AB Practice Exam 1 — AP Calculus AB/BC

Section II, Part A · Graphing calculator required

Question 1 9 points CalculatorDifficulty 3/5

A municipal reservoir is monitored over a 12-hour period, where $t$ is measured in hours and $0 \le t \le 12$. Water flows **into** the reservoir at a rate of $R(t)$ gallons per hour, where $R$ is a differentiable function whose values at selected times are given in the table below. | $t$ (hours) | 0 | 2 | 5 | 8 | 12 | |---|---|---|---|---|---| | $R(t)$ (gal/hr) | 0 | 24 | 60 | 48 | 12 | During the same period, water is released **out** of the reservoir at a rate modeled by $$D(t) = 30 + 18\sin\!\left(\tfrac{t}{4}\right) \text{ gallons per hour},$$ for $0 \le t \le 12$. At time $t = 0$ the reservoir contains $400$ gallons of water. (a) Using a trapezoidal sum with the four subintervals indicated by the table, estimate the total amount of water that flows into the reservoir over the 12-hour period. Show the computation that leads to your answer. (b) Find the value of $\displaystyle\int_0^{12} D(t)\,dt$. Using correct units, explain the meaning of this value in the context of the problem. (c) The average value of the inflow rate $R$ over the interval $0 \le t \le 12$ is $36$ gallons per hour. For $5 \le t \le 8$, must there be a time $t = c$ at which the instantaneous rate of change of $R$ equals $-4$ gallons per hour per hour? Justify your answer. (d) Let $W(t)$ be the total amount of water (in gallons) in the reservoir at time $t$. Find $W'(8)$. Using correct units, interpret whether the amount of water in the reservoir is increasing or decreasing at $t = 8$.

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Part (a)2 PTS

Using a trapezoidal sum with the four subintervals $[0,2],[2,5],[5,8],[8,12]$ from the table, estimate the total amount of water that flows into the reservoir over $0 \le t \le 12$.

The total inflow is $\int_0^{12} R(t)\,dt$, estimated by a trapezoidal sum over the four (unequal) subintervals. Each trapezoid contributes $(\Delta t)\cdot\frac{R(\text{left})+R(\text{right})}{2}$: $$\int_0^{12} R(t)\,dt \approx 2\cdot\frac{0+24}{2} + 3\cdot\frac{24+60}{2} + 3\cdot\frac{60+48}{2} + 4\cdot\frac{48+12}{2}.$$ Evaluating each term: $$= 24 + 126 + 162 + 120 = 432.$$ The estimated total inflow over the 12 hours is $\boxed{432}$ gallons.

<span class="math-block">\[\int_0^{12} R(t)\,dt \approx \sum (\Delta t)\,\frac{R_{\text{left}}+R_{\text{right}}}{2}\]</span>

<span class="math-block">\[= 2\cdot\tfrac{0+24}{2} + 3\cdot\tfrac{24+60}{2} + 3\cdot\tfrac{60+48}{2} + 4\cdot\tfrac{48+12}{2}\]</span>

<span class="math-block">\[= 24 + 126 + 162 + 120 = 432 \text{ gallons}\]</span>

2 points: 1 for a correct trapezoidal sum setup using the four table subintervals (correct widths 2,3,3,4 and averaged heights); 1 for the answer 432 gallons. Key error to avoid: using equal widths (e.g. treating all subintervals as width 3) instead of the actual spacing from the table, or dropping the 1/2 averaging factor (which would give a right- or left-Riemann sum, not a trapezoidal sum).

Part (b)2 PTS

Find $\int_0^{12} D(t)\,dt$ and explain its meaning with correct units.

Using a calculator to evaluate the definite integral of the outflow model, $$\int_0^{12} D(t)\,dt = \int_0^{12}\!\left(30 + 18\sin\tfrac{t}{4}\right)dt = 432 - 72\cos(3) \approx 503.279.$$ **Meaning:** Because $D(t)$ is the rate (gallons per hour) at which water is released, integrating it accumulates a quantity of water. The value $\int_0^{12} D(t)\,dt \approx 503.279$ gallons is the **total amount of water released out of the reservoir** during the 12-hour period $0 \le t \le 12$.

<span class="math-block">\[\int_0^{12}\left(30+18\sin\tfrac{t}{4}\right)dt = \Big[30t - 72\cos\tfrac{t}{4}\Big]_0^{12}\]</span>

<span class="math-block">\[= (360 - 72\cos 3) - (0 - 72) = 432 - 72\cos 3 \approx 503.279\]</span>

<span class="math-block">\[\text{Interpretation: total gallons released out of the reservoir over } 0\le t\le 12\]</span>

2 points: 1 for the numerical value $\approx 503.279$ gallons (calculator); 1 for the interpretation as the total volume of water released out of the reservoir over the 12 hours, with units of gallons. Key error to avoid: calling it a rate (it is an accumulated amount, not gal/hr) or stating it as a net change in the reservoir (it is only the outflow, not inflow minus outflow).

