Calculus 2 Quiz: Accumulation Functions In Context
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Accumulation Functions In ContextQuestion 1 of 7

A particle moves with acceleration a(t)=6t−12a(t) = 6t - 12 m/s² starting from rest at the origin. The particle's kinetic energy is given by KE=12mv2KE = \frac{1}{2}mv^2 where m=2m = 2 kg. At what time does the particle first return to having zero kinetic energy after t=0t = 0?

t=2t = 2 seconds
t=4t = 4 seconds
t=6t = 6 seconds
The particle never returns to zero kinetic energy
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Calculus 2 Quiz

Calculus 2 Quiz: Accumulation Functions In Context

Practice Accumulation Functions In Context in Calculus 2 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 Accumulation Functions In Context, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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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.

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Question 1

A particle moves with acceleration a(t)=6t−12a(t) = 6t - 12 m/s² starting from rest at the origin. The particle's kinetic energy is given by KE=12mv2KE = \frac{1}{2}mv^2 where m=2m = 2 kg. At what time does the particle first return to having zero kinetic energy after t=0t = 0?

  1. t=2t = 2 seconds
  2. t=4t = 4 seconds (correct answer)
  3. t=6t = 6 seconds
  4. The particle never returns to zero kinetic energy
Explanation: The correct answer is B. Starting from rest, v(t)=∫0ta(τ)dτ=∫0t(6τ−12)dτ=3t2−12tv(t) = \int_0^t a(\tau) d\tau = \int_0^t (6\tau - 12) d\tau = 3t^2 - 12t. Kinetic energy is zero when velocity is zero: 3t2−12t=03t^2 - 12t = 0, so 3t(t−4)=03t(t - 4) = 0. This gives t=0t = 0 (initial condition) or t=4t = 4. The particle first returns to zero kinetic energy at t=4t = 4 seconds. Choice A gives the time when acceleration is zero, not velocity. Choice C might result from incorrectly setting acceleration equal to zero instead of velocity. Choice D fails to recognize that the particle can return to rest.

Question 2

The rate of change of atmospheric pressure with respect to altitude is dPdh=−0.12P\frac{dP}{dh} = -0.12P where PP is pressure in atmospheres and hh is altitude in kilometers. If pressure is 1 atmosphere at sea level, the average pressure between sea level and 5 km altitude is:

  1. 15∫05e−0.12hdh\frac{1}{5}\int_0^5 e^{-0.12h} dh (correct answer)
  2. 15∫05(1−0.12h)dh\frac{1}{5}\int_0^5 (1 - 0.12h) dh
  3. e−0.6−1−0.6\frac{e^{-0.6} - 1}{-0.6}
  4. e−0.6e^{-0.6}
Explanation: The correct answer is A. The differential equation dPdh=−0.12P\frac{dP}{dh} = -0.12P has solution P(h)=e−0.12hP(h) = e^{-0.12h} (using initial condition P(0)=1P(0) = 1). The average value of a function over [a,b][a,b] is 1b−a∫abf(x)dx\frac{1}{b-a}\int_a^b f(x) dx. Therefore, average pressure is 15∫05e−0.12hdh\frac{1}{5}\int_0^5 e^{-0.12h} dh. Choice B incorrectly assumes linear change rather than exponential decay. Choice C gives the result of evaluating the integral but not the average (missing the 15\frac{1}{5} factor). Choice D gives the pressure at 5 km, not the average pressure.

Question 3

A manufacturing plant's energy consumption rate varies cyclically throughout a 24-hour period according to the function E(t)=800+400cos⁡(πt12)E(t) = 800 + 400\cos(\frac{\pi t}{12}) kilowatts, where tt is hours after midnight. The plant operates continuously, and energy costs vary by time of day: $0.10 per kWh from midnight to 6 AM, $0.15 per kWh from 6 AM to 6 PM, and $0.12 per kWh from 6 PM to midnight.

What expression represents the total energy cost for one complete day of operation?

  1. 0.10∫06E(t)dt+0.15∫618E(t)dt+0.12∫1824E(t)dt0.10\int_0^6 E(t) dt + 0.15\int_6^{18} E(t) dt + 0.12\int_{18}^{24} E(t) dt (correct answer)
  2. ∫024E(t)⋅C(t)dt\int_0^{24} E(t) \cdot C(t) dt where C(t)C(t) is the piecewise cost function
  3. 0.10⋅6⋅E(3)+0.15⋅12⋅E(12)+0.12⋅6⋅E(21)0.10 \cdot 6 \cdot E(3) + 0.15 \cdot 12 \cdot E(12) + 0.12 \cdot 6 \cdot E(21)
  4. (0.10+0.15+0.12)∫024E(t)dt(0.10 + 0.15 + 0.12)\int_0^{24} E(t) dt
Explanation: The correct answer is A. Energy cost equals rate × time × unit cost. Since consumption rate E(t)E(t) varies continuously and cost per kWh changes at specific times, we integrate E(t)E(t) over each time period and multiply by the corresponding rate. Choice B is conceptually equivalent but less explicit about the piecewise nature. Choice C incorrectly uses average values instead of integrating the continuous consumption. Choice D incorrectly adds the rates instead of applying them to their respective time periods.

Question 4

The rate at which heat flows through a window varies throughout the day according to H(t)=50+30sin⁡(πt12)H(t) = 50 + 30\sin(\frac{\pi t}{12}) BTU per hour, where tt is hours after sunrise. If the cumulative heat transfer from sunrise to hour TT must not exceed 400 BTU, what is the latest time TT (in hours after sunrise) this condition can be satisfied?

