Accumulation Functions in Context - Calculus 2

1. A particle moves with acceleration $a(t) = 6t - 12$ m/s² starting from rest at the origin. The particle's kinetic energy is given by $KE = \frac{1}{2}mv^2$ where $m = 2$ kg. At what time does the particle first return to having zero kinetic energy after $t = 0$?
   Answer: $t = 4$ seconds The correct answer is B. Starting from rest, $v(t) = \int_0^t a(\tau) d\tau = \int_0^t (6\tau - 12) d\tau = 3t^2 - 12t$. Kinetic energy is zero when velocity is zero: $3t^2 - 12t = 0$, so $3t(t - 4) = 0$. This gives $t = 0$ (initial condition) or $t = 4$. The particle first returns to zero kinetic energy at $t = 4$ 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.
2. A reservoir's water level changes according to the net inflow rate $N(t) = 100\cos(\frac{\pi t}{6}) - 20$ cubic feet per minute, where $t$ is time in minutes. Starting with 5000 cubic feet at $t = 0$, during which time interval is the water level increasing most rapidly?
   Answer: When $N(t)$ is maximized The correct answer is A. The water level at time t is $W(t) = 5000 + \int_0^t N(\tau) d\tau$. The rate of change of water level is $W'(t) = N(t)$ by the Fundamental Theorem of Calculus. Therefore, the water level increases most rapidly when $N(t)$ 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).
3. The rate of change of atmospheric pressure with respect to altitude is $\frac{dP}{dh} = -0.12P$ where $P$ is pressure in atmospheres and $h$ is altitude in kilometers. If pressure is 1 atmosphere at sea level, the average pressure between sea level and 5 km altitude is:
   Answer: $\frac{1}{5}\int_0^5 e^{-0.12h} dh$ The correct answer is A. The differential equation $\frac{dP}{dh} = -0.12P$ has solution $P(h) = e^{-0.12h}$ (using initial condition $P(0) = 1$). The average value of a function over $[a,b]$ is $\frac{1}{b-a}\int_a^b f(x) dx$. Therefore, average pressure is $\frac{1}{5}\int_0^5 e^{-0.12h} dh$. 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 $\frac{1}{5}$ factor). Choice D gives the pressure at 5 km, not the average pressure.
4. The rate at which heat flows through a window varies throughout the day according to $H(t) = 50 + 30\sin(\frac{\pi t}{12})$ BTU per hour, where $t$ is hours after sunrise. If the cumulative heat transfer from sunrise to hour $T$ must not exceed 400 BTU, what is the latest time $T$ (in hours after sunrise) this condition can be satisfied?
   Answer: $T$ such that $\int_0^T [50 + 30\sin(\frac{\pi t}{12})] dt \leq 400$ and equality holds The correct answer is D. The cumulative heat transfer from sunrise to hour T is given by the integral $\int_0^T H(t) dt$. 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.
5. A population of bacteria grows at a rate proportional to $P(t) = 1000 + 200\cos(\frac{\pi t}{12})$ bacteria per hour, where $t$ 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?
   Answer: $5000 + \int_0^6 0.02[1000 + 200\cos(\frac{\pi t}{12})] dt$ The correct answer is C. The phrase 'rate proportional to P(t)' means the actual growth rate is $0.02 \cdot P(t) = 0.02[1000 + 200\cos(\frac{\pi t}{12})]$ 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.
6. What expression represents the total energy cost for one complete day of operation?
   Answer: $0.10\int_0^6 E(t) dt + 0.15\int_6^{18} E(t) dt + 0.12\int_{18}^{24} E(t) dt$ The correct answer is A. Energy cost equals rate × time × unit cost. Since consumption rate $E(t)$ varies continuously and cost per kWh changes at specific times, we integrate $E(t)$ 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.
7. The velocity of oil leaking from a tank is $v(t) = \frac{60}{\sqrt{t+4}}$ gallons per minute for $t \geq 0$ 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 $t_0$ such that cleanup protocols are required for the interval $[t_0, t_0 + 8]$?
   Answer: $t_0$ such that $\int_{t_0}^{t_0+8} \frac{60}{\sqrt{t+4}} dt = 200$ The correct answer is A. We need to find when the total leakage over an 8-minute period first exceeds 200 gallons. Since $v(t) = \frac{60}{\sqrt{t+4}}$ is decreasing (larger denominators give smaller values), the maximum leakage occurs in the earliest 8-minute interval $[0,8]$. However, the question asks for the minimum $t_0$ where cleanup is required, meaning we want the earliest time where the 8-minute accumulation equals 200. Choice B assumes this occurs at $t_0 = 0$ without verification. Choice C confuses instantaneous rate with accumulated amount. Choice D makes an assumption about the integral value without calculation.

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