Exploring Accumulations of Change - AP Calculus BC

A leak changes oil amount at rate $A'(t)=-5\sqrt{t}$ liters/hour for $0\le t\le 4$ ; what is net change in oil?
Answer: $-\frac{80}{3}$ liters This problem uses accumulation reasoning to find net change in oil from the leak rate. The net change equals ∫A'(t)dt = ∫(-5√t)dt from t = 0 to t = 4. Computing: ∫(- $5t^{1/2}$ )dt = -5( $2t^{3/2}$ /3) = - $10t^{3/2}$ /3 evaluated from 0 to 4 gives -10(8)/3 - 0 = -80/3 liters. Choice C incorrectly gives a rate (liters/hour) rather than the accumulated volume change. When integrating power functions with fractional exponents, apply the power rule carefully and track negative signs.
A battery's charge changes at rate $q'(t)=4\cos(2t)$ coulombs/min for $0\le t\le\frac{\pi}{2}$ ; find net change.
Answer: $0$ coulombs This problem uses accumulation reasoning to find net charge change. The net change equals ∫₀^(π/2) 4cos(2t)dt = [2sin(2t)]₀^(π/2) = 2sin(π) - 2sin(0) = 0 - 0 = 0 coulombs. The rate q'(t) = 4cos(2t) oscillates with period π, starting at 4cos(0) = 4 and ending at 4cos(π) = -4, completing exactly one full cycle. Choice C (4 coulombs) might tempt students who evaluate only at t=0, missing the accumulation concept. When a periodic rate function completes full cycles, the net accumulation is zero.
A runner's speed is $v(t)=2+\cos t$ m/s for $0\le t\le2\pi$ ; what distance is traveled?
Answer: $4\pi$ meters This problem assesses the skill of exploring accumulations of change by calculating the total distance traveled from a speed function. The speed is $v(t) = 2 + \cos t$ meters per second over $0$ to $2\pi$ seconds. Since speed is always positive, the distance is the integral of $v(t)$ from $0$ to $2\pi$ , accumulating the rate of distance over time. This gives $[2t + \sin t]$ from $0$ to $2\pi$ = $4\pi + 0 = 4\pi$ meters. A tempting distractor is $2\pi$ meters, from integrating only the constant 2 and ignoring $\cos t$ , but the full function must be included. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.
The rate of change of profit is $R'(t)=t^2-4t$ thousand dollars/month for $0 \le t \le 4$ ; find net change in profit.
Answer: $-\frac{32}{3}$ This problem tests the skill of exploring accumulations of change by using integration to find the net change in profit from a rate of change. The net change is the definite integral of $R'(t) = t^2 - 4t$ from t=0 to t=4, accumulating the monthly profit variations. The antiderivative $\frac{1}{3}t^3 - 2t^2$ evaluated gives $\left( \frac{64}{3} - 32 \right) - 0 = -\frac{32}{3}$ thousand dollars. This negative accumulation results from the rate being negative initially and not fully compensated later. A tempting distractor like 0 might stem from finding where the rate is zero instead of integrating, but it fails as the integral captures the overall deficit. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.
A battery's charge changes at rate $Q'(t)=\frac{6}{t+1}$ coulombs/hour for $0\le t\le 2$ ; find net charge change.
Answer: $6\ln 3$ coulombs This problem requires accumulation reasoning to find net charge change from the charging rate. The net change equals ∫Q'(t)dt = ∫6/(t+1)dt from t = 0 to t = 2. Computing: ∫6/(t+1)dt = 6ln|t+1| evaluated from 0 to 2 gives 6ln(3) - 6ln(1) = 6ln(3) - 0 = 6ln(3) coulombs. Choice E incorrectly includes units of rate (coulombs/hour) rather than accumulated charge (coulombs). To find accumulated change from a rational function rate, integrate using logarithms and evaluate at the bounds.
A fish population changes at rate $F'(t)=t-2$ hundred/month for $1\le t\le5$ ; what is net change?
Answer: $4$ hundred This problem uses accumulation reasoning to find net population change. The net change equals ∫₁⁵(t-2)dt = [t²/2-2t]₁⁵ = (25/2-10)-(1/2-2) = 25/2-10-1/2+2 = 24/2-8 = 12-8 = 4 hundred. The rate F'(t) = t-2 is negative for 1≤t<2 (population decreases) and positive for 2<t≤5 (population increases). Choice B (0) might seem plausible if the negative and positive changes canceled, but the positive accumulation outweighs the negative. When accumulating signed rates, integrate over the entire interval to capture both increases and decreases.
Water flows with rate $r(t)=\frac{6}{t+1}$ L/min for $0\le t\le3$ ; how many liters enter total?
Answer: $6\ln 4$ This problem tests the skill of exploring accumulations of change by using integration to find the total water volume from a flow rate. The total liters entering is the definite integral of r(t) = 6/(t+1) from t=0 to t=3, accumulating the flow over the interval. The antiderivative 6 ln|t+1| evaluated gives 6(ln4 - ln1) = 6 ln4 liters. This reflects the decreasing flow rate still yielding a positive accumulation. A tempting distractor like ln4 might arise from omitting the coefficient 6, but it fails as the integral requires multiplying by that factor. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.
