Aimathic – Accumulation of change

Accumulation of change

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Difficulty: 1 – 5

54905212

The temperature of a kiln changes at rate $T'(t)$, measured in degrees Fahrenheit per minute. A technician calculates $\int_{12}^{20}T'(t)\,\text{d}t=-36$ and reports, “The kiln cooled at $-36$ degrees Fahrenheit per minute.” Correct the technician's interpretation, including the proper units.

Hints

- Track how the time unit in the differential interacts with the rate unit. - Decide whether the integral gives an instantaneous quantity or a change over an interval. - Use the negative sign to describe direction of change.

Solution

1. Integrating a rate measured in degrees Fahrenheit per minute over time measured in minutes produces degrees Fahrenheit. 2. The integral is the net temperature change from minute $12$ to minute $20$. 3. The value $-36$ means the kiln's temperature decreases by $36$ degrees Fahrenheit over that interval.

Answer

The kiln's temperature has a net change of $-36\,^{\circ}\text{F}$; equivalently, it cools by $36\,^{\circ}\text{F}$ from minute $12$ to minute $20$. The answer is not a rate.

54905912

Over a $12$-minute interval, the average net rate of change of a laboratory freezer's temperature is $-0.25\,^{\circ}\text{F}$ per minute. The temperature is $18\,^{\circ}\text{F}$ at the start. Find the net temperature change and the final temperature.

Hints

- Relate an average rate over an interval to the accumulated change over that interval. - Use the sign of the average rate when updating the initial temperature. - Check that the rate and time units combine to produce a temperature change.

Solution

1. Average net rate times elapsed time gives the total net change: $(-0.25)(12)=-3\,^{\circ}\text{F}$. 2. Add the change to the initial temperature: $18-3=15\,^{\circ}\text{F}$.

Answer

The net change is $-3\,^{\circ}\text{F}$, and the final temperature is $15\,^{\circ}\text{F}$.

52489512

Suppose $\int_0^6f(x)\,\text{d}x=24$. Interpret the value of the integral in each context. a) The function $f$ models the instantaneous rate of change of the number of bacteria in a petri dish, where $x$ is measured in hours and $f(x)$ is measured in millions of bacteria per hour. b) The function $f$ models the electrical power used by a factory, where $x$ is measured in hours and $f(x)$ is measured in megawatts. c) A planar region is bounded by the x-axis, the lines $x=0$ and $x=6$, and the graph of $f$. The graph lies above the x-axis on $[0, 6]$, and both axes are measured in centimeters.

Hints

- Multiply the units of the function values by the units of the independent variable. - Integrating a rate over time gives an accumulated change. - Power integrated over time gives energy. - A definite integral can represent geometric area when the graph lies above the axis.

Solution

1. a) The integral gives the net change in the bacteria population during the first $6$ hours. The population increases by $24$ million bacteria. 2. b) Integrating power over time gives energy. The factory uses $24\,\text{MWh}$ during the $6$-hour period. 3. c) Because the graph is above the x-axis, the integral gives the geometric area of the region, which is $24\,\text{cm}^2$.

Answer

a) The bacteria population increases by $24$ million during the first $6$ hours. b) The factory uses $24\,\text{MWh}$ of electrical energy. c) The region has area $24\,\text{cm}^2$.

52489612

Suppose $\int_2^5g(x)\,\text{d}x=-1.5$. Explain the meaning of the equation in each context. a) The function $g$ is the rate of change of the water level in a reservoir, where $x$ is measured in hours and $g(x)$ is measured in meters per hour. b) The function $g$ is a diver's vertical velocity, where $x$ is measured in seconds and $g(x)$ is measured in meters per second. Positive values indicate upward motion. State what the integral tells you and one quantity it does not determine. c) The function $g$ is the rate of change of air pressure during a weather shift, where $x$ is measured in hours and $g(x)$ is measured in hectopascals per hour.

Hints

- Use the sign of the integral to determine the direction of the net change. - Multiply the rate units by the time units to identify the accumulated quantity. - Velocity integrated over time gives displacement, not total distance unless the direction is known. - The integral describes net change, so positive and negative contributions may cancel.

Solution

1. a) The integral gives the net change in water level from hour $2$ to hour $5$. The negative value means the level decreases by $1.5\,\text{m}$. 2. b) The integral gives the diver's vertical displacement from $2$ to $5$ seconds. The value $-1.5$ means the diver ends $1.5\,\text{m}$ lower than at $t=2$. It does not determine the total distance traveled, because the diver may have changed direction. 3. c) The integral gives the net change in air pressure from hour $2$ to hour $5$. The pressure decreases by $1.5\,\text{hPa}$.

Answer

a) The water level decreases by $1.5\,\text{m}$. b) The diver's vertical displacement is $-1.5\,\text{m}$; the total distance traveled cannot be determined. c) The air pressure decreases by $1.5\,\text{hPa}$.

52660912

The function $f$ is the instantaneous rate of change of the mass of a pollutant in a lake. Time $t$ is measured in weeks, and $f(t)$ is measured in kilograms per week. At $t=0$, the lake contains $500\,\text{kg}$ of pollutant. Interpret each statement in context. a) $f(t)<0$ for $t\in[5, 8]$ b) $\int_0^4f(t)\,\text{d}t=-100$ c) $500+\int_0^Tf(t)\,\text{d}t=250$

Hints

- Interpret the sign of the rate as increase or decrease. - The integral of a rate gives net change. - Identify the initial value in part c). - Use initial value plus accumulated change to interpret the final value.

