Accumulations of Change - Calculus 1

1. What is the height of the water, in cm, at time $t=8$ hours?
   Answer: 45 The height at $t=8$ is the height at $t=3$ plus the net change in height from $t=3$ to $t=8$. This is calculated as $h(8) = h(3) + \int_3^8 h'(t) dt$. We have $h(3) = 25$. The integral is $\int_3^8 10(t+1)^{-1/2} dt = [20(t+1)^{1/2}]_3^8 = 20\sqrt{9} - 20\sqrt{4} = 20(3) - 20(2) = 60 - 40 = 20$. Therefore, $h(8) = 25 + 20 = 45$ cm.
2. What is the meaning of the expression $\frac{1}{4} \int_{0}^{4} C'(t) dt$?
   Answer: The average rate of coffee consumption, in liters per hour, between 8:00 AM and 12:00 PM. The expression $\frac{1}{b-a} \int_a^b f(x) dx$ defines the average value of the function $f(x)$ on the interval $[a, b]$. In this context, $\int_{0}^{4} C'(t) dt$ represents the total liters of coffee consumed in the 4-hour period from $t=0$ (8:00 AM) to $t=4$ (12:00 PM). Dividing by the length of the interval, 4 hours, gives the average rate of consumption in liters per hour over that period.
3. A function $f(x)$ has derivative $f'(x) = \cos(\pi x)$. If $f(1) = 5$, what is the value of $f(2)$?
   Answer: 5 According to the Fundamental Theorem of Calculus, the net change of a function is the integral of its derivative: $f(2) - f(1) = \int_1^2 f'(x) dx$. We can find $f(2)$ using $f(2) = f(1) + \int_1^2 \cos(\pi x) dx$. The integral is $\int_1^2 \cos(\pi x) dx = [\frac{1}{\pi}\sin(\pi x)]_1^2 = \frac{1}{\pi}\sin(2\pi) - \frac{1}{\pi}\sin(\pi) = \frac{1}{\pi}(0) - \frac{1}{\pi}(0) = 0$. Therefore, $f(2) = f(1) + 0 = 5 + 0 = 5$.
4. A particle's velocity is given by $v(t) = 2e^t - 4$ for $t \ge 0$. What is the total distance traveled by the particle on the interval $[0, 2]$?
   Answer: $2e^2 + 8\ln 2 - 14$ Total distance traveled is the integral of the speed, $\int_0^2 |v(t)| dt$. First, find where the velocity is zero: $2e^t - 4 = 0 \implies e^t = 2 \implies t = \ln 2$. On the interval $[0, \ln 2)$, $v(t)$ is negative. On the interval $(\ln 2, 2]$, $v(t)$ is positive. Therefore, we must split the integral: Distance = $\int_0^{\ln 2} -(2e^t - 4) dt + \int_{\ln 2}^2 (2e^t - 4) dt$. The antiderivative of $2e^t - 4$ is $2e^t - 4t$. The first integral is $-[2e^t - 4t]_0^{\ln 2} = -((2e^{\ln 2} - 4\ln 2) - (2e^0 - 0)) = -((4 - 4\ln 2) - 2) = -(2 - 4\ln 2) = 4\ln 2 - 2$. The second integral is $[2e^t - 4t]_{\ln 2}^2 = (2e^2 - 8) - (2e^{\ln 2} - 4\ln 2) = (2e^2 - 8) - (4 - 4\ln 2) = 2e^2 - 12 + 4\ln 2$. Total distance = $(4\ln 2 - 2) + (2e^2 - 12 + 4\ln 2) = 2e^2 + 8\ln 2 - 14$.
5. Snow begins to fall at midnight ($t=0$). From $t=0$ to $t=6$ hours, the rate of snowfall is a constant 0.5 inches per hour. From $t=6$ to $t=10$, the rate increases linearly from 0.5 inches per hour to 2.5 inches per hour. What is the total accumulation of snow at $t=10$ hours?
