Accumulation Functions in Context - Calculus 1

1. The population of a certain species of fish in a lake is changing at a rate of $r(t)$ fish per month. At time $t=3$ months, the population is measured to be 1,500 fish. Which expression represents the fish population at time $t=12$ months?
   Answer: $1500 + \int_{3}^{12} r(t) dt$ Let $P(t)$ be the population of fish at time $t$. We are given that $P'(t) = r(t)$ and $P(3) = 1500$. According to the Fundamental Theorem of Calculus, the net change in population from $t=3$ to $t=12$ is $\int_{3}^{12} r(t) dt = P(12) - P(3)$. To find the population at $t=12$, we can rearrange this equation: $P(12) = P(3) + \int_{3}^{12} r(t) dt$. Substituting the known value $P(3) = 1500$, we get $P(12) = 1500 + \int_{3}^{12} r(t) dt$.
2. Sand is added to a beach at a rate of $S(t) = 60 - 2t$ tons per hour. The tide removes sand at a rate of $R(t) = 15 + \sin(\frac{\pi t}{6})$ tons per hour, for $0 \le t \le 6$. Let $A(t)$ be the total amount of sand on the beach. At $t=3$, which of the following is true?
   Answer: The amount of sand is increasing, and the rate of accumulation is decreasing. The net rate of change of the amount of sand is $A'(t) = S(t) - R(t) = (60 - 2t) - (15 + \sin(\frac{\pi t}{6})) = 45 - 2t - \sin(\frac{\pi t}{6})$. At $t=3$, $A'(3) = 45 - 6 - \sin(\frac{\pi}{2}) = 39 - 1 = 38$. Since $A'(3) > 0$, the amount of sand is increasing. To determine if the rate of accumulation is increasing or decreasing, we check the sign of $A''(t)$. $A''(t) = -2 - \frac{\pi}{6}\cos(\frac{\pi t}{6})$. At $t=3$, $A''(3) = -2 - \frac{\pi}{6}\cos(\frac{\pi}{2}) = -2 - 0 = -2$. Since $A''(3) < 0$, the rate of accumulation is decreasing.
3. A chemical reaction produces a substance whose amount is $A(t)$. The rate of production is $P(t) = 12 - t^2$ grams per minute for $0 \le t \le 5$. Simultaneously, the substance is consumed at a constant rate of 3 grams per minute. At what time $t$ on the interval $0 \le t \le 5$ is the amount of the substance at its absolute maximum?
   Answer: $t = 3$ Let $A(t)$ be the amount of the substance. The net rate of change of the amount is $A'(t) = P(t) - 3 = (12 - t^2) - 3 = 9 - t^2$. To find the maximum amount, we find critical points by setting $A'(t) = 0$, which gives $9 - t^2 = 0$, so $t=3$ (since $t \ge 0$). We must check the sign of $A'(t)$. For $0 \le t < 3$, $A'(t) > 0$, so the amount is increasing. For $3 < t \le 5$, $A'(t) < 0$, so the amount is decreasing. Therefore, the absolute maximum occurs at $t=3$.
4. The marginal cost of producing $x$ widgets is given by the function $M(x) = 10 + \frac{500}{x+25}$ dollars per widget. The total cost to produce the first 100 widgets is $5000. Which expression represents the total cost to produce 300 widgets?
   Answer: $5000 + \int_{100}^{300} M(x) dx$ Let $C(x)$ be the total cost to produce $x$ widgets. The marginal cost is the derivative of the total cost, so $C'(x) = M(x)$. We are given that $C(100) = 5000$. We want to find $C(300)$. The net change in cost from producing 100 widgets to 300 widgets is given by the definite integral $\int_{100}^{300} M(x) dx$. This represents the additional cost to produce the next 200 widgets after the first 100. By the Fundamental Theorem of Calculus, $\int_{100}^{300} M(x) dx = C(300) - C(100)$. To find the total cost for 300 widgets, we rearrange the formula: $C(300) = C(100) + \int_{100}^{300} M(x) dx$. Substituting the given cost, we get $C(300) = 5000 + \int_{100}^{300} M(x) dx$.
5. A population of insects increases at a rate of $R(t) = 200 + 6t^2$ insects per day. If the initial population at $t=0$ is 1000, what is the population at $t=3$ days?