Part (c)2 PTS

On $5 \le t \le 8$, must there be a time $t=c$ where $R'(c) = -4$ gal/hr per hr? Justify.

Yes. $R$ is differentiable on $[0,12]$, so $R$ is continuous on $[5,8]$ and differentiable on $(5,8)$. The average rate of change of $R$ over $[5,8]$ is $$\frac{R(8)-R(5)}{8-5} = \frac{48-60}{3} = \frac{-12}{3} = -4 \text{ gal/hr per hr}.$$ Because the hypotheses of the **Mean Value Theorem** are satisfied, there exists a time $c$ with $5 < c < 8$ such that $R'(c)$ equals this average rate of change, namely $R'(c) = -4$ gallons per hour per hour. So such a time must exist.

<span class="math-block">\[R \text{ differentiable on } [0,12] \Rightarrow \text{continuous on } [5,8],\ \text{differentiable on } (5,8)\]</span>

<span class="math-block">\[\frac{R(8)-R(5)}{8-5} = \frac{48-60}{3} = -4\]</span>

<span class="math-block">\[\text{MVT} \Rightarrow \exists\, c\in(5,8): R'(c) = -4 \text{ gal/hr per hr}\]</span>

2 points: 1 for stating that $R$ is differentiable/continuous on $[5,8]$ so the Mean Value Theorem applies; 1 for computing the average rate of change $\frac{48-60}{3}=-4$ and concluding such a $c$ exists. Key error to avoid: invoking the IVT on $R$ itself (the question is about the derivative's value, which requires the MVT), or omitting the hypothesis check that $R$ is differentiable.

Part (d)3 PTS

Find $W'(8)$ where $W(t)$ is the volume in the reservoir, and state whether the volume is increasing or decreasing at $t=8$.

The reservoir's volume changes at the net rate inflow minus outflow: $$W'(t) = R(t) - D(t).$$ At $t = 8$, the table gives $R(8) = 48$ gal/hr, and the model gives $$D(8) = 30 + 18\sin\!\left(\tfrac{8}{4}\right) = 30 + 18\sin(2) \approx 46.367 \text{ gal/hr}.$$ Therefore $$W'(8) = R(8) - D(8) = 48 - (30 + 18\sin 2) = 18 - 18\sin 2 \approx 1.633 \text{ gallons per hour}.$$ Since $W'(8) \approx 1.633 > 0$, the amount of water in the reservoir is **increasing** at time $t = 8$ (water is flowing in faster than it is being released).

<span class="math-block">\[W'(t) = R(t) - D(t)\]</span>

<span class="math-block">\[R(8) = 48,\quad D(8) = 30 + 18\sin 2 \approx 46.367\]</span>

<span class="math-block">\[W'(8) = 48 - (30+18\sin 2) = 18 - 18\sin 2 \approx 1.633 &gt; 0 \Rightarrow \text{increasing}\]</span>

3 points: 1 for the relationship $W'(t)=R(t)-D(t)$ (rate in minus rate out); 1 for the value $W'(8) \approx 1.633$ gal/hr (using $R(8)=48$ and $D(8)\approx 46.367$); 1 for the correct conclusion 'increasing' with reason $W'(8)>0$. Key error to avoid: using $W'(8)=R(8)+D(8)$ (adding instead of subtracting outflow), or concluding the trend from the sign of $R$ or $D$ alone rather than from the sign of their difference.

Question 2 9 points CalculatorDifficulty 4/5

A particle moves along a horizontal line for $0 \le t \le 8$, where $t$ is measured in seconds. The velocity of the particle is given by $$v(t) = (t - 3)\,e^{-t/4} \text{ meters per second}.$$ At time $t = 0$, the particle is at position $x(0) = 10$ meters. A calculator may be used on this problem. (a) For $0 \le t \le 8$, find all times $t$ at which the particle is at rest, and determine the time interval(s) on which the particle is moving in the negative direction. Justify your answer. (b) Find the acceleration of the particle at time $t = 5$. Is the speed of the particle increasing or decreasing at $t = 5$? Give a reason for your answer. (c) Find the displacement of the particle over the interval $0 \le t \le 8$, and use it to find the particle's position $x(8)$. (d) Find the total distance traveled by the particle over the interval $0 \le t \le 8$.

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Part (a)2 PTS

Find all rest times on $0 \le t \le 8$ and the interval(s) where the particle moves in the negative direction; justify.