For $0\le t\le 3$ , a population changes at rate $P'(t)=2e^{-t}$ thousand/day; what is total population increase?
Answer: $2(1-e^{-3})$ thousand This problem requires accumulation reasoning to find total population increase from the rate of change. The total increase equals ∫P'(t)dt = ∫ $2e^{-t}$ dt from t = 0 to t = 3. Computing the integral: ∫ $2e^{-t}$ dt = - $2e^{-t}$ evaluated from 0 to 3 gives - $2e^{-3}$ - ( $-2e^0$ ) = - $2e^{-3}$ + 2 = 2(1 - $e^{-3}$ ) thousand. Choice C incorrectly gives a rate (thousand/day) rather than the accumulated population change. To find total change from an exponential rate, integrate carefully and evaluate at the bounds.
A battery's charge changes at $Q'(t)=-\frac{3}{t+1}$ coulombs/min for $0\le t\le3$ ; what is $Q(3)-Q(0)$ ?
Answer: $-3\ln 4$ coulombs This problem assesses the skill of exploring accumulations of change by finding the net change in charge from a rate function. The rate is Q'(t) = -3/(t+1) coulombs per minute over 0 to 3 minutes. The net change Q(3) - Q(0) is the integral of Q'(t) from 0 to 3, accumulating the rate over time. This gives -3 [ln(t+1)] from 0 to 3 = -3(ln4 - ln1) = -3 ln4 coulombs. A tempting distractor is 3 ln4, from omitting the negative sign in the rate, but the sign indicates discharge. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.
A lake's depth changes at $h'(t)=t-2$ cm/day for $0\le t\le5$ ; what is the net change in depth?
Answer: $\dfrac{5}{2}$ cm This problem tests the skill of exploring accumulations of change by calculating the net change in depth from a rate function. The rate is h'(t) = t - 2 cm per day over 0 to 5 days. The net change is the integral of h'(t) from 0 to 5, accumulating the rate over the interval. This yields [t²/2 - 2t] from 0 to 5 = 12.5 - 10 = 2.5 cm, or 5/2 cm. A tempting distractor is 0 cm, from evaluating h'(5) = 3 instead of integrating, but integration captures the full accumulation. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.
A machine's output rate is $R(t)=\frac{10}{1+t^2}$ units/hour for $0\le t\le3$ ; how many units are produced?
Answer: $10\arctan(3)$ units This problem requires accumulation reasoning to find total production. The total units produced equal ∫₀³ 10/(1+t²)dt = 10∫₀³ 1/(1+t²)dt = 10[arctan(t)]₀³ = 10(arctan(3) - arctan(0)) = 10arctan(3) units. The rate R(t) = 10/(1+t²) decreases from 10 at t=0 to 10/10 = 1 at t=3. Choice C incorrectly evaluates the rate at t=3 rather than accumulating over time. To find total accumulation from a rate, integrate the rate function—recognizing 1/(1+t²) as the derivative of arctan(t) is key.
A particle's acceleration is $a(t)=2t$ m/s $^2$ and $v(0)=-3$ m/s; what is change in velocity on $[0,2]$ ?
Answer: $4$ m/s This problem uses accumulation reasoning to find velocity change from acceleration. The change in velocity equals ∫₀² 2t dt = [t²]₀² = 4 - 0 = 4 m/s. Since acceleration a(t) = 2t is always positive on [0,2], velocity increases throughout the interval. The initial velocity v(0) = -3 m/s affects the final velocity (which would be -3 + 4 = 1 m/s) but not the change in velocity. Choice B (1 m/s) incorrectly gives the final velocity rather than the change. To find change in a quantity from its rate of change, integrate the rate—initial conditions don't affect the accumulated change.
A car's velocity is $v(t)=t-2$ m/s for $0 \le t \le 5$ ; what is net displacement over the interval?
Answer: $2.5$ This problem tests the skill of exploring accumulations of change by using integration to find the net displacement from a velocity function. The net displacement is the definite integral of $v(t) = t - 2$ from $t=0$ to $t=5$ , accumulating position changes over the interval. The antiderivative $\frac{1}{2}t^2 - 2t$ evaluated gives $(\frac{25}{2} - 10) - 0 = 2.5$ meters. This positive net results despite initial backward movement, as later forward motion dominates. A tempting distractor like $-10$ might come from evaluating at a point instead of integrating, but it fails because the full integral accounts for the entire interval. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.
Water flows into a tank at rate $r(t)=6-2t$ liters/min for $0\le t\le4$ ; what is net change in volume?
Answer: $8$ liters This problem requires using accumulation reasoning to find the net change in water volume. The net change equals the integral of the rate function: ∫₀⁴(6-2t)dt = [6t-t²]₀⁴ = (24-16)-(0) = 8 liters. Since the rate r(t) = 6-2t starts positive (at t=0, r=6) and becomes negative after t=3, water flows in then out. A common error is computing 6(4)-2(4) = 16, which incorrectly evaluates the rate at one instant rather than accumulating over time. To find net change from a rate, always integrate the rate function over the given interval.