Solution

1. a) A negative rate means the pollutant mass is decreasing, so it decreases throughout weeks $5$ through $8$. 2. b) The integral is the net change during the first $4$ weeks. The pollutant mass decreases by $100\,\text{kg}$. 3. c) The expression on the left is the pollutant mass at time $T$. The equation says that the mass is $250\,\text{kg}$, half the initial amount, at time $T$.

Answer

a) The pollutant mass decreases from week $5$ through week $8$. b) The pollutant mass decreases by $100\,\text{kg}$ during the first $4$ weeks. c) At time $T$, the lake contains $250\,\text{kg}$ of pollutant.

53392912

The graph of the derivative $f^{\prime}$ is shown. The graph of $f$ passes through $P(0, 1)$. Use geometry and the graph to find $f(2)$.

Hints

- The signed area under a derivative gives the change in the original function. - Identify the geometric shape under the graph from $x=0$ to $x=2$. - Add the change to the given initial value.

Solution

1. The change in $f$ from $x=0$ to $x=2$ is $f(2)-f(0)=\int_0^2 f^{\prime}(x)\,dx$. 2. The region under $f^{\prime}$ is a trapezoid with parallel sides of lengths $1$ and $2$ and width $2$. Its area is $\frac{1+2}{2}\cdot2=3$. 3. Therefore, $f(2)=f(0)+3=1+3=4$.

Answer

$f(2)=4$

54903012

A library's automated counter records the net rate at which books enter its sorting room. The rate is $r(t)$ books per hour, where $t$ is hours after 8:00 a.m. At 8:00 a.m., $1840$ books are in the room, and $\int_0^6 r(t)\,\text{d}t=275$. How many books are in the sorting room at 2:00 p.m.? Explain what the sign of the integral means in this context.

Hints

- Identify whether the integral represents an amount present or a change in that amount. - Connect the sign of the accumulated rate to a net gain or net loss. - Combine the initial amount with the accumulated change.

Solution

1. The integral is the net change in the number of books from 8:00 a.m. to 2:00 p.m. 2. Add the net change to the initial amount: $1840+275=2115$. 3. Because the integral is positive, more books entered than left during the interval.

Answer

There are $2115$ books in the sorting room at 2:00 p.m. The positive integral means the room had a net gain of $275$ books.

54904112

A report states that a museum's visitor-count rate $r(t)$ satisfies $r(t)\ge0$ for every $t\in[1, 4]$, but also states that $\int_1^4 r(t)\,\text{d}t=-120$. Can both statements be true? Give a mathematical explanation in context.

Hints

- Connect the sign of a rate at every time with the sign of its accumulated change. - Consider whether any negative signed area is possible under the stated condition. - Express the contradiction as an inequality involving the integral.

Solution

1. A nonnegative rate on the entire interval has nonnegative accumulated change. 2. Therefore, $\int_1^4r(t)\,\text{d}t\ge0$. 3. The claimed value $-120$ contradicts this consequence, so the two statements cannot both be true.

Answer

No. If $r(t)\ge0$ throughout $[1, 4]$, then the accumulated visitor-count change must be nonnegative, so the integral cannot equal $-120$.

54904912

A robot transfers components from Bin A to Bin B at rate $q(t)$ components per minute, and no components enter or leave the two-bin system. Initially, Bin A contains $74$ components and Bin B contains $31$ components. During the transfer, $\int_0^{8}q(t)\,\text{d}t=18$. Find the number of components in each bin after $8$ minutes and verify that the total number is conserved.

Hints

- The same transferred amount affects the two bins with opposite signs. - Use the integral as the number of components moved, not as a rate at the final time. - Check conservation by comparing the initial and final totals.

Solution

1. Bin A loses the accumulated transfer, so it contains $74-18=56$ components. 2. Bin B gains the same accumulated transfer, so it contains $31+18=49$ components. 3. Initially the total is $74+31=105$ components, and finally it is $56+49=105$ components.

Answer

Bin A contains $56$ components, and Bin B contains $49$ components. The total remains $105$ components.

54905412

A cold-storage bin contains $120$ pounds of ice at $t=2$ hours. From $t=2$ to $t=5$, the net accumulated change is $+18$ pounds. Beginning at $t=5$, ice leaves at a constant net rate of $6$ pounds per hour. After $t=5$, at what time does the bin contain $90$ pounds?

Hints

- First determine the amount present when the constant-rate phase begins. - Find how much must be removed to reach the target. - Convert that required change into elapsed time using the constant rate.

Solution

1. At $t=5$, the bin contains $120+18=138$ pounds. 2. To reach $90$ pounds, it must lose $138-90=48$ pounds. 3. At $6$ pounds per hour, the loss takes $48\div6=8$ hours. 4. The bin reaches $90$ pounds at $t=5+8=13$ hours.

Answer

After $t=5$, the bin contains $90$ pounds at $t=13\,\text{h}$.

54905512

At a festival entrance, $420$ wristbands are issued and $135$ are returned during a two-hour period. The booth begins with $900$ wristbands. a) What is the net accumulated change in the booth's inventory? b) How many wristbands remain? c) How many wristband transactions occurred in total?

Hints

- Assign signs according to whether each action adds to or removes from booth inventory. - Net change and total activity treat opposite-direction actions differently. - Use the net change, not the transaction count, to update inventory.