   Answer: 9.0 inches The total accumulation is the integral of the rate of snowfall. We can calculate this by finding the area under the rate graph. For the first interval, $0 \le t \le 6$, the accumulation is the area of a rectangle: $6 \text{ hours} \times 0.5 \text{ in/hr} = 3$ inches. For the second interval, $6 \le t \le 10$, the rate graph is a trapezoid with base $10-6=4$ hours and heights 0.5 and 2.5 in/hr. The area is $\frac{1}{2}(0.5+2.5)(4) = \frac{1}{2}(3)(4) = 6$ inches. The total accumulation is the sum of the amounts from both intervals: $3 + 6 = 9$ inches.
6. The number of people entering a concert venue is modeled by the function $E(t) = -t^2 + 20t + 100$ people per hour. The number of people leaving the venue is modeled by $L(t) = 30t + 20$ people per hour. The venue opens at $t=0$ with 50 people already inside. Which of the following expressions gives the number of people in the venue at $t=4$ hours?
   Answer: $50 + \int_0^4 (-t^2 - 10t + 80) dt$ The total number of people is the initial amount plus the net accumulation. The net rate of change of people in the venue is the rate of people entering minus the rate of people leaving: $N(t) = E(t) - L(t) = (-t^2 + 20t + 100) - (30t + 20) = -t^2 - 10t + 80$. The total change in the number of people from $t=0$ to $t=4$ is the integral of this net rate, $\int_0^4 N(t) dt$. The number of people at $t=4$ is the initial number (50) plus this net change. So, the expression is $50 + \int_0^4 (-t^2 - 10t + 80) dt$.
7. If $F(x) = \int_{\sin(x)}^{1} \sqrt{1+t^2} dt$, what is $F'(x)$?
   Answer: $-\cos(x)\sqrt{1+\sin^2(x)}$ First, flip the limits of integration to get the variable in the upper limit: $F(x) = - \int_{1}^{\sin(x)} \sqrt{1+t^2} dt$. By the Fundamental Theorem of Calculus, Part 1, combined with the Chain Rule, we have $F'(x) = -\left(\sqrt{1+(\sin(x))^2} \cdot \frac{d}{dx}(\sin(x))\right)$. This simplifies to $F'(x) = -\sqrt{1+\sin^2(x)} \cdot \cos(x)$.
8. Which of the following expressions represents the total amount of water, in liters, in the tank at time $t=5$ minutes?
   Answer: $100 + \int_0^5 (20e^{-0.1t} - 4) dt$ The total amount of water at time $t=5$ is the initial amount plus the net change. The initial amount is 100 liters. The net rate of change is the rate in minus the rate out, $R(t) - L(t) = 20e^{-0.1t} - 4$. The net change over the first 5 minutes is the integral of this net rate from 0 to 5. Therefore, the total amount is $100 + \int_0^5 (20e^{-0.1t} - 4) dt$.
9. A company's bank account balance is affected by deposits and withdrawals. The rate of deposits is given by $D(t) = 3t^2 - 12t + 15$ and the rate of withdrawals is given by $W(t) = 15 - t$, both in thousands of dollars per month for $0 \le t \le 6$. At what time $t$ on the interval $[0, 6]$ is the amount of money in the account at an absolute minimum?
   Answer: $t = 11/3$ To find the minimum amount, we need to find where the rate of change of the amount is zero. Let $M(t)$ be the amount of money. The net rate of change is $M'(t) = D(t) - W(t) = (3t^2 - 12t + 15) - (15 - t) = 3t^2 - 11t$. We set $M'(t)=0$ to find critical points: $t(3t - 11) = 0$, which gives $t=0$ and $t=11/3$. To find the absolute minimum on $[0, 6]$, we check the critical points in the interval and the endpoints. $M'(t)$ is negative for $t \in (0, 11/3)$ and positive for $t \in (11/3, 6)$. This means the amount $M(t)$ decreases until $t=11/3$ and then increases. Therefore, the absolute minimum occurs at $t=11/3$.
10. What is the total net revenue, in thousands of dollars, for the first 12 months?
    Answer: 108 The total net revenue is the integral of the rate function over the entire interval. Since the rate is piecewise, we split the integral: $\int_0^{12} r(t) dt = \int_0^6 2t dt + \int_6^{12} 12 dt$. The first integral is $[t^2]_0^6 = 36$. The second integral is $[12t]_6^{12} = 12(12) - 12(6) = 144 - 72 = 72$. The total net revenue is the sum of these two amounts: $36 + 72 = 108$ thousand dollars.
