Exploring Accumulations of Change - AP Calculus AB

A temperature changes at rate $T'(t)=\frac{1}{t^2}$ for $1\le t\le 2$ . What is $T(2)-T(1)$ ?
Answer: $\tfrac{1}{2}$ This problem applies accumulation reasoning to find temperature change from a reciprocal rate function. The rate $T'(t) = \frac{1}{t^2}$ decreases rapidly over $[1,2]$ , so we integrate $\int_1^2 \frac{1}{t^2} \, dt = \int_1^2 t^{-2} \, dt$ . This equals $[-t^{-1}]_1^2 = -\frac{1}{2} - (-\frac{1}{1}) = -\frac{1}{2} + 1 = \frac{1}{2}$ . The negative reciprocal antiderivative captures the decreasing contribution as t increases. Students might choose 1 by evaluating $1/1^2$ at the left endpoint, but this gives the initial rate, not the accumulated change. When integrating power functions with negative exponents, apply the power rule with careful attention to signs.
A runner's velocity is $v(t)=2$ for $0\le t\le 3$ and $v(t)=-1$ for $3\le t\le 5$ . Net displacement?
Answer: $4$ units This problem requires accumulation reasoning to find net displacement from piecewise velocity. The runner moves at v(t) = 2 for 3 seconds, accumulating 2 × 3 = 6 units forward, then at v(t) = -1 for 2 seconds, accumulating -1 × 2 = -2 units backward. The net displacement is 6 + (-2) = 4 units forward. Students might incorrectly choose 8 by adding the absolute values |6| + |2| = 8, but this gives total distance traveled rather than net displacement. For piecewise constant velocity, accumulate each piece separately then sum algebraically to find net displacement.
A function satisfies $f'(x)=-1$ for $-3\le x\le 1$ and $f(-3)=5$ . What is $f(1)$ ?
Answer: $1$ This problem applies accumulation reasoning with an initial condition to find a function value. Given f'(x) = -1 is constant over [-3,1] and f(-3) = 5, the change f(1) - f(-3) = -1×(1-(-3)) = -1×4 = -4. Therefore, f(1) = f(-3) + (-4) = 5 - 4 = 1. The negative constant rate produces linear decrease from the initial value. Students might choose -1 by confusing the rate with the final value, but f(1) equals the initial value plus the accumulated change. When rates of change are negative constants, subtract the accumulated change magnitude from the initial value.
A company's profit changes at rate $R(t)=5$ thousand dollars/month for $0\le t\le3$ and $R(t)=-5$ for $3\le t\le6$ . Net change?
Answer: $0$ thousand dollars This problem uses accumulation reasoning to find the net profit change from a rate function that switches between positive and negative values. For 0 ≤ t ≤ 3: profit increases at 5 thousand dollars/month, so ∫5 dt = 5t evaluated from 0 to 3 gives 15 thousand dollars gain. For 3 ≤ t ≤ 6: profit decreases at 5 thousand dollars/month, so ∫(-5)dt = -5t evaluated from 3 to 6 gives -5(6) - (-5(3)) = -15 thousand dollars loss. The net change is 15 + (-15) = 0 thousand dollars. Choice A (30 thousand) might tempt students who add the magnitudes instead of considering the opposite signs. When rates of equal magnitude but opposite sign act for equal time periods, the net accumulation is zero.
Water flows into a tank at rate $r(t)=3t-6$ liters/min for $0\le t\le4$ . What is the net change in volume?
Answer: $0$ liters This problem requires using accumulation reasoning to find the net change in water volume from the given rate function. The net change equals the integral of r(t) = 3t - 6 from 0 to 4, which represents the total accumulation over the time interval. Computing ∫(3t - 6)dt = (3t²/2 - 6t) evaluated from 0 to 4 gives (3(16)/2 - 24) - 0 = 24 - 24 = 0 liters. The rate starts negative (water flowing out), becomes zero at t = 2, then positive (water flowing in), with equal amounts flowing out and in. Choice B (12 liters) might tempt students who only consider the positive accumulation after t = 2, ignoring the negative accumulation before. When a rate function changes sign, always integrate over the entire interval to find the net accumulation, as positive and negative changes may cancel.
A particle's velocity is $v(t)=\begin{cases}1,&0\le t\le 2\\-3,&2\le t\le 3\end{cases}$ . Net displacement?
Answer: $-1$ unit This problem requires accumulation reasoning to find net displacement from piecewise constant velocity. Over [0,2], the particle moves at v = 1, accumulating 1×2 = 2 units forward. Over [2,3], it moves at v = -3, accumulating -3×1 = -3 units backward. The net displacement is 2 + (-3) = -1 unit, indicating final position is 1 unit behind the starting point. Students might choose 1 by taking the absolute value, but displacement preserves direction unlike distance. For piecewise constant velocities, accumulate each segment separately then combine algebraically.
A tank drains at rate $r(t)=-3$ L/min for $2\le t\le 6$ . What is the net change in volume?