Solution

1. Issuing removes wristbands and returns add them, so the net change is $-420+135=-285$. 2. The ending inventory is $900-285=615$. 3. Total transactions count both directions: $420+135=555$.

Answer

a) $-285$ wristbands b) $615$ wristbands c) $555$ transactions

54906012

A data-processing buffer contains $90$ megabytes at $t=1$ hour, and its net data rate is $r(t)$ megabytes per hour. Two students propose formulas: Student A: $B(t)=90+\int_0^t r(s)\,\text{d}s$ Student B: $B(t)=90+\int_1^t r(s)\,\text{d}s$ a) Which formula is valid from the given information? Explain. b) If $\int_1^4r(s)\,\text{d}s=-12$, find $B(4)$.

Hints

- Match the lower limit of accumulation to the time at which the amount is known. - Test each formula by substituting the reference time. - Use the given integral as the change from the reference time to $t=4$.

Solution

1. The known amount is anchored at $t=1$, so accumulated change must begin at $1$. Student B's formula is valid. 2. Student A's formula would require $B(0)=90$, which is not given. 3. Using Student B's formula, $B(4)=90-12=78$ megabytes.

Answer

a) Student B's formula b) $B(4)=78\,\text{MB}$

52658712

A rechargeable battery has $E(t)$ watt-hours of stored energy after $t$ hours. Its charging power is $p(t)=E'(t)$ watts. Match each description with all applicable expressions from the list. a) After $2$ hours, the battery stores exactly $65\,\text{Wh}$. b) At $t=2$, the battery is charging at $20\,\text{W}$. c) From $t=2$ to $t=5$, the stored energy increases by $15\,\text{Wh}$. d) The battery starts with $40\,\text{Wh}$ and stores $70\,\text{Wh}$ after $5$ hours. Expressions: A. $E(2)=65$ B. $p(2)=20$ C. $E'(2)=20$ D. $E(5)-E(2)=15$ E. $\int_2^5p(t)\,\text{d}t=15$ F. $40+\int_0^5p(t)\,\text{d}t=70$ G. $\int_0^5p(t)\,\text{d}t=30$ H. $E(5)=E(2)+15$

Hints

- Distinguish an amount $E(t)$ from its rate of change $p(t)=E'(t)$. - The integral of charging power over time gives the change in stored energy. - A change can be expressed as final value minus initial value or as final value equals initial value plus change. - Equivalent equations may represent the same physical statement.

Solution

1. a) The statement gives the stored-energy value at $t=2$, so A applies. 2. b) Charging power is $p(t)$, and $p(t)=E'(t)$. Therefore, both B and C apply. 3. c) The increase can be written as $E(5)-E(2)=15$, as the accumulated charging power $\int_2^5p(t)\,\text{d}t=15$, or as $E(5)=E(2)+15$. Thus, D, E, and H apply. 4. d) Initial energy plus accumulated charging power gives final energy, so F applies. Subtracting $40$ from both sides gives an accumulated change of $30\,\text{Wh}$, so G also applies.

Answer

a) A b) B, C c) D, E, H d) F, G

52658812

The net flow rate of water in a reservoir is modeled for $0\le t\le8$ by $k(t)=100(t^2-10t+21)$, where $t$ is measured in hours and $k(t)$ is measured in cubic meters per hour. A negative value means water is leaving the reservoir. At $t=0$, the reservoir contains $5000\,\text{m}^3$ of water. a) Explain the meaning of $\int_3^7k(t)\,\text{d}t$ in context. b) Evaluate $\int_0^3k(t)\,\text{d}t$ and interpret the result. c) Write an expression for the water volume $V(8)$ after $8$ hours.

Hints

- Factor $t^2-10t+21$ to determine the sign of the flow rate on the relevant interval. - Integrating a flow rate over time gives a volume change. - Add accumulated net change to the initial volume to obtain the stored volume. - Check that $\text{m}^3/\text{h}$ multiplied by hours gives cubic meters.

Solution

1. a) Since $k(t)=100(t-3)(t-7)$, the rate is negative on $(3, 7)$. Therefore, $\int_3^7k(t)\,\text{d}t$ is the net change in reservoir volume during that interval. It is negative, and its absolute value is the volume of water that leaves. 2. b) $\int_0^3k(t)\,\text{d}t=100\left[\frac{1}{3}t^3-5t^2+21t\right]_0^3=2700\,\text{m}^3$. Thus, the reservoir gains $2700\,\text{m}^3$ during the first $3$ hours. 3. c) Add the accumulated net change to the initial volume: $V(8)=5000+\int_0^8k(t)\,\text{d}t$.

Answer

a) The integral is the net volume change from hour $3$ to hour $7$. It is negative, and its absolute value is the volume that leaves. b) $2700\,\text{m}^3$, a net increase during the first $3$ hours c) $V(8)=5000+\int_0^8k(t)\,\text{d}t$

52661012

A solar farm supplies electrical energy to the grid. The function $P(t)$ gives its power output in kilowatts, where $t$ is measured in hours after midnight. Write an appropriate mathematical expression, equation, or inequality for each question. a) How much energy is supplied between 8:00 a.m. and noon? b) At what first time $t$, after 6:00 a.m., has the solar farm supplied $1000\,\text{kWh}$ of energy since 6:00 a.m.? c) Which $2$-hour intervals contain more than $400\,\text{kWh}$ of supplied energy?