11. What is the increase in total revenue when the number of units sold increases from 50 to 100?
    Answer: $7,500 The increase in total revenue is the net change, which is the definite integral of the marginal revenue function. We need to calculate $\int_{50}^{100} (300 - 2x) dx$. The antiderivative is $300x - x^2$. Evaluating at the limits: $[300x - x^2]_{50}^{100} = (300(100) - 100^2) - (300(50) - 50^2) = (30000 - 10000) - (15000 - 2500) = 20000 - 12500 = 7500$.
12. At what rate is the number of cars in the garage changing at 9:05 AM (i.e., at $t=5$)?
    Answer: $r(5)$ The question asks for the rate of change of the number of cars, which is $C'(t)$. According to the Fundamental Theorem of Calculus, Part 1, the derivative of an integral function $\int_a^t f(x) dx$ with respect to $t$ is $f(t)$. Therefore, $C'(t) = \frac{d}{dt} \left(50 + \int_0^t r(x) dx\right) = 0 + r(t) = r(t)$. The rate of change at $t=5$ is simply $r(5)$.
13. What is the balloon's net change in altitude during the first 6 minutes?
    Answer: 36 meters The net change in altitude is the definite integral of the rate of change from $t=0$ to $t=6$. $\int_0^6 (15 - 3t) dt = [15t - \frac{3t^2}{2}]_0^6 = (15(6) - \frac{3(36)}{2}) - 0 = 90 - 54 = 36$ meters. Note that this is different from total distance traveled, which would require integrating the absolute value of the rate.
14. A quantity $Q(t)$ changes over time, with its initial value being $Q(0)$. The rate of change of the quantity is given by the function $R(t)$. If it is known that $\int_0^{10} R(t) dt < 0$, which of the following statements must be true?
    Answer: $Q(10) < Q(0)$ The Fundamental Theorem of Calculus states that the net change in a quantity $Q(t)$ from $t=a$ to $t=b$ is the definite integral of its rate of change, $R(t)$. Thus, $Q(10) - Q(0) = \int_0^{10} R(t) dt$. Given that $\int_0^{10} R(t) dt < 0$, it follows directly that $Q(10) - Q(0) < 0$, which simplifies to $Q(10) < Q(0)$. Choice A is incorrect because $Q(10)$ could be positive if $Q(0)$ is large enough. Choice C is a sufficient but not necessary condition; the integral can be negative even if $R(t)$ is positive for some parts of the interval. Choice D is incorrect because $Q(t)$ is decreasing only when its derivative, $R(t)$, is negative. We only know the net accumulation is negative, not that the rate is always negative.
15. Sand is being added to a pile at a rate of $A(t) = 8\cos(0.2t)$ cubic feet per hour. At the same time, sand is being removed by erosion at a rate of $E(t) = 2t$ cubic feet per hour. The pile contains 50 cubic feet of sand at time $t=0$. Which expression gives the total amount of sand in the pile at time $t=4$ hours?
    Answer: $50 + \int_0^4 (8\cos(0.2t) - 2t) dt$ The net rate of change of the amount of sand is the rate in minus the rate out, which is $A(t) - E(t) = 8\cos(0.2t) - 2t$. To find the total amount of sand at $t=4$, we need to add the net change in sand over the interval $[0, 4]$ to the initial amount. The net change is the integral of the net rate of change from $t=0$ to $t=4$, which is $\int_0^4 (8\cos(0.2t) - 2t) dt$. The initial amount is 50 cubic feet. Therefore, the total amount at $t=4$ is $50 + \int_0^4 (8\cos(0.2t) - 2t) dt$.
16. If the population is 20,000 bacteria at $t=3$ hours, what was the population at $t=1$ hour, in thousands of bacteria?