The particle is at rest when $v(t) = 0$. Since $e^{-t/4} > 0$ for all $t$, we need $t - 3 = 0$, so the only rest time on $[0,8]$ is $$t = 3 \text{ seconds}.$$ The sign of $v(t)$ is the sign of the factor $(t-3)$ (because $e^{-t/4}>0$): - For $0 \le t < 3$: $\;t-3 < 0$, so $v(t) < 0$ — the particle moves in the **negative** direction. - For $3 < t \le 8$: $\;t-3 > 0$, so $v(t) > 0$ — the particle moves in the positive direction. Therefore the particle moves in the negative direction on the interval $0 \le t < 3$, and it is at rest only at $t = 3$.

<span class="math-block">\[v(t) = (t-3)e^{-t/4} = 0 \Rightarrow t-3 = 0 \ (\text{since } e^{-t/4}&gt;0)\]</span>

<span class="math-block">\[t = 3 \text{ s (only rest time on } [0,8])\]</span>

<span class="math-block">\[v(t) &lt; 0 \text{ for } 0\le t&lt;3 \Rightarrow \text{negative direction on } [0,3)\]</span>

2 points: 1 for the rest time $t=3$ (with the observation that $e^{-t/4}\neq 0$, so only the $(t-3)$ factor matters); 1 for stating the particle moves in the negative direction on $0\le t<3$ with the justification $v(t)<0$ there. Key error to avoid: claiming additional rest times from the exponential factor, or reversing the direction (the sign of $v$, not of position, determines direction).

Part (b)3 PTS

Find $a(5)$ and state whether the speed is increasing or decreasing at $t=5$, with reasoning.

Differentiate $v(t) = (t-3)e^{-t/4}$ using the product rule: $$a(t) = v'(t) = e^{-t/4} + (t-3)\!\left(-\tfrac14 e^{-t/4}\right) = e^{-t/4}\!\left[1 - \tfrac14(t-3)\right] = \frac{(7 - t)}{4}\,e^{-t/4}.$$ At $t = 5$: $$a(5) = \frac{7-5}{4}e^{-5/4} = \tfrac12 e^{-5/4} \approx 0.143 \text{ m/s}^2.$$ To decide whether **speed** is increasing, compare the signs of $v(5)$ and $a(5)$: $$v(5) = (5-3)e^{-5/4} = 2e^{-5/4} \approx 0.573 > 0, \qquad a(5) \approx 0.143 > 0.$$ Since $v(5)$ and $a(5)$ have the **same sign** (both positive), the speed of the particle is **increasing** at $t = 5$.

<span class="math-block">\[a(t) = v'(t) = e^{-t/4} - \tfrac14(t-3)e^{-t/4} = \tfrac{7-t}{4}e^{-t/4}\]</span>

<span class="math-block">\[a(5) = \tfrac{2}{4}e^{-5/4} = \tfrac12 e^{-5/4} \approx 0.143 \text{ m/s}^2\]</span>

<span class="math-block">\[v(5)=2e^{-5/4}\approx 0.573&gt;0,\ a(5)&gt;0 \Rightarrow \text{same sign} \Rightarrow \text{speed increasing}\]</span>

3 points: 1 for a correct acceleration expression $a(t)=\frac{7-t}{4}e^{-t/4}$ (product rule); 1 for $a(5)\approx 0.143\,\text{m/s}^2$; 1 for the conclusion 'speed increasing' justified by $v(5)$ and $a(5)$ having the same sign. Key error to avoid: concluding from $a(5)>0$ alone that speed is increasing — speed increases only when velocity and acceleration share a sign, so the sign of $v(5)$ must also be checked.

Part (c)2 PTS

Find the displacement over $0\le t\le 8$ and the position $x(8)$.

Displacement is the integral of velocity: $$\int_0^8 v(t)\,dt = \int_0^8 (t-3)e^{-t/4}\,dt = 4 - 36e^{-2} \approx -0.872 \text{ meters}.$$ (The exact antiderivative is $\int (t-3)e^{-t/4}\,dt = -4(t+1)e^{-t/4}$; a calculator gives the value directly.) The displacement is negative, so the particle ends slightly to the left of where it started. The position at $t=8$ is $$x(8) = x(0) + \int_0^8 v(t)\,dt = 10 + (4 - 36e^{-2}) \approx 10 + (-0.872) = 9.128 \text{ meters}.$$

<span class="math-block">\[\text{displacement} = \int_0^8 (t-3)e^{-t/4}\,dt = 4 - 36e^{-2} \approx -0.872 \text{ m}\]</span>

<span class="math-block">\[x(8) = x(0) + \int_0^8 v(t)\,dt = 10 + (4-36e^{-2})\]</span>

<span class="math-block">\[\approx 10 + (-0.872) = 9.128 \text{ m}\]</span>

2 points: 1 for the displacement integral $\int_0^8 v(t)\,dt \approx -0.872$ m; 1 for $x(8)=x(0)+\int_0^8 v\,dt \approx 9.128$ m. Key error to avoid: integrating $|v(t)|$ here (that gives total distance, not displacement), or forgetting to add the initial position $x(0)=10$ when finding $x(8)$.