   Answer: 1654 The population at $t=3$ is the initial population plus the total increase from $t=0$ to $t=3$. The total increase is the integral of the rate: $\int_0^3 (200 + 6t^2) dt = [200t + 2t^3]_0^3 = (200(3) + 2(3^3)) - 0 = 600 + 2(27) = 600 + 54 = 654$. The total population is $1000 + 654 = 1654$.
6. Let $G(x)$ be an accumulation function defined by $G(x) = \int_1^x (t^2 - 8t + 12) dt$. For what open interval of $x$ values is the graph of $G(x)$ concave down?
   Answer: $(-\infty, 4)$ The concavity of $G(x)$ is determined by the sign of its second derivative, $G''(x)$. Using the Fundamental Theorem of Calculus, $G'(x) = x^2 - 8x + 12$. Differentiating again, we get $G''(x) = 2x - 8$. The graph of $G(x)$ is concave down when $G''(x) < 0$. So, we solve $2x - 8 < 0$, which gives $2x < 8$, or $x < 4$. This corresponds to the interval $(-\infty, 4)$.
7. Let $A(t)$ be the amount of a certain drug in a patient's bloodstream, in milligrams, $t$ hours after an injection. The rate of absorption of the drug is $R(t)$ mg/hour, while the rate at which the drug is eliminated is $E(t)$ mg/hour. If $\int_0^4 (R(t) - E(t)) dt < 0$, what can be concluded?
   Answer: The amount of the drug in the bloodstream is lower at $t=4$ than at $t=0$. The net rate of change of the drug in the bloodstream is $A'(t) = R(t) - E(t)$. The integral $\int_0^4 (R(t) - E(t)) dt$ represents the net change in the amount of the drug from $t=0$ to $t=4$. If this value is negative, it means $A(4) - A(0) < 0$, which implies $A(4) < A(0)$. Therefore, the amount of the drug at $t=4$ is less than the amount at $t=0$.
8. A tank initially contains $W_0$ gallons of water. For $t \ge 0$, water is pumped into the tank at a constant rate of $C$ gallons per minute. Simultaneously, water leaks out of the tank at a rate of $L(t)$ gallons per minute, where $L(t)$ is a differentiable function. Which of the following expressions represents the amount of water in the tank at time $T$?
   Answer: $W_0 + \int_{0}^{T} (C - L(t)) dt$ Let $W(t)$ be the amount of water in the tank at time $t$. The rate of change of the amount of water is the rate in minus the rate out. So, $W'(t) = C - L(t)$. To find the total amount of water at time $T$, we must add the initial amount, $W_0$, to the net change in water from $t=0$ to $t=T$. The net change is the integral of the net rate of change. Therefore, the amount of water at time $T$ is given by $W(T) = W(0) + \int_{0}^{T} W'(t) dt = W_0 + \int_{0}^{T} (C - L(t)) dt$.
9. A population of bacteria is 10,000 at time $t=0$. The population's growth rate in bacteria per hour is modeled by $g(t) = 100t - 50t^2$ for $t \ge 0$. At what time $t > 0$ does the population first return to its initial size of 10,000?
   Answer: $t=3$ The population returns to its initial size when the net change in population is zero. We need to find $t > 0$ such that $\int_0^t g(\tau) d\tau = 0$. We calculate the integral: $\int_0^t (100\tau - 50\tau^2) d\tau = [50\tau^2 - \frac{50}{3}\tau^3]_0^t = 50t^2 - \frac{50}{3}t^3$. Setting this to zero gives $50t^2(1 - \frac{t}{3}) = 0$. The solutions are $t=0$ and $t=3$. Since the question asks for $t>0$, the answer is $t=3$.
10. A water tank contains 80 gallons of water at time $t=0$. For the next 4 hours, water is pumped out of the tank at a rate of $P(t) = 5t+3$ gallons per hour. How much water, in gallons, is in the tank at time $t=4$ hours?
    Answer: 28 The amount of water at $t=4$ is the initial amount minus the total amount pumped out from $t=0$ to $t=4$. The total amount pumped out is the integral of the rate: $\int_0^4 (5t+3) dt = [\frac{5}{2}t^2 + 3t]_0^4 = (\frac{5}{2}(4^2) + 3(4)) - 0 = (\frac{5}{2}(16) + 12) = 40 + 12 = 52$ gallons. The amount remaining in the tank is $80 - 52 = 28$ gallons.
