = P(6) - P(3), which is the total change in profit from the end of month 3 to the end of month 6.

Question 18

A reservoir initially contains 1000 cubic meters of water. For the first six hours of a day (0≤t≤60 \le t \le 6), the rate of change of the volume of water is given by V′(t)=20−t2V'(t) = 20 - t^2 cubic meters per hour. What is the total volume of water in the reservoir at t=4t=4 hours?

  1. 31763\frac{3176}{3} cubic meters (correct answer)
  2. 1763\frac{176}{3} cubic meters
  3. 1016 cubic meters
  4. 4 cubic meters
Explanation: The total volume at time t=4t=4 is the initial volume plus the net change from t=0t=0 to t=4t=4. The net change is the definite integral of the rate function. V(4)=V(0)+∫04V′(t)dt=1000+∫04(20−t2)dtV(4) = V(0) + \int_0^4 V'(t) dt = 1000 + \int_0^4 (20 - t^2) dt. The integral evaluates to [20t−t33]04=(20(4)−433)−0=80−643=1763[20t - \frac{t^3}{3}]_0^4 = (20(4) - \frac{4^3}{3}) - 0 = 80 - \frac{64}{3} = \frac{176}{3}. The total volume is 1000+1763=3000+1763=317631000 + \frac{176}{3} = \frac{3000+176}{3} = \frac{3176}{3}.

Question 19

A chemical reaction produces a substance whose amount is A(t)A(t). The rate of production is P(t)=12−t2P(t) = 12 - t^2 grams per minute for 0≤t≤50 \le t \le 5. Simultaneously, the substance is consumed at a constant rate of 3 grams per minute. At what time tt on the interval 0≤t≤50 \le t \le 5 is the amount of the substance at its absolute maximum?

  1. t=23t = 2\sqrt{3}
  2. t=5t = 5
  3. t=3t = 3 (correct answer)
  4. t=0t = 0
Explanation: Let A(t)A(t) be the amount of the substance. The net rate of change of the amount is A′(t)=P(t)−3=(12−t2)−3=9−t2A'(t) = P(t) - 3 = (12 - t^2) - 3 = 9 - t^2. To find the maximum amount, we find critical points by setting A′(t)=0A'(t) = 0, which gives 9−t2=09 - t^2 = 0, so t=3t=3 (since t≥0t \ge 0). We must check the sign of A′(t)A'(t). For 0≤t<30 \le t < 3, A′(t)>0A'(t) > 0, so the amount is increasing. For 3<t≤53 < t \le 5, A′(t)<0A'(t) < 0, so the amount is decreasing. Therefore, the absolute maximum occurs at t=3t=3.

Question 20

A particle moves along the x-axis with velocity v(t)=cos⁡(πt)v(t) = \cos(\pi t). The particle is at position x=5x=5 at time t=0t=0. What is the correct interpretation of the expression 5+∫02cos⁡(πt)dt5 + \int_0^2 \cos(\pi t) dt?

  1. The particle's displacement from t=0t=0 to t=2t=2.
  2. The particle's total distance traveled from t=0t=0 to t=2t=2.
  3. The particle's velocity at t=2t=2.
  4. The particle's position at t=2t=2. (correct answer)
Explanation: The position of the particle at time tt, denoted x(t)x(t), is given by the initial position plus the net change in position (displacement). The displacement from t=0t=0 to t=2t=2 is ∫02v(t)dt=∫02cos⁡(πt)dt\int_0^2 v(t) dt = \int_0^2 \cos(\pi t) dt. Therefore, the expression x(0)+∫02v(t)dt=5+∫02cos⁡(πt)dtx(0) + \int_0^2 v(t) dt = 5 + \int_0^2 \cos(\pi t) dt represents the particle's position at t=2t=2.