Part (d)2 PTS

Find the total distance traveled over $0\le t\le 8$.

Total distance is the integral of speed, $\int_0^8 |v(t)|\,dt$. Since $v(t)<0$ on $[0,3)$ and $v(t)>0$ on $(3,8]$ (from part (a)), split at the rest time $t=3$: $$\int_0^8 |v(t)|\,dt = -\int_0^3 v(t)\,dt + \int_3^8 v(t)\,dt.$$ Evaluating each piece (by calculator): $$\int_0^3 v(t)\,dt \approx -3.558, \qquad \int_3^8 v(t)\,dt \approx 2.686.$$ Therefore $$\int_0^8 |v(t)|\,dt \approx |-3.558| + 2.686 = 3.558 + 2.686 = 6.244 \text{ meters}.$$ The total distance traveled is approximately $6.244$ meters.

<span class="math-block">\[\text{total distance} = \int_0^8 |v(t)|\,dt = -\int_0^3 v\,dt + \int_3^8 v\,dt\]</span>

<span class="math-block">\[\int_0^3 v\,dt \approx -3.558,\quad \int_3^8 v\,dt \approx 2.686\]</span>

<span class="math-block">\[= 3.558 + 2.686 = 6.244 \text{ m}\]</span>

2 points: 1 for setting up total distance as $\int_0^8 |v(t)|\,dt$ (or equivalently splitting at $t=3$ with the correct signs); 1 for the value $\approx 6.244$ m. Key error to avoid: reporting the displacement ($\approx -0.872$ m) as the total distance — the sub-interval where $v<0$ must contribute its absolute value, not its signed value.

Section II, Part B · No calculator

Question 3 9 points No calculatorDifficulty 4/5

Consider the differential equation $\dfrac{dy}{dx}=\dfrac{x(y+1)}{2}$. Let $y=f(x)$ be the particular solution to this differential equation with the initial condition $f(2)=1$. (a) On the axes provided, a slope field for the differential equation is drawn at selected points in the $xy$-plane. Describe the slope field: state along which line every slope is $0$, state the sign of the slope in the region where $x>0$ and $y>-1$, and explain how the slopes behave as $x$ increases (with $y>-1$ fixed). (b) Write an equation for the line tangent to the graph of $f$ at the point $(2,1)$, and use it to approximate $f(2.1)$. (c) Find $\dfrac{d^{2}y}{dx^{2}}$ in terms of $x$ and $y$. Determine whether the approximation found in part (b) is an overestimate or an underestimate of $f(2.1)$, and justify your answer. (d) Find the particular solution $y=f(x)$ to the differential equation with the initial condition $f(2)=1$.

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Part (a)2 PTS

Describe the slope field: where all slopes are 0, the sign of the slope for $x>0,\,y>-1$, and how slopes change as $x$ increases.

Because $\dfrac{dy}{dx}=\dfrac{x(y+1)}{2}$, the slope is $0$ exactly when the numerator is $0$, i.e. when $x=0$ (the entire $y$-axis) or when $y=-1$ (the horizontal line $y=-1$). So along the $y$-axis and along the line $y=-1$ every drawn segment is horizontal. In the region where $x>0$ and $y>-1$, both factors $x$ and $(y+1)$ are positive, so $\dfrac{dy}{dx}>0$; the segments rise from left to right (positive slope). For a fixed $y>-1$, as $x$ increases the factor $x$ grows, so $\dfrac{dy}{dx}$ increases: the segments become steeper (more steeply positive) as you move to the right.

<span class="math-block">\[\frac{dy}{dx}=0 \iff x=0 \text{ or } y=-1\]</span>

<span class="math-block">\[x&gt;0,\ y&gt;-1 \Rightarrow x(y+1)&gt;0 \Rightarrow \frac{dy}{dx}&gt;0\]</span>

<span class="math-block">\[\text{fix } y&gt;-1:\ \frac{dy}{dx}=\tfrac{(y+1)}{2}\,x \text{ increases as } x \text{ increases}\]</span>

2 points: 1 for identifying both zero-slope loci ($x=0$ and $y=-1$); 1 for the correct sign (positive) in the stated region together with the steepening-as-$x$-increases behavior. Naming only one zero-slope locus, or giving the wrong sign, loses a point.