  1. TT such that ∫0T[50+30sin⁡(πt12)]dt=400\int_0^T [50 + 30\sin(\frac{\pi t}{12})] dt = 400
  2. TT such that 50T+30sin⁡(πT12)=40050T + 30\sin(\frac{\pi T}{12}) = 400
  3. TT such that 50+30sin⁡(πT12)≤400T50 + 30\sin(\frac{\pi T}{12}) \leq \frac{400}{T}
  4. TT such that ∫0T[50+30sin⁡(πt12)]dt≤400\int_0^T [50 + 30\sin(\frac{\pi t}{12})] dt \leq 400 and equality holds (correct answer)
Explanation: The correct answer is D. The cumulative heat transfer from sunrise to hour T is given by the integral ∫0TH(t)dt\int_0^T H(t) dt. We want the latest time T such that this doesn't exceed 400 BTU, which means we want the largest T where the integral equals 400 (the boundary of the constraint). Choice A is mathematically equivalent but doesn't capture that we want the latest such time. Choice B confuses the rate function with the cumulative amount. Choice C incorrectly relates the instantaneous rate to the average rate constraint.

Question 5

The velocity of oil leaking from a tank is v(t)=60t+4v(t) = \frac{60}{\sqrt{t+4}} gallons per minute for t≥0t \geq 0 minutes. Environmental regulations require that if more than 200 gallons leak in any 8-minute period, cleanup protocols must be initiated. What is the minimum value of t0t_0 such that cleanup protocols are required for the interval [t0,t0+8][t_0, t_0 + 8]?

  1. t0t_0 such that ∫t0t0+860t+4dt=200\int_{t_0}^{t_0+8} \frac{60}{\sqrt{t+4}} dt = 200 (correct answer)
  2. t0=0t_0 = 0 since leakage is highest initially
  3. t0t_0 such that 60t0+4=25\frac{60}{\sqrt{t_0+4}} = 25
  4. No cleanup protocols are needed since ∫0860t+4dt<200\int_0^8 \frac{60}{\sqrt{t+4}} dt < 200
Explanation: The correct answer is A. We need to find when the total leakage over an 8-minute period first exceeds 200 gallons. Since v(t)=60t+4v(t) = \frac{60}{\sqrt{t+4}} is decreasing (larger denominators give smaller values), the maximum leakage occurs in the earliest 8-minute interval [0,8][0,8]. However, the question asks for the minimum t0t_0 where cleanup is required, meaning we want the earliest time where the 8-minute accumulation equals 200. Choice B assumes this occurs at t0=0t_0 = 0 without verification. Choice C confuses instantaneous rate with accumulated amount. Choice D makes an assumption about the integral value without calculation.

Question 6

A population of bacteria grows at a rate proportional to P(t)=1000+200cos⁡(πt12)P(t) = 1000 + 200\cos(\frac{\pi t}{12}) bacteria per hour, where tt is time in hours after midnight. If there are 5000 bacteria at midnight, approximately how many bacteria will there be at 6 AM if the proportionality constant is 0.02?

  1. 5000+0.02∫06[1000+200cos⁡(πt12)]dt5000 + 0.02\int_0^6 [1000 + 200\cos(\frac{\pi t}{12})] dt
  2. 5000e0.02∫06[1000+200cos⁡(πt12)]dt5000e^{0.02\int_0^6 [1000 + 200\cos(\frac{\pi t}{12})] dt}
  3. 5000+∫060.02[1000+200cos⁡(πt12)]dt5000 + \int_0^6 0.02[1000 + 200\cos(\frac{\pi t}{12})] dt (correct answer)
  4. 5000e0.02⋅6[1000+200cos⁡(π⋅312)]5000e^{0.02 \cdot 6[1000 + 200\cos(\frac{\pi \cdot 3}{12})]}
Explanation: The correct answer is C. The phrase 'rate proportional to P(t)' means the actual growth rate is 0.02⋅P(t)=0.02[1000+200cos⁡(πt12)]0.02 \cdot P(t) = 0.02[1000 + 200\cos(\frac{\pi t}{12})] bacteria per hour. The population is the initial amount plus the accumulation of this rate over 6 hours. Choice A omits the proportionality constant from inside the integral. Choice B incorrectly uses exponential growth (which would apply if the rate were proportional to the current population). Choice D incorrectly evaluates the trigonometric function at a single point rather than integrating over the interval.

Question 7

A reservoir's water level changes according to the net inflow rate N(t)=100cos⁡(πt6)−20N(t) = 100\cos(\frac{\pi t}{6}) - 20 cubic feet per minute, where tt is time in minutes. Starting with 5000 cubic feet at t=0t = 0, during which time interval is the water level increasing most rapidly?

  1. When N(t)N(t) is maximized (correct answer)
  2. When N′(t)N'(t) is maximized
  3. When ∫0tN(τ)dτ\int_0^t N(\tau) d\tau is maximized
  4. When ∣N(t)∣|N(t)| is maximized
Explanation: The correct answer is A. The water level at time t is W(t)=5000+∫0tN(τ)dτW(t) = 5000 + \int_0^t N(\tau) d\tau. The rate of change of water level is W′(t)=N(t)W'(t) = N(t) by the Fundamental Theorem of Calculus. Therefore, the water level increases most rapidly when N(t)N(t) is maximized. Choice B finds when the rate of change is accelerating most, not when it's fastest. Choice C finds when the total accumulated change is greatest, not the instantaneous rate. Choice D considers both positive and negative rates equally, missing that we want increasing (positive rate).