    Answer: $20 - \frac{6}{\ln 2}$ By the Fundamental Theorem of Calculus, $P(3) - P(1) = \int_1^3 P'(t) dt$. We can solve for the initial population, $P(1) = P(3) - \int_1^3 2^t dt$. The integral is $\int_1^3 2^t dt = [\frac{2^t}{\ln 2}]_1^3 = \frac{2^3}{\ln 2} - \frac{2^1}{\ln 2} = \frac{6}{\ln 2}$. Since $P(3) = 20$ (in thousands), the population at $t=1$ was $P(1) = 20 - \frac{6}{\ln 2}$ thousands.
17. The rate at which a machine produces widgets is given by a differentiable function $R(t)$, where $R$ is measured in widgets per minute and $t$ is measured in minutes from the start of a shift. What is the best interpretation of the expression $\frac{1}{60} \int_0^{60} R(t) dt$?
    Answer: The average rate of widget production, in widgets per minute, during the first hour of the shift. The expression $\frac{1}{b-a} \int_a^b f(x) dx$ gives the average value of the function $f(x)$ on the interval $[a, b]$. In this context, $f(x)$ is the rate function $R(t)$, and the interval is $[0, 60]$ minutes, which is the first hour of the shift. Therefore, $\frac{1}{60-0} \int_0^{60} R(t) dt$ represents the average value of the rate of production. The units of $R(t)$ are widgets per minute, so the average value will also have units of widgets per minute. Thus, the expression is the average rate of widget production during the first hour.
18. The velocity of a particle moving on a line is given by $v(t) = t^2 - 3t + 2$ for $0 \le t \le 3$. What is the total distance traveled by the particle on this interval?
    Answer: $\frac{11}{6}$ Total distance is the integral of the speed, $\int_0^3 |v(t)| dt$. First, find where $v(t)$ is negative. $v(t) = (t-1)(t-2)$ is negative for $1 < t < 2$. The distance is $\int_0^1 v(t) dt + \int_1^2 -v(t) dt + \int_2^3 v(t) dt$. $\int(t^2-3t+2)dt = \frac{t^3}{3} - \frac{3t^2}{2} + 2t$. $\int_0^1 v(t)dt = \frac{5}{6}$. $\int_1^2 v(t)dt = -\frac{1}{6}$. $\int_2^3 v(t)dt = \frac{5}{6}$. The total distance is $\frac{5}{6} + |-\frac{1}{6}| + \frac{5}{6} = \frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6}$.
19. A particle moves along a line with velocity $v(t) = t^2 - 8t + 12$ for $t \ge 0$. What is the total distance traveled by the particle during the time interval $0 \le t \le 6$?
    Answer: $64/3$ Total distance traveled is the integral of the speed, which is the absolute value of velocity: $\int_0^6 |v(t)| dt$. First, find where $v(t) = 0$: $t^2 - 8t + 12 = (t-2)(t-6) = 0$, so $t=2$ and $t=6$. On the interval $[0, 2]$, $v(t) > 0$. On $[2, 6]$, $v(t) < 0$. The integral must be split: $\int_0^2 (t^2 - 8t + 12) dt + \int_2^6 -(t^2 - 8t + 12) dt$. The antiderivative is $F(t) = \frac{t^3}{3} - 4t^2 + 12t$. $\int_0^2 v(t) dt = F(2) - F(0) = (\frac{8}{3} - 16 + 24) - 0 = 32/3$. $\int_2^6 v(t) dt = F(6) - F(2) = (\frac{216}{3} - 144 + 72) - 32/3 = 0 - 32/3 = -32/3$. The total distance is $|32/3| + |-32/3| = 64/3$.
20. Let $h(x) = \int_0^{x^2} e^{-t^2} dt$. For which values of $x$ is $h(x)$ increasing?
    Answer: For $x > 0$ A function is increasing when its derivative is positive. We find $h'(x)$ using the Fundamental Theorem of Calculus and the Chain Rule: $h'(x) = e^{-(x^2)^2} \cdot \frac{d}{dx}(x^2) = e^{-x^4} \cdot 2x$. We need to determine where $h'(x) > 0$. The term $e^{-x^4}$ is always positive for any real $x$. Therefore, the sign of $h'(x)$ is determined by the sign of $2x$. $2x > 0$ when $x > 0$. Thus, $h(x)$ is increasing for $x > 0$.

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