Hints

- Integrating power over time gives energy. - Use the clock times as bounds after converting them to hours after midnight. - For an amount accumulated from a fixed starting time to an unknown time, place the unknown in the upper bound. - Represent a general $2$-hour interval as $[t, t+2]$ and restrict $t$ so the interval remains within one day.

Solution

1. a) Integrating power over time gives energy, so the expression is $\int_8^{12}P(t)\,\text{d}t$. 2. b) Energy supplied from 6:00 a.m. to time $t$ is $\int_6^tP(x)\,\text{d}x$. The required time is the least $t\ge6$ satisfying $\int_6^tP(x)\,\text{d}x=1000$. 3. c) A $2$-hour interval beginning at $t$ is $[t, t+2]$. The condition is $\int_t^{t+2}P(x)\,\text{d}x>400$, with $0\le t\le22$.

Answer

a) $\int_8^{12}P(t)\,\text{d}t$ b) The least $t\ge6$ satisfying $\int_6^tP(x)\,\text{d}x=1000$ c) $\int_t^{t+2}P(x)\,\text{d}x>400$, with $0\le t\le22$

53391612

The graph of the derivative $f'$ of a continuous function $f$ is shown, and $f(0)=2$. 1. Find the net change in $f$ on $[0, 4]$ and on $[4, 6]$. 2. Use the net changes to find $f(4)$ and $f(6)$. 3. At what x-value on $[0, 6]$ does $f$ attain its absolute maximum? Explain using the graph of $f'$.

Hints

- Read net change as signed area under the derivative graph. - Areas above the x-axis are positive; areas below it are negative. - Add each net change to the known function value. - The sign of $f'$ tells where $f$ increases or decreases.

Solution

1. On $[0, 4]$, the region under $f'$ consists of two triangles, each with area $\frac12\cdot2\cdot2=2$. Thus, $\int_0^4f'(x)\,\text{d}x=4$. On $[4, 6]$, the triangular region lies below the x-axis, so $\int_4^6f'(x)\,\text{d}x=-\frac12\cdot2\cdot2=-2$. 2. By accumulation of change, $f(4)=f(0)+4=6$. Then $f(6)=f(4)-2=4$. 3. The derivative is positive on $(0, 4)$ and negative on $(4, 6)$. Therefore, $f$ increases through $x=4$ and decreases afterward, so its absolute maximum occurs at $x=4$.

Answer

1. On $[0, 4]$: $4$; on $[4, 6]$: $-2$ 2. $f(4)=6$ and $f(6)=4$ 3. $x=4$

54903112

A theater's costume department tracks the net rate $c(t)$, in costume pieces per day, at which pieces are added to or removed from storage. Over a $5$-day inventory period, $\int_0^5 c(t)\,\text{d}t=-145$. At the end of the period, $960$ costume pieces remain. How many pieces were in storage at the beginning of the period?

Hints

- Write a relationship among the initial amount, the net change, and the final amount. - A negative accumulated change means the final amount is smaller than the initial amount. - Use an unknown for the beginning inventory and solve the resulting equation.

Solution

1. The integral gives the net change: final amount minus initial amount is $-145$. 2. Let $C_0$ be the initial amount. Then $960-C_0=-145$. 3. Solving gives $C_0=1105$.

Answer

The costume department began with $1105$ pieces.

54903212

The graph shows the rate $r(t)$, in degrees Celsius per hour, at which the air temperature in a greenhouse changes for $0\le t\le6$, where $t$ is measured in hours. At $t=0$, the temperature is $12\,^\circ\text{C}$. a) Find the temperature at $t=4$. b) Find the temperature at $t=6$. c) At what time is the temperature highest?

Hints

- Treat signed area under the rate graph as change in temperature. - Look for cancellation when equal regions lie on opposite sides of the horizontal axis. - Add each accumulated change to the initial temperature. - The highest temperature occurs when the rate changes from positive to negative.

Solution

1. From $t=0$ to $t=2$, the signed area is $2\cdot2=4$. From $t=2$ to $t=4$, the positive and negative triangular areas cancel, so the accumulated change on that interval is $0$. Thus, the temperature at $t=4$ is $12+4=16\,^\circ\text{C}$. 2. From $t=4$ to $t=6$, the signed area is $2(-2)=-4$. Thus, the temperature at $t=6$ is $16-4=12\,^\circ\text{C}$. 3. The temperature increases while $r(t)>0$ and decreases while $r(t)<0$. The sign changes from positive to negative at $t=3$, so the highest temperature occurs at $t=3$.

Answer

a) $16\,^\circ\text{C}$ b) $12\,^\circ\text{C}$ c) $t=3\,\text{h}$

54903312

A library records the net rate $r(t)$ at which books enter its sorting room. Positive values mean books enter; negative values mean books leave. During three consecutive one-hour intervals, $\int_0^1 r(t)\,\text{d}t=18$, $\int_1^2 r(t)\,\text{d}t=-11$, and $\int_2^3 r(t)\,\text{d}t=7$. A student claims that $14$ books were moved because $18-11+7=14$. Explain the error. State the net change in the number of books in the room and the total number of books moved into or out of the room, assuming $r$ does not change sign within any of the three intervals.

Hints

- Distinguish the change in the number present from the total number that crossed the doorway. - Use the signs when finding net change. - Use magnitudes when counting all books moved. - The no-sign-change condition lets each integral's magnitude represent the activity on that interval.