Part (b)2 PTS

Write the tangent line to $f$ at $(2,1)$ and approximate $f(2.1)$.

At $(2,1)$ the slope is $$\left.\frac{dy}{dx}\right|_{(2,1)}=\frac{2(1+1)}{2}=2.$$ The tangent line is $$y=1+2(x-2).$$ Using it to approximate $f(2.1)$: $$f(2.1)\approx 1+2(2.1-2)=1+2(0.1)=1.2.$$

<span class="math-block">\[\left.\frac{dy}{dx}\right|_{(2,1)}=\frac{2(1+1)}{2}=2\]</span>

<span class="math-block">\[y=1+2(x-2)\]</span>

<span class="math-block">\[f(2.1)\approx 1+2(0.1)=1.2\]</span>

2 points: 1 for the correct tangent-line equation (slope $2$ at $(2,1)$); 1 for the approximation $f(2.1)\approx 1.2$. A slope error cascades; forgetting to evaluate at $x=2.1$ loses the approximation point.

Part (c)3 PTS

Find $d^2y/dx^2$ in terms of $x,y$; decide over/under-estimate and justify.

Differentiate $\dfrac{dy}{dx}=\dfrac{x(y+1)}{2}=\dfrac12\big(xy+x\big)$ with respect to $x$, using the product rule on $xy$ and remembering $y'=\dfrac{dy}{dx}$: $$\frac{d^{2}y}{dx^{2}}=\frac12\left(y+x\frac{dy}{dx}+1\right)=\frac12\left(y+1+x\cdot\frac{x(y+1)}{2}\right)=\frac{(y+1)}{2}\left(1+\frac{x^{2}}{2}\right).$$ At $(2,1)$: $$\frac{d^{2}y}{dx^{2}}=\frac{(1+1)}{2}\left(1+\frac{4}{2}\right)=1\cdot 3=3>0.$$ Near $x=2$ the solution is concave up ($y''>0$), so its graph lies **above** its tangent line. Therefore the tangent-line value $1.2$ is an **underestimate** of $f(2.1)$.

<span class="math-block">\[\frac{d^{2}y}{dx^{2}}=\frac12\!\left(y+x\,\frac{dy}{dx}+1\right)=\frac{(y+1)}{2}\!\left(1+\frac{x^{2}}{2}\right)\]</span>

<span class="math-block">\[\left.\frac{d^{2}y}{dx^{2}}\right|_{(2,1)}=1\cdot 3=3&gt;0\]</span>

<span class="math-block">\[y''&gt;0 \Rightarrow \text{concave up} \Rightarrow \text{tangent line below curve} \Rightarrow \text{underestimate}\]</span>

3 points: 1 for correctly differentiating (product rule on $xy$ and substituting $y'$); 1 for a correct simplified $d^2y/dx^2$ and showing it is positive at $(2,1)$; 1 for concluding 'underestimate' with the concavity justification. Treating $y$ as constant (omitting the $x\,y'$ term) loses the first point and usually cascades.

Part (d)2 PTS

Solve the IVP by separation of variables.

Separate variables (valid where $y+1\neq 0$; here $y(2)=1>-1$): $$\frac{dy}{y+1}=\frac{x}{2}\,dx.$$ Integrate both sides: $$\ln|y+1|=\frac{x^{2}}{4}+C.$$ Apply $f(2)=1$: $\ln|1+1|=\dfrac{4}{4}+C$, so $\ln 2=1+C$ and $C=\ln 2-1$. Then $$\ln(y+1)=\frac{x^{2}}{4}+\ln 2-1.$$ Exponentiate (with $y+1>0$): $$y+1=e^{\frac{x^{2}}{4}+\ln 2-1}=2\,e^{\frac{x^{2}}{4}-1},\qquad\boxed{\,y=f(x)=2\,e^{\frac{x^{2}-4}{4}}-1\,}.$$ Check: $f(2)=2e^{0}-1=1.$

<span class="math-block">\[\frac{dy}{y+1}=\frac{x}{2}\,dx\]</span>

<span class="math-block">\[\ln|y+1|=\frac{x^{2}}{4}+C\]</span>

<span class="math-block">\[\ln 2=1+C \Rightarrow C=\ln 2-1\]</span>

<span class="math-block">\[y=2\,e^{\frac{x^{2}-4}{4}}-1\]</span>

2 points: 1 for separating and antidifferentiating correctly to $\ln|y+1|=x^2/4+C$ (with constant of integration); 1 for using the initial condition and solving for $y$ explicitly. Dropping $+C$, or leaving the answer implicit, loses the second point.