Answer: $-12$ liters This problem applies accumulation reasoning to find volume change from a constant draining rate. The tank drains at r(t) = -3 L/min over the 4-minute interval [2,6], so we accumulate this negative rate over time. The net change equals -3 L/min × (6-2) min = -3 × 4 = -12 liters, indicating the tank loses 12 liters. A student might choose -3 by confusing the rate with the total change, ignoring the 4-minute duration. When dealing with constant negative rates (outflow/draining), multiply the rate by the time interval to find total volume lost.
A runner's velocity is $v(t)=4$ for $0\le t\le2$ and $v(t)=-4$ for $2\le t\le5$ . What is the net displacement on $[0,5]$ ?
Answer: $-4$ This problem applies accumulation reasoning to find net displacement from velocity intervals. From 0 to 2, velocity 4 gives displacement 4 × 2 = 8. From 2 to 5, velocity -4 gives displacement -4 × 3 = -12. The net displacement is 8 + (-12) = -4. A tempting error is to compute 4 × 2 - 4 × 3 = -4 without recognizing this is the same calculation. To find net change from rates, multiply each rate by its time duration and add all results.
A savings account changes at rate $S'(t)=-2$ dollars/week for $0\le t\le5$ . What is the net change in savings?
Answer: $-10$ dollars This problem requires accumulation reasoning to find the net change in savings from a constant negative rate. The net change equals the integral of S'(t) = -2 from 0 to 5, representing the total accumulation (which is negative) over the time period. Computing ∫(-2)dt = -2t evaluated from 0 to 5 gives -2(5) - (-2(0)) = -10 dollars. The negative rate means money is being withdrawn at a constant rate of 2 dollars per week for 5 weeks. Choice B (10 dollars) might attract students who forget the negative sign, thinking of the magnitude only. When the rate of change is constant and negative, the accumulation is simply the rate times the time interval, keeping the negative sign to indicate a decrease.
A function satisfies $f'(x)=\frac{1}{\sqrt{x}}$ for $1\le x\le 4$ . What is $f(4)-f(1)$ ?
Answer: $2$ This problem applies accumulation reasoning to find function change from a square root reciprocal derivative. The rate f'(x) = 1/√x = $x^{-1/2}$ over [1,4] integrates to ∫₁⁴ $x^{-1/2}$ dx = [ $2x^{1/2}$ ]₁⁴ = 2√4 - 2√1 = 2(2) - 2(1) = 4 - 2 = 2. The power rule for integration converts the negative fractional exponent to a positive one with coefficient adjustment. Students might choose 3 by incorrectly applying the power rule, but $x^{-1/2}$ integrates to $2x^{1/2}$ , not $3x^{1/2}$ . When integrating fractional powers, apply the power rule: ∫ $x^n$ dx = $x^{n+1}$ /(n+1) for n ≠ -1.
A car's velocity is $v(t)=3$ for $0\le t\le 2$ and $v(t)=-1$ for $2\le t\le 6$ . What is displacement on $[0,6]$ ?
Answer: $2$ units This problem uses accumulation reasoning to find total displacement from a piecewise velocity function. Displacement equals the integral of velocity, which we compute in two parts: from t = 0 to t = 2, the car travels 3(2 - 0) = 6 units forward, and from t = 2 to t = 6, it travels -1(6 - 2) = -4 units (backward). The net displacement is 6 + (-4) = 2 units forward. A tempting error is to add the distances traveled (6 + 4 = 10), but displacement accounts for direction, not just total distance. To find displacement from velocity, integrate each piece separately and add the signed results to get the net change in position.
A function satisfies $f'(x)=2x-4$ on $0 \le x \le 4$ . What is $f(4)-f(0)$ ?
Answer: $0$ This problem uses accumulation reasoning to find function change from a quadratic derivative that crosses zero. The rate $f'(x) = 2x - 4 = 2(x-2)$ is negative for $x < 2$ and positive for $x > 2$ over $[0,4]$ . We integrate $\int_0^4 (2x-4) \, dx = [x^2 - 4x]_0^4 = (16 - 16) - 0 = 0$ . The symmetric quadratic creates equal negative and positive contributions that cancel exactly. Students might choose -8 by evaluating only the negative part or 8 by considering only the positive part, but the full interval yields zero net change. When quadratic rates are symmetric about their zero, the accumulated change can be zero despite significant variation.
The rate of change of temperature is $T'(t)=4$ °C/hr for $0\le t\le2$ and $T'(t)=-3$ °C/hr for $2\le t\le4$ . Net change?
Answer: $2$ °C This problem requires accumulation reasoning to find the net temperature change from a piecewise constant rate function. For 0 ≤ t ≤ 2: the temperature increases at 4°C/hr, so ∫4 dt = 4t evaluated from 0 to 2 gives 8°C increase. For 2 ≤ t ≤ 4: the temperature decreases at 3°C/hr, so ∫(-3)dt = -3t evaluated from 2 to 4 gives -3(4) - (-3(2)) = -6°C. The net change is 8 + (-6) = 2°C. Choice A (-6°C) might attract students who only consider the cooling phase or make a sign error. For problems with heating and cooling phases, integrate each phase separately and sum algebraically to find the net temperature change.