Solution

1. Adding the signed changes gives the net inventory change: $18-11+7=14$ books. 2. The total number moved counts the magnitude of the change on each interval: $18+11+7=36$ books. 3. The student's calculation gives net change, not total activity, because the $11$ books that left were subtracted instead of counted as books moved.

Answer

The net change is an increase of $14$ books, and $36$ books were moved into or out of the room. The student confused signed accumulation with accumulated magnitude.

54903412

Two school clubs track their cash balances using net cash-flow rates $a(t)$ and $b(t)$, in dollars per week. At week $0$, Club A has $\$450$ and Club B has $\$470$. The following accumulated changes are known: $\int_0^4 a(t)\,\text{d}t=80$, $\int_4^7 a(t)\,\text{d}t=-35$, $\int_0^2 b(t)\,\text{d}t=-10$, and $\int_2^7 b(t)\,\text{d}t=45$. Which club has more money at week $7$, and by how much?

Hints

- Combine accumulated changes over adjacent intervals for each club separately. - Remember that each integral changes the initial balance rather than replacing it. - Compare the two resulting final balances.

Solution

1. Club A's week-7 balance is $450+80-35=495$ dollars. 2. Club B's week-7 balance is $470-10+45=505$ dollars. 3. The difference is $505-495=10$ dollars, so Club B has more.

Answer

Club B has more money at week $7$, by $\$10$.

54903512

An event hall contains $2000$ people when admission monitoring begins. For $0\le t\le8$, where $t$ is measured in minutes, the net entry rate is $r(t)=300-50t$ people per minute. a) At what time is the number of people in the hall greatest? b) Find the greatest number of people in the hall. c) Find the number of people in the hall at $t=8$.

Hints

- The sign of the net entry rate determines whether attendance is increasing or decreasing. - Find where the rate changes sign. - Integrate the rate from the initial time to obtain the net change. - Add the accumulated change to the initial attendance.

Solution

1. The number of people increases while $r(t)>0$ and decreases when $r(t)<0$. Since $300-50t=0$ at $t=6$, the maximum occurs at $t=6$. 2. The accumulated change through $t=6$ is $\int_0^6(300-50t)\,\text{d}t=900$ people. The greatest number is $2000+900=2900$. 3. The accumulated change through $t=8$ is $\int_0^8(300-50t)\,\text{d}t=800$ people. Therefore, the hall contains $2000+800=2800$ people at $t=8$.

Answer

a) $t=6\,\text{min}$ b) $2900$ people c) $2800$ people

54903712

During a $5$-minute calibration, the rate $s(t)$ at which a greenhouse sensor's reading drifts satisfies $-0.8\le s(t)\le1.3$ for every $t\in[0, 5]$, where $s(t)$ is measured in parts per million per minute. Without knowing a formula for $s$, give the tightest guaranteed interval for the net change in the sensor reading over the calibration period.

Hints

- Compare the unknown rate with constant rates that bound it at every moment. - Ask what changes would result if the rate stayed at either extreme for the entire interval. - Keep the rate units and time units together when interpreting the bounds.

Solution

1. Accumulating the lower rate bound for $5$ minutes gives $\int_0^5(-0.8)\,\text{d}t=-4$. 2. Accumulating the upper rate bound gives $\int_0^5 1.3\,\text{d}t=6.5$. 3. Therefore, $-4\le\int_0^5s(t)\,\text{d}t\le6.5$, so the net reading change lies in that interval.

Answer

The net change is guaranteed to lie in $[-4, 6.5]$ parts per million.

54903812

The graph shows the net rate $r(t)$, in gigabytes per minute, at which data is added to or removed from a server's processing queue. Time $t$ is measured in minutes, and the queue contains $50\,\text{GB}$ at $t=0$. a) When is the amount of queued data smallest? b) What is that minimum amount? c) After $t=0$, when does the queue first return to its initial amount?

Hints

- Use the sign of the net rate to determine when the queued amount decreases or increases. - Compute signed areas from the graph to find the accumulated change. - To find the return time, determine when the total accumulated change becomes $0$.

Solution

1. The queued amount decreases while $r<0$ and begins increasing when $r$ changes from negative to positive. The sign change occurs at $t=3$, so the minimum occurs then. 2. From $0$ to $2$, the signed area is $-6$. From $2$ to $3$, the triangular signed area is $-1.5$. The minimum amount is $50-7.5=42.5\,\text{GB}$. 3. The accumulated change through $t=4$ is $-6\,\text{GB}$. From $4$ onward, the rate is $3\,\text{GB/min}$, so two more minutes add $6\,\text{GB}$. The queue first returns to $50\,\text{GB}$ at $t=6$.

Answer

a) $t=3\,\text{min}$ b) $42.5\,\text{GB}$ c) $t=6\,\text{min}$

54903912

The net rate $m(t)$, in kilograms per hour, describes the change in material on a recycling conveyor. Measurements show $\int_0^{10}m(t)\,\text{d}t=-42$, $\int_0^3m(t)\,\text{d}t=18$, and $\int_7^{10}m(t)\,\text{d}t=-25$. Find $\int_3^7m(t)\,\text{d}t$ and interpret its value.

Hints

- Break the total accumulated change into changes over adjacent time intervals. - Keep the negative sign attached to a net loss. - After finding the missing value, translate its units and sign back into the setting.