The rate of change of profit is $P'(t)=\frac{10}{t+2}-3$ thousand dollars/week for $0\le t\le 4$ ; find net profit change.
Answer: $10\ln 3-12$ thousand dollars This problem uses accumulation reasoning to find net profit change from a rate function. The net change equals the integral of P'(t) = 10/(t+2) - 3 from t = 0 to t = 4. Computing ∫[10/(t+2) - 3]dt = 10ln|t+2| - 3t, and evaluating from 0 to 4 gives [10ln(6) - 12] - [10ln(2) - 0] = 10ln(6) - 12 - 10ln(2) = 10ln(6/2) - 12 = 10ln(3) - 12 thousand dollars. The answer 10ln(6) - 12 (choice B) forgets to subtract the initial value 10ln(2) when evaluating the definite integral. When computing definite integrals, always evaluate at both limits and subtract: F(b) - F(a).
A lake's water level changes at rate $h'(t)=2\cos(\pi t)$ cm/day for $0\le t\le 1$ ; what is net level change?
Answer: $0$ cm This problem uses accumulation reasoning to find net water level change from the rate. The net change equals $\int_0^1 h'(t) \, dt = \int_0^1 2 \cos(\pi t) \, dt$ . Computing: $\int 2 \cos(\pi t) \, dt = \frac{2}{\pi} \sin(\pi t)$ evaluated from 0 to 1 gives $\frac{2}{\pi} \sin(\pi) - \frac{2}{\pi} \sin(0) = 0 - 0 = 0$ cm. Choice C incorrectly gives a rate (cm/day) rather than the accumulated level change. When integrating trigonometric functions, the result over a complete period often yields zero net change, reflecting oscillatory behavior.
Water flows into a tank at rate $r(t)=6-2t$ liters/min for $0\le t\le4$ ; what is the net change in volume?
Answer: $8$ liters This problem assesses the skill of exploring accumulations of change by calculating the net change in volume from a given flow rate. The flow rate is r(t) = 6 - 2t liters per minute over the interval from t=0 to t=4. The net change in volume is the definite integral of r(t) from 0 to 4, which accumulates the rate over time. Computing this integral yields [6t - t²] from 0 to 4 = 24 - 16 = 8 liters. A tempting distractor is 24 liters, which comes from integrating only the constant term 6 and ignoring the -2t term, but the full rate function must be integrated. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.
For $0\le t\le6$ , a machine's production rate is $p(t)=\begin{cases}2t,&0\le t\le3\\12-2t,&3\le t\le6\end{cases}$ ; how many items are produced?
Answer: $18$ items This problem evaluates the skill of exploring accumulations of change by determining total production from a piecewise rate function. The production rate is 2t for 0 to 3 and 12 - 2t for 3 to 6. Total items produced is the integral over each piece, accumulating the rate over time. This gives [t²] from 0 to 3 = 9, plus [12t - t²] from 3 to 6 = 36 - 27 = 9, totaling 18 items. A tempting distractor is 36 items, from doubling the first integral without computing the second properly, but both pieces must be integrated separately. In general, to find the net accumulation of a quantity from its rate of change, compute the definite integral of the rate function over the specified interval.
For $2 \le t \le 5$ , temperature changes at rate $T'(t) = \frac{1}{t} - 2$ °C/min; what is net temperature change?
Answer: $\ln\!\left(\frac{5}{2}\right) - 6$ °C This problem requires accumulation reasoning to find net temperature change from the rate. The net change equals $\int T'(t) \, dt = \int \left( \frac{1}{t} - 2 \right) dt$ from t = 2 to t = 5. Computing: $\int \left( \frac{1}{t} - 2 \right) dt = \ln |t| - 2t$ evaluated from 2 to 5 gives $(\ln 5 - 10) - (\ln 2 - 4) = \ln 5 - \ln 2 - 6 = \ln\left( \frac{5}{2} \right) - 6$ °C. Choice C incorrectly gives units of rate (°C/min) rather than accumulated temperature change. To find net change, integrate each term separately and combine the results at the bounds.
A reservoir's outflow rate is $q(t)=2+e^{-t}$ m $^3$ /hr for $0\le t\le2$ ; how much water leaves in total?
Answer: $4+(1-e^{-2})$ This problem tests the skill of exploring accumulations of change by using integration to find the total water outflow from a rate. The total water leaving is the definite integral of q(t) = 2 + $e^{-t}$ from t=0 to t=2, accumulating the volume over the interval. The antiderivative 2t - $e^{-t}$ evaluated gives (4 - $e^{-2}$ ) - (0 - 1) = 4 - $e^{-2}$ + 1 = 5 - $e^{-2}$ , or equivalently 4 + (1 - $e^{-2}$ ) m³. This accumulation reflects the decreasing but positive outflow rate. A tempting distractor like 4 + $e^{-2}$ might result from signing the exponential integral incorrectly, but it fails as the proper antiderivative yields the subtraction. In general, to find the net accumulation over an interval, compute the definite integral of the rate function from the lower to upper limit.

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