Question 4 9 points No calculatorDifficulty 4/5

The continuous function $f$ is defined on the closed interval $-2\le x\le 8$. The graph of $f$ consists of three line segments and one semicircle, as described below (all coordinates are exact): - a line segment from $(-2,0)$ up to $(0,3)$; - the **lower** half of the circle of radius $2$ centered at $(2,0)$, for $0\le x\le 4$ (this semicircle lies on or below the $x$-axis, reaching its lowest point $(2,-2)$); - a line segment from $(4,0)$ up to $(6,2)$; - a line segment from $(6,2)$ down to $(8,0)$. Thus $f(-2)=0$, $f(0)=3$, $f(2)=-2$, $f(4)=0$, $f(6)=2$, and $f(8)=0$. Let $g$ be the function defined by $g(x)=\displaystyle\int_{0}^{x} f(t)\,dt$. (a) Find $g(-2)$ and $g(6)$. Show the computations using signed areas. (b) Find $g'(2)$ and $g''(2)$. Justify your answers. (c) For $-2< x< 8$, find the $x$-coordinate of each point at which $g$ has a relative (local) maximum and each at which $g$ has a relative minimum. Justify. Also find all $x$-values at which the graph of $g$ has a point of inflection, with justification. (d) Find the absolute minimum value of $g$ on the closed interval $-2\le x\le 8$, and the $x$-value where it occurs. Justify.

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Part (a)2 PTS

Evaluate $g(-2)$ and $g(6)$ using signed areas.

**$g(-2)$:** $g(-2)=\displaystyle\int_{0}^{-2} f(t)\,dt=-\int_{-2}^{0} f(t)\,dt.$ On $[-2,0]$ the graph is the segment from $(-2,0)$ to $(0,3)$, a triangle above the axis with base $2$ and height $3$, area $\tfrac12(2)(3)=3$. So $\int_{-2}^{0}f=3$ and $$g(-2)=-3.$$ **$g(6)$:** $g(6)=\displaystyle\int_{0}^{6} f(t)\,dt$. Split into $[0,4]$ (lower semicircle, signed area negative) and $[4,6]$ (triangle above axis). - $[0,4]$: the lower half-disk of radius $2$ has area $\tfrac12\pi(2)^2=2\pi$, but it is below the axis, so its signed area is $-2\pi$. - $[4,6]$: triangle from $(4,0)$ to $(6,2)$, base $2$, height $2$, area $\tfrac12(2)(2)=2$ (above axis). Thus $$g(6)=-2\pi+2=2-2\pi.$$

<span class="math-block">\[g(-2)=-\int_{-2}^{0} f=-\tfrac12(2)(3)=-3\]</span>

<span class="math-block">\[\int_{0}^{4} f=-\tfrac12\pi(2)^2=-2\pi\]</span>

<span class="math-block">\[\int_{4}^{6} f=\tfrac12(2)(2)=2\]</span>

<span class="math-block">\[g(6)=-2\pi+2=2-2\pi\]</span>

2 points: 1 for $g(-2)=-3$ (note the reversed limits give a sign flip); 1 for $g(6)=2-2\pi$ (semicircle area $2\pi$ taken as negative, plus triangle $2$). Forgetting the sign flip on $g(-2)$, or using $\pi r^2$ instead of $\tfrac12\pi r^2$ for the half-disk, loses a point.

Part (b)2 PTS

Find $g'(2)$ and $g''(2)$ with justification.

By the Fundamental Theorem of Calculus, $g'(x)=f(x)$. The point $(2,-2)$ is the bottom of the lower semicircle, so $$g'(2)=f(2)=-2.$$ Also $g''(x)=f'(x)$. At $x=2$ the semicircle has a horizontal tangent (the lowest point of the arc), so $f'(2)=0$, giving $$g''(2)=f'(2)=0.$$

<span class="math-block">\[g'(x)=f(x)\ \text{(FTC)} \Rightarrow g'(2)=f(2)=-2\]</span>

<span class="math-block">\[g''(x)=f'(x) \Rightarrow g''(2)=f'(2)=0\ \text{(bottom of semicircle: horizontal tangent)}\]</span>

2 points: 1 for $g'(2)=f(2)=-2$ via FTC; 1 for $g''(2)=f'(2)=0$ because the semicircle's tangent at its lowest point is horizontal. Confusing $g''$ with $f$ (instead of $f'$) loses the second point.

Part (c)3 PTS

Locate relative extrema of $g$ and inflection points of $g$, with justification.