Solution

1. Split the full interval into adjacent parts: $\int_0^{10}m(t)\,\text{d}t=\int_0^3m(t)\,\text{d}t+\int_3^7m(t)\,\text{d}t+\int_7^{10}m(t)\,\text{d}t$. 2. Substitute the known values: $-42=18+\int_3^7m(t)\,\text{d}t-25$. 3. Solving gives $\int_3^7m(t)\,\text{d}t=-35$. 4. The conveyor's material amount has a net decrease of $35$ kilograms from hour $3$ to hour $7$.

Answer

$\int_3^7m(t)\,\text{d}t=-35$. The material on the conveyor decreases by a net $35\,\text{kg}$ during that interval.

54904012

The graph shows a portable generator's power output $P(t)$, in kilowatts, during a $7$-hour test. a) How much electrical energy does the generator produce during the test? b) At what time has the generator produced exactly half of that energy?

Hints

- Interpret area under a power-versus-time graph as energy. - Decompose the graph into familiar geometric regions. - Locate the halfway point by accumulating area from left to right, not by halving the time interval.

Solution

1. Energy is accumulated power. The total area is two triangles and a rectangle: $\frac{1}{2}(2)(4)+(3)(4)+\frac{1}{2}(2)(4)=20$ kilowatt-hours. 2. Half the total is $10$ kilowatt-hours. The first two hours produce $4$ kilowatt-hours. At the constant output of $4$ kilowatts, another $6$ kilowatt-hours require $1.5$ hours. 3. Half the energy has been produced at $t=2+1.5=3.5$ hours.

Answer

a) $20\,\text{kWh}$ b) $t=3.5\,\text{h}$

54904212

Sand moves onto and off a weighing platform at the net rate $r(t)=k-2t$ pounds per minute for $0\le t\le6$, where $t$ is measured in minutes and $k$ is constant. The platform is required to contain the same amount of sand at $t=6$ as at $t=0$. Find $k$ and state its units.

Hints

- Same initial and final amounts imply a particular value for the net accumulated change. - Treat the parameter as a constant while accumulating the rate. - Use the units of the entire rate expression to identify the parameter’s units.

Solution

1. Returning to the initial amount requires zero accumulated change: $\int_0^6(k-2t)\,\text{d}t=0$. 2. Evaluating gives $6k-36=0$. 3. Solving gives $k=6$. Since $k$ is part of a rate, its units are pounds per minute.

Answer

$k=6\,\text{lb/min}$

54904312

An online retailer has $80$ unfilled orders at $t=0$. During the next $3$ hours, where $t$ is measured in hours, new orders arrive at $12+2t$ orders per hour and employees complete orders at $5+t^2$ orders per hour. a) Write one integral for the net change in the number of unfilled orders. b) Find the number of unfilled orders at $t=3$.

Hints

- Form the net rate by subtracting the completion rate from the arrival rate. - Integrate the net rate over the three-hour interval. - Add the accumulated change to the initial backlog.

Solution

1. The net rate is the arrival rate minus the completion rate: $(12+2t)-(5+t^2)=7+2t-t^2$. 2. The net change is $\int_0^3(7+2t-t^2)\,\text{d}t=21$ orders. 3. Add the change to the initial backlog: $80+21=101$ orders.

Answer

a) $\int_0^3\big[(12+2t)-(5+t^2)\big]\,\text{d}t$ b) $101$ unfilled orders

54904412

A sculpture's mass is $30\,\text{kg}$ at time $t=2$ hours. Its mass changes at rate $r(t)$ kilograms per hour. a) Write a formula for the mass $M(t)$ that is valid for times both before and after $t=2$. b) Interpret $\int_5^2 r(t)\,\text{d}t$ in terms of the sculpture's mass.

Hints

- Anchor the accumulation at the time where the amount is known. - Use a different letter inside the integral so the upper-limit variable remains clear. - Reversing the order of the limits reverses the direction of the change.

Solution

1. Starting from the known value at $t=2$, the mass is $M(t)=30+\int_2^t r(u)\,\text{d}u$. Reversed limits automatically handle $t<2$. 2. Since $\int_5^2r(t)\,\text{d}t=M(2)-M(5)$, the integral is the signed difference between the mass at time $2$ and the mass at time $5$.

Answer

a) $M(t)=30+\int_2^t r(u)\,\text{d}u$ b) $\int_5^2 r(t)\,\text{d}t=M(2)-M(5)$, the change in mass when moving backward from time $5$ to time $2$.

54904512

A temporary data buffer contains $120\,\text{GB}$. During an $8$-second processing burst, its net input rate $r(t)$ satisfies $4\le r(t)\le7$ gigabytes per second. The buffer's capacity is $170\,\text{GB}$. Is exceeding the capacity impossible, possible but not guaranteed, or guaranteed? Justify your classification using accumulated-change bounds.

Hints

- Convert the rate bounds into bounds on the accumulated input. - Add both accumulated-change bounds to the initial amount. - Compare the resulting interval with the buffer capacity.

Solution

1. The accumulated input is at least $4\cdot8=32\,\text{GB}$ and at most $7\cdot8=56\,\text{GB}$. 2. Therefore, the final amount lies between $120+32=152\,\text{GB}$ and $120+56=176\,\text{GB}$. 3. Since the interval of possible final amounts includes values below and above $170\,\text{GB}$, exceeding capacity is possible but not guaranteed.