11. The rate at which items are produced at a factory is given by $p(t)$ items per hour, where $t$ is the number of hours since the beginning of the workday at 8:00 AM. Let $A(x) = \int_{2}^{x} p(t) dt$. If $A(5)=150$, what is the correct interpretation of this statement?
    Answer: The net number of items produced between 10:00 AM and 1:00 PM is 150. The function $A(x) = \int_{2}^{x} p(t) dt$ represents the net accumulation of items produced from time $t=2$ to time $t=x$. The variable $t$ represents hours since 8:00 AM. Therefore, $t=2$ corresponds to 10:00 AM, and $t=5$ corresponds to 1:00 PM. The statement $A(5)=150$ means that the value of the integral from $t=2$ to $t=5$ is 150. In context, this is the net number of items produced during the time interval from 10:00 AM to 1:00 PM.
12. A company's total savings $S(t)$ is modeled by an accumulation function $S(t) = \int_0^t (50 - 2^\tau) d\tau$, where $t$ is in months. At what time $t$ does the company's savings stop increasing and start decreasing?
    Answer: $\log_2(50)$ When you encounter an accumulation function like $S(t) = \int_0^t (50 - 2^\tau) d\tau$, you're looking at a scenario where the rate of change is given by the integrand. To find when savings stop increasing and start decreasing, you need to determine when the rate of savings becomes zero. The rate of change of savings is $S'(t) = 50 - 2^t$ (by the Fundamental Theorem of Calculus). The company's savings increase when $S'(t) > 0$ and decrease when $S'(t) < 0$. The transition point occurs when $S'(t) = 0$. Setting the derivative equal to zero: $50 - 2^t = 0$, which gives us $2^t = 50$. Taking the logarithm base 2 of both sides: $t = \log_2(50)$. This is answer choice D. Now let's examine why the other answers are incorrect. Choice A (50) comes from mistakenly thinking the answer is simply the constant term in the integrand. Choice B (25) likely results from incorrectly dividing 50 by 2, perhaps confusing this with a linear function. Choice C ($\sqrt{50}$) might arise from taking the square root instead of the logarithm, possibly confusing exponential and quadratic relationships. Remember this key strategy: when dealing with accumulation functions, the derivative tells you the instantaneous rate of change. Critical points where this rate equals zero indicate where the function changes from increasing to decreasing (or vice versa). Always differentiate first, then solve for when that derivative equals zero.
13. Let $f(t)$ be a continuous function and define $G(x) = \int_3^x f(t) dt$. Which of the following statements is always true?
    Answer: $G'(x) = f(x)$ This question directly tests the Fundamental Theorem of Calculus, Part 1. The theorem states that if $G(x) = \int_a^x f(t) dt$, then $G'(x) = f(x)$. Choice A is incorrect because $G(x)$ is an antiderivative, but not necessarily the antiderivative; antiderivatives can differ by a constant. Choice B is incorrect; by definition, $G(3) = \int_3^3 f(t) dt = 0$. Choice D makes an unsupported assumption.
14. The population of a beehive is changing at a rate of $N'(t) = 100\cos(t) + 50$ bees per week over a 10-week period. If the initial population is $N(0) = 5000$, during which of the following intervals is the population of bees decreasing?
    Answer: $\pi/2 < t < 3\pi/2$ The population is decreasing when its rate of change, $N'(t)$, is negative. We need to solve the inequality $100\cos(t) + 50 < 0$. This simplifies to $100\cos(t) < -50$, or $\cos(t) < -1/2$. On the unit circle, $\cos(t) = -1/2$ at $t = 2\pi/3$ and $t = 4\pi/3$. The inequality $\cos(t) < -1/2$ holds for $t$ between these values. Since $\pi/2 \approx 1.57$, $2\pi/3 \approx 2.09$, $3\pi/2 \approx 4.71$, and $4\pi/3 \approx 4.19$, the interval $\pi/2 < t < 3\pi/2$ contains the entire region where the population is decreasing.
15. The marginal cost to produce $q$ items is given by the function $C'(q)$, in dollars per item. If the company has fixed costs of $5000 (the cost to produce 0 items), what does the expression $5000 + \int_0^{100} C'(q) dq$ represent?