Since $g'=f$, extrema of $g$ occur where $f$ changes sign. Reading the graph: $f>0$ on $(-2,0)$, $f<0$ on $(0,4)$, $f>0$ on $(4,8)$. - At $x=0$, $f$ changes from $+$ to $-$, so $g$ has a **relative maximum at $x=0$**. - At $x=4$, $f$ changes from $-$ to $+$, so $g$ has a **relative minimum at $x=4$**. Inflection points of $g$ occur where $g''=f'$ changes sign, i.e. where the slope of $f$ changes sign: - At $x=0$: just left, $f$ is the line of slope $\tfrac32$ ($f'>0$); just right, the semicircle has $f'<0$. Sign change $+\to-$, so $g$ has an inflection point at $x=0$. - At $x=2$: along the semicircle $f'<0$ for $x<2$ and $f'>0$ for $x>2$ (it bottoms out at $x=2$). Sign change $-\to+$, inflection at $x=2$. - At $x=6$: just left, segment $(4,6)$ has slope $+1$; just right, segment $(6,8)$ has slope $-1$. Sign change $+\to-$, inflection at $x=6$. (At $x=4$, $f'$ goes from $+\infty$-like along the semicircle to $+1$ on the next segment — it stays positive, so there is **no** inflection at $x=4$.) So: relative max at $x=0$, relative min at $x=4$; inflection points at $x=0,\ 2,\ 6$.

<span class="math-block">\[g'=f:\ f:+\,(-2,0),\ -\,(0,4),\ +\,(4,8)\]</span>

<span class="math-block">\[x=0:\ f:+\to- \Rightarrow \text{rel. max of } g\]</span>

<span class="math-block">\[x=4:\ f:-\to+ \Rightarrow \text{rel. min of } g\]</span>

<span class="math-block">\[g''=f' \text{ sign change at } x=0,\,2,\,6 \Rightarrow \text{inflection of } g\]</span>

3 points: 1 for relative max at $x=0$ (with $f$ sign-change justification); 1 for relative min at $x=4$; 1 for the complete inflection set $x=0,2,6$ justified by $g''=f'$ sign changes (and correctly excluding $x=4$). Listing extrema without the sign-change reasoning, or missing $x=2$ in the inflection set, costs the relevant point.

Part (d)2 PTS

Find the absolute minimum of $g$ on $[-2,8]$ and where it occurs.

On a closed interval the absolute minimum of $g$ occurs at an endpoint or an interior critical point (where $g'=f=0$ with $f$ changing $-\to+$). The interior candidate from part (c) is the relative minimum $x=4$; compare with the endpoints $x=-2$ and $x=8$. Compute the values (using signed areas as in part (a)): - $g(-2)=-3$. - $g(4)=\displaystyle\int_0^4 f=-2\pi$ (the full lower semicircle). - $g(8)=g(6)+\displaystyle\int_6^8 f=(2-2\pi)+\tfrac12(2)(2)=4-2\pi$. Numerically $-2\pi\approx-6.283$, $4-2\pi\approx-2.283$, and $-3$. The smallest is $-2\pi$. Therefore the absolute minimum value of $g$ on $[-2,8]$ is $$g(4)=-2\pi,\quad\text{occurring at } x=4.$$

<span class="math-block">\[g(-2)=-3,\quad g(4)=\int_0^4 f=-2\pi,\quad g(8)=(2-2\pi)+2=4-2\pi\]</span>

<span class="math-block">\[-2\pi\approx-6.283 &lt; 4-2\pi\approx-2.283 &lt; -3? \Rightarrow \min = -2\pi\]</span>

<span class="math-block">\[\text{absolute min } g(4)=-2\pi \text{ at } x=4\]</span>

2 points: 1 for identifying and evaluating all candidates ($x=-2,4,8$) with correct values $-3,\,-2\pi,\,4-2\pi$; 1 for selecting the absolute minimum $g(4)=-2\pi$ at $x=4$. Omitting an endpoint candidate, or comparing $g'$ values instead of $g$ values, loses a point.

Question 5 9 points No calculatorDifficulty 3/5

Let $R$ be the region in the first quadrant bounded above by the parabola $y=6x-x^{2}$ and below by the line $y=x$. The two graphs intersect where $6x-x^{2}=x$, that is $x^{2}-5x=0$, so $x=0$ and $x=5$; on the interval $0\le x\le 5$ the parabola lies on or above the line. (a) Find the area of region $R$. (b) Region $R$ is the base of a solid. For this solid, each cross section perpendicular to the $x$-axis is a square. Find the volume of the solid. (c) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when region $R$ is revolved about the $x$-axis.

Show worked solution

Part (a)3 PTS

Find the area of $R$.