Answer

Exceeding the capacity is possible but not guaranteed. The final amount can lie anywhere from $152\,\text{GB}$ to $176\,\text{GB}$ under the stated bounds.

54904712

A tide-controlled observation chamber has water depth $2\,\text{m}$ at $t=0$. Over one $12$-hour cycle, the depth changes at rate $r(t)=3\sin\left(\frac{\pi t}{6}\right)$ meters per hour. a) When is the depth greatest? b) Find the greatest depth exactly. c) Find the depth at $t=12$.

Hints

- Use the sign of the rate over each half of the cycle. - Accumulate only up to the point where the rate switches from positive to negative to find the maximum. - Use the symmetry of one full sine cycle to assess the total net change.

Solution

1. The rate is positive on $(0, 6)$ and negative on $(6, 12)$, so the greatest depth occurs at $t=6$. 2. The change through $t=6$ is $\int_0^6 3\sin\left(\frac{\pi t}{6}\right)\,\text{d}t=\frac{36}{\pi}$. The greatest depth is $2+\frac{36}{\pi}$ meters. 3. The positive accumulation on $[0, 6]$ is canceled by the negative accumulation on $[6, 12]$, so the net change over the cycle is $0$. The depth at $t=12$ is $2$ meters.

Answer

a) $t=6\,\text{h}$ b) $\left(2+\frac{36}{\pi}\right)\,\text{m}$ c) $2\,\text{m}$

54904812

A food pantry begins Monday with $108$ sealed meal boxes. The net changes over three consecutive periods are shown. <table><tr><th>Period</th><th>Accumulated net change</th></tr><tr><td>Monday to Tuesday</td><td>$+15$ boxes</td></tr><tr><td>Tuesday to Thursday</td><td>$-8$ boxes</td></tr><tr><td>Thursday to Friday</td><td>$+20$ boxes</td></tr></table> a) At which listed checkpoint does the pantry first have more than $120$ boxes? b) How many boxes are present Friday?

Hints

- Update the amount chronologically rather than combining all changes immediately. - Check the threshold after each listed period. - Preserve the sign of each accumulated change.

Solution

1. Tuesday's amount is $108+15=123$, which is the first listed checkpoint above $120$. 2. Thursday's amount is $123-8=115$. 3. Friday's amount is $115+20=135$.

Answer

a) Tuesday b) $135$ boxes

54905012

During a $4$-hour event, two donation kiosks have net collection rates $r_A(t)$ and $r_B(t)$, in dollars per hour, with $r_A(t)=1.5r_B(t)+20$ at every time. Together, the kiosks collect a net $\$1250$ over the event. How much net money is collected by each kiosk?

Hints

- Translate each rate into a total by accumulating it over the full four hours. - The constant $20$ in the rate relation also accumulates over time. - Use the combined event total after expressing one kiosk's accumulation in terms of the other.

Solution

1. Let $B=\int_0^4 r_B(t)\,\text{d}t$. Then $\int_0^4 r_A(t)\,\text{d}t=\int_0^4[1.5r_B(t)+20] \,\text{d}t=1.5B+80$. 2. The combined accumulation gives $(1.5B+80)+B=1250$, so $2.5B=1170$ and $B=468$. 3. Kiosk A collects $1.5(468)+80=782$ dollars.

Answer

Kiosk A collects a net $\$782$, and Kiosk B collects a net $\$468$.

54905312

A precision scale's drift rate $r(t)$, in grams per hour, where $t$ is measured in hours, satisfies $r(4+h)=-r(4-h)$ whenever both times lie in $[0, 8]$. The scale reads $75$ grams at $t=0$. What does the scale read at $t=8$? Explain without finding a formula for $r$.

Hints

- Pair times that are equally far from the midpoint of the interval. - Consider how opposite rate values affect the total signed accumulation. - Apply the resulting net change to the initial reading.

Solution

1. The rate values at times equally spaced from $t=4$ are opposites. 2. Their accumulated contributions on $[0, 4]$ and $[4, 8]$ cancel, so $\int_0^8r(t)\,\text{d}t=0$. 3. The reading therefore remains $75+0=75$ grams at $t=8$.

Answer

The scale reads $75\,\text{g}$ at $t=8$.

54905612

A theater's reserve fund changes at rate $r(t)$ dollars per month. Over a six-month season, $\int_0^6r(t)\,\text{d}t=-240$. The theater requires at least $\$500$ in the fund at the end of the season. What is the least possible starting balance that meets the requirement?

Hints

- Translate the accumulated rate into the change from starting balance to ending balance. - Express the ending-balance requirement as an inequality. - Find the smallest initial value that satisfies that inequality.

Solution

1. Let $B$ be the starting balance. The ending balance is $B-240$. 2. The requirement is $B-240\ge500$. 3. Thus $B\ge740$, so the least starting balance is $\$740$.

Answer

The least starting balance is $\$740$.

54905712

Let $Q(t)=Q(0)+\int_0^t r(s)\,\text{d}s$, where $r$ is continuous. Determine whether each statement must be true. Justify each answer. a) If $r(3)=0$, then $Q(3)=Q(0)$. b) If $\int_0^5r(s)\,\text{d}s=0$, then $Q(5)=Q(0)$. c) If $r(t)<0$ for $4<t<6$, then $Q(6)<Q(4)$.

Hints

- Distinguish an instantaneous rate statement from an accumulated-change statement. - Rewrite each comparison of quantities as an integral over the corresponding interval. - Use continuity and the sign of the rate over an entire interval where relevant.