For $0\le t\le6$ , the rate of change is $R(t)=2$ on $[0,1]$ , $-2$ on $[1,4]$ , and $1$ on $[4,6]$ . What is the net change?
Answer: $-2$ This problem tests accumulation reasoning with multiple rate intervals to find net change. On [0,1], rate 2 gives change 2 × 1 = 2. On [1,4], rate -2 gives change -2 × 3 = -6. On [4,6], rate 1 gives change 1 × 2 = 2. The net change is 2 + (-6) + 2 = -2. Students might incorrectly compute 2 + 2 + 1 = 5 by adding rates without considering durations. For piecewise rates, multiply each rate by its interval length and sum algebraically.
A particle's velocity is $v(t)=|t-2|$ for $0 \le t \le 4$ . What is its displacement?
Answer: $4$ This problem requires accumulation reasoning to find displacement from an absolute value velocity function. Since v(t) = |t-2|, we split at t = 2: for t ∈ [0,2], v(t) = 2-t, and for t ∈ [2,4], v(t) = t-2. We integrate $\int_0^2 (2-t) \, dt + \int_2^4 (t-2) \, dt = [2t - \frac{t^2}{2}]_0^2 + [\frac{t^2}{2} - 2t]_2^4 = (4-2) + (8-8-2+4) = 2 + 2 = 4$ . Both pieces contribute positive displacement since the absolute value ensures non-negative velocity. Students might choose 0 expecting cancellation, but absolute value prevents backward motion. When velocity involves absolute values, split the integral at zeros of the expression inside the absolute value.
A function satisfies $f'(x)=2x$ for $-1 \le x \le 2$ . What is $f(2)-f(-1)$ ?
Answer: $3$ This problem uses accumulation reasoning to find function change from a simple quadratic derivative. The rate $f'(x) = 2x$ over $[-1,2]$ integrates to $\int_{-1}^2 2x \, dx = [x^2]_{-1}^2 = 4 - 1 = 3$ . The quadratic accumulation results from the linear rate, with the definite integral evaluating at both positive and negative endpoints. Students might choose 2 by evaluating the rate at x = 1, but this ignores both the integration process and the full interval extent. When integrating linear functions over intervals that include negative values, evaluate the antiderivative at both endpoints.
A quantity $Q$ changes at rate $Q'(t)=-1$ for $0\le t\le3$ and $Q'(t)=3$ for $3\le t\le4$ . What is $Q(4)-Q(0)$ ?
Answer: $0$ This problem uses accumulation reasoning to find total change from rates over different intervals. From 0 to 3, rate -1 gives change -1 × 3 = -3. From 3 to 4, rate 3 gives change 3 × 1 = 3. The net change is -3 + 3 = 0. Students might incorrectly compute -1 + 3 = 2 by adding rates without considering time durations. For accumulation problems, always multiply rate by duration for each interval before summing.
A tank's net flow rate is $r(t)=6-2t$ gal/min for $0\le t\le 4$ . Net change in volume?
Answer: $8$ gallons This problem requires accumulation reasoning to find volume change from a linear flow rate. The rate r(t) = 6 - 2t changes linearly from 6 gal/min at t = 0 to -2 gal/min at t = 4, so we integrate ∫₀⁴ (6-2t) dt. This equals [6t - t²]₀⁴ = (24 - 16) - 0 = 8 gallons net increase. The positive inflow initially exceeds the growing outflow, resulting in net gain. Students might choose 12 by evaluating only 6t at t = 4, ignoring the negative -t² term that reduces the accumulation. For linear rates, integrate the entire expression to capture the changing flow dynamics.
A particle's velocity is $v(t)=2$ for $0\le t\le 1$ and $v(t)=-2$ for $1\le t\le 3$ . What is displacement on $[0,3]$ ?
Answer: $-2$ units This problem requires accumulation reasoning to find displacement from a piecewise constant velocity. From t = 0 to t = 1, the particle moves 2(1 - 0) = 2 units forward, and from t = 1 to t = 3, it moves -2(3 - 1) = -4 units (backward). The net displacement is 2 + (-4) = -2 units, indicating the particle ends up 2 units behind its starting position. A common error is to add the distances traveled (2 + 4 = 6) without considering direction, but displacement is a signed quantity. For piecewise constant velocities, compute the displacement for each interval and sum the signed results.
A function satisfies $f'(x)=\sec^2 x$ for $0\le x\le \tfrac{\pi}{4}$ . What is $f(\tfrac{\pi}{4})-f(0)$ ?
Answer: $1$ This problem uses accumulation reasoning to find function change from the tangent derivative. The rate f'(x) = sec² x is the derivative of tan(x), so we integrate ∫₀^(π/4) sec² x dx = [tan(x)]₀^(π/4) = tan(π/4) - tan(0) = 1 - 0 = 1. This is another standard integral that appears frequently in trigonometric calculus. Students might choose π/4 by confusing the upper limit with the result, but the integral evaluates to the difference of tangent values. When recognizing standard trigonometric derivative forms, use the corresponding antiderivative to evaluate definite integrals.

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