    Answer: The total cost to produce the first 100 items. The expression represents the initial (fixed) cost plus the total variable cost for producing the first 100 items. The integral of the marginal cost, $\int_0^{100} C'(q) dq$, gives the total change in cost (the total variable cost) of producing items 1 through 100. Adding this to the fixed cost $C(0) = 5000$ gives the total cost $C(100)$.
16. If $F'(x) = e^{-x^2}$ and $F(2) = 7$, which of the following is an expression for $F(5)$?
    Answer: $7 + \int_2^5 e^{-t^2} dt$ By the Net Change Theorem, $\int_2^5 F'(t) dt = F(5) - F(2)$. Substituting the given information, we have $\int_2^5 e^{-t^2} dt = F(5) - 7$. To solve for $F(5)$, we add 7 to both sides: $F(5) = 7 + \int_2^5 e^{-t^2} dt$. The variable of integration can be any letter, so using $x$ or $t$ doesn't change the value.
17. A rocket burns fuel at a rate of $r(t) = 15\sqrt{t}$ 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?
    Answer: 10,000 The amount of fuel remaining is the initial amount minus the total fuel burned. The total fuel burned in the first 100 seconds is $\int_0^{100} r(t) dt = \int_0^{100} 15t^{1/2} dt$. The integral evaluates to $[15 \cdot \frac{2}{3}t^{3/2}]_0^{100} = [10t^{3/2}]_0^{100} = 10(100^{3/2}) - 0 = 10(10^3) = 10,000$ kg. The remaining fuel is the initial amount minus the amount burned: $20,000 - 10,000 = 10,000$ kg.
18. A glacier's volume is decreasing due to melting. At the beginning of a year ($t=0$), its volume is $V_0$. The rate at which it loses volume is modeled by $M(t)$ cubic kilometers per year, where $t$ is in years. Which expression represents the volume of the glacier remaining after $T$ years?
    Answer: $V_0 - \int_0^T M(t) dt$ The total volume lost over $T$ years is the integral of the rate of loss, $\int_0^T M(t) dt$. The remaining volume is the initial volume, $V_0$, minus the total volume lost. Therefore, the remaining volume is $V_0 - \int_0^T M(t) dt$.
19. A car starts at a position 10 miles east of a landmark. The car's velocity is given by $v(t) = 15t - 3t^2$ miles per hour, where $t$ is in hours and positive velocity indicates eastward travel. What is the car's position relative to the landmark after 2 hours?
    Answer: 32 miles east The position $x(t)$ is the initial position plus the displacement. The initial position is $x(0)=10$. The displacement over the first 2 hours is $\int_0^2 v(t) dt = \int_0^2 (15t - 3t^2) dt$. This integral evaluates to $[\frac{15}{2}t^2 - t^3]_0^2 = (\frac{15}{2}(4) - 2^3) - 0 = 30 - 8 = 22$ miles. The final position is the initial position plus the displacement: $x(2) = x(0) + 22 = 10 + 22 = 32$ miles east of the landmark.
20. A particle's velocity is given by $v(t) = 3t^2 - 12$ for $t \ge 0$. The particle's initial position is $x(0) = 5$. What is the minimum position of the particle for $t \ge 0$?
    Answer: -11 When you're given a particle's velocity function and need to find its minimum position, you're working with the relationship between velocity, position, and motion analysis. The key insight is that position is the antiderivative of velocity, and minimum position occurs when the particle stops moving downward and starts moving upward. First, find the position function by integrating the velocity: $x(t) = \int (3t^2 - 12) dt = t^3 - 12t + C$. Using the initial condition $x(0) = 5$, we get $C = 5$, so $x(t) = t^3 - 12t + 5$. To find the minimum position, determine when the velocity equals zero: $3t^2 - 12 = 0$, which gives $t^2 = 4$, so $t = 2$ (since $t \geq 0$). This is where the particle momentarily stops. At $t = 2$: $x(2) = 2^3 - 12(2) + 5 = 8 - 24 + 5 = -11$. Looking at the wrong answers: (A) 5 is the initial position, not the minimum. (B) -16 likely comes from miscalculating $x(2)$ or using wrong integration constants. (C) 2 is the time when minimum position occurs, not the position value itself. Study tip: For particle motion problems, remember the three-step process: integrate velocity to get position (don't forget the constant!), set velocity to zero to find critical times, then evaluate position at those times. Always distinguish between when something happens versus where it happens.

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