Solution

1. a) False. A zero rate at one instant does not determine the accumulated change over $[0, 3]$. 2. b) True. The defining equation gives $Q(5)-Q(0)=\int_0^5r(s)\,\text{d}s=0$. 3. c) True. A negative rate throughout $(4, 6)$ gives $\int_4^6r(t)\,\text{d}t<0$, so $Q(6)-Q(4)<0$.

Answer

a) False b) True c) True

54903612

A fabrication lab begins a shift with $640$ usable parts. The net rate of change in usable inventory is $p(t)$ parts per hour, where rejected parts make $p(t)$ negative. Write, but do not solve, each mathematical statement. a) The inventory after $T$ hours. b) The first time the inventory reaches $800$ parts. c) The condition that the inventory never falls below $500$ parts during the first $10$ hours.

Hints

- Build each statement from an initial amount and a change accumulated up to a variable time. - Distinguish an equation for reaching a level from an inequality for staying above a level. - Words such as “first” and “during the first ten hours” impose conditions on the time variable.

Solution

1. Add accumulated net production to the initial inventory: $640+\int_0^T p(t)\,\text{d}t$. 2. The first hitting time satisfies $640+\int_0^T p(t)\,\text{d}t=800$, with $T$ chosen as the least nonnegative solution. 3. For every $T\in[0, 10]$, require $640+\int_0^T p(t)\,\text{d}t\ge500$.

Answer

a) $640+\int_0^T p(t)\,\text{d}t$ b) $640+\int_0^T p(t)\,\text{d}t=800$, where $T$ is the least nonnegative solution c) $640+\int_0^T p(t)\,\text{d}t\ge500$ for every $T\in[0, 10]$

54904612

Two battery-storage systems begin with the same amount of stored energy. Their net charging rates, $r_A(t)$ and $r_B(t)$, are measured in kilowatts and are shown for $0\le t\le4$, where $t$ is measured in hours. a) Which system contains more stored energy for $0<t<4$? b) Compare their stored energy at $t=4$.

Hints

- Express the difference between the stored-energy amounts as an integral of the rate difference. - Determine the sign of that accumulated difference for $0<t<4$. - At $t=4$, compare the negative and positive signed areas.

Solution

1. The difference in stored energy is $\int_0^t[r_A(s)-r_B(s)]\,\text{d}s=\int_0^t(s-2)\,\text{d}s=\frac{t^2}{2}-2t$. 2. For $0<t<4$, $\frac{t^2}{2}-2t=\frac{t(t-4)}{2}<0$, so System A contains less energy and System B contains more. 3. At $t=4$, the difference is $0$. The negative and positive triangular signed areas under $r_A$ cancel, so the systems contain equal amounts of stored energy.

Answer

a) System B contains more stored energy for every $0<t<4$. b) The systems contain equal amounts of stored energy at $t=4$.

54905112

The graph shows the rate of change $q'(t)$, in degrees Celsius per minute squared, of a reactor's heating rate $q(t)$, where $t$ is measured in minutes. At $t=0$, $q(0)=-2\,^\circ\text{C}/\text{min}$. a) Find $q(4)$. b) Find all times in $[0, 6]$ when $q(t)=0$. c) At which of those times does the reactor switch between cooling and heating?

Hints

- Accumulating $q'(t)$ gives the change in the heating rate $q(t)$. - Begin with the given value $q(0)$ before adding signed areas. - Find where the accumulated heating rate reaches zero. - A switch between cooling and heating occurs only when $q(t)$ changes sign.

Solution

1. The signed area under $q'$ from $0$ to $4$ is $4+0=4$. Thus, $q(4)=-2+4=2\,^\circ\text{C}/\text{min}$. 2. On $[0, 2]$, $q(t)=-2+2t$, so $q(t)=0$ at $t=1$. From $t=4$ onward, $q(t)=2-2(t-4)$, so $q(t)=0$ at $t=5$. 3. The heating rate changes from negative to positive at $t=1$, so the reactor switches from cooling to heating. It changes from positive to negative at $t=5$, so the reactor switches from heating to cooling. Therefore, a switch occurs at both times.

Answer

a) $2\,^\circ\text{C}/\text{min}$ b) $t=1\,\text{min}$ and $t=5\,\text{min}$ c) The reactor switches between cooling and heating at both times.

54905812

A campus shuttle stop has $50$ waiting passengers at $t=0$. Time $t$ is measured in minutes. Its net passenger rate $r(t)$, in passengers per minute, is continuous and positive for $0\le t\le5$. Measurements give $\int_0^2r(t)\,\text{d}t=8$ and $\int_0^5r(t)\,\text{d}t=23$. Show that there is a time $T$ with $2<T<5$ when exactly $70$ passengers are waiting.

Hints

- Find the passenger count at each endpoint of the requested time interval. - Compare the target count with those two endpoint counts. - Use the continuity of an accumulated quantity to justify reaching every intermediate value.

Solution

1. At $t=2$, the number waiting is $50+8=58$. 2. At $t=5$, the number waiting is $50+23=73$. 3. The accumulated passenger count is continuous because $r$ is continuous. 4. Since $70$ lies between $58$ and $73$, the count equals $70$ at some $T\in(2, 5)$.

Answer

There is at least one $T\in(2, 5)$ for which $50+\int_0^T r(t)\,\text{d}t=70$.

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