On $[0,5]$ the top curve is $y=6x-x^{2}$ and the bottom is $y=x$, so the area is $$A=\int_{0}^{5}\big[(6x-x^{2})-x\big]\,dx=\int_{0}^{5}\big(5x-x^{2}\big)\,dx.$$ Antidifferentiate: $$A=\left[\frac{5x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{5}=\frac{5(25)}{2}-\frac{125}{3}=\frac{125}{2}-\frac{125}{3}=\frac{375-250}{6}=\frac{125}{6}.$$ The area of $R$ is $\dfrac{125}{6}\ \big(\approx 20.833\big).$

<span class="math-block">\[A=\int_{0}^{5}\big((6x-x^{2})-x\big)\,dx=\int_{0}^{5}(5x-x^{2})\,dx\]</span>

<span class="math-block">\[=\left[\tfrac{5x^{2}}{2}-\tfrac{x^{3}}{3}\right]_{0}^{5}\]</span>

<span class="math-block">\[=\tfrac{125}{2}-\tfrac{125}{3}=\frac{125}{6}\]</span>

3 points: 1 for the integrand (top $-$ bottom $=5x-x^2$); 1 for a correct antiderivative; 1 for the value $125/6$. Reversing top and bottom, or using wrong limits, loses the integrand point and cascades.

Part (b)4 PTS

Volume of the solid with square cross sections perpendicular to the $x$-axis.

For a square cross section, the side length equals the vertical extent of $R$ at $x$, namely $s(x)=(6x-x^{2})-x=5x-x^{2}$. The cross-sectional area is $$A(x)=s(x)^{2}=(5x-x^{2})^{2}.$$ The volume is $$V=\int_{0}^{5}(5x-x^{2})^{2}\,dx=\int_{0}^{5}\big(25x^{2}-10x^{3}+x^{4}\big)\,dx.$$ Antidifferentiate: $$V=\left[\frac{25x^{3}}{3}-\frac{10x^{4}}{4}+\frac{x^{5}}{5}\right]_{0}^{5}=\left[\frac{25x^{3}}{3}-\frac{5x^{4}}{2}+\frac{x^{5}}{5}\right]_{0}^{5}.$$ At $x=5$: $\dfrac{25(125)}{3}-\dfrac{5(625)}{2}+\dfrac{3125}{5}=\dfrac{3125}{3}-\dfrac{3125}{2}+625.$ Combine the first two: $\dfrac{3125}{3}-\dfrac{3125}{2}=\dfrac{6250-9375}{6}=-\dfrac{3125}{6}$. Then $$V=-\frac{3125}{6}+625=\frac{-3125+3750}{6}=\frac{625}{6}.$$ The volume is $\dfrac{625}{6}\ \big(\approx 104.167\big).$

<span class="math-block">\[s(x)=(6x-x^{2})-x=5x-x^{2},\quad A(x)=(5x-x^{2})^{2}\]</span>

<span class="math-block">\[V=\int_{0}^{5}(5x-x^{2})^{2}\,dx=\int_{0}^{5}(25x^{2}-10x^{3}+x^{4})\,dx\]</span>

<span class="math-block">\[=\left[\tfrac{25x^{3}}{3}-\tfrac{5x^{4}}{2}+\tfrac{x^{5}}{5}\right]_{0}^{5}\]</span>

<span class="math-block">\[=\frac{625}{6}\]</span>

4 points: 1 for the side length $s=5x-x^2$; 1 for the integrand $(5x-x^2)^2$ (squaring the side, not the difference of squares); 1 for a correct antiderivative; 1 for the value $625/6$. Using $\pi$ (treating it as a circle/disk) forfeits the integrand point.

Part (c)2 PTS

Set up (do not evaluate) the volume of revolution about the $x$-axis.

Revolving $R$ about the $x$-axis produces washers. At each $x$ the outer radius reaches the parabola, $R_{\text{out}}=6x-x^{2}$, and the inner radius reaches the line, $R_{\text{in}}=x$ (both nonnegative on $[0,5]$). The volume is $$V=\pi\int_{0}^{5}\Big[(6x-x^{2})^{2}-(x)^{2}\Big]\,dx.$$ The problem asks only for this expression, not its value. (As a check, evaluating gives $\dfrac{625\pi}{3}\approx 654.498$.)

<span class="math-block">\[R_{\text{out}}=6x-x^{2},\quad R_{\text{in}}=x\]</span>

<span class="math-block">\[V=\pi\int_{0}^{5}\Big[(6x-x^{2})^{2}-x^{2}\Big]\,dx\]</span>

2 points: 1 for the washer form $\pi\big[(\text{outer})^2-(\text{inner})^2\big]$ with outer $=6x-x^2$ and inner $=x$; 1 for the correct limits $0$ to $5$ and the $\pi$ factor. A common error is $\pi\int\big((6x-x^2)-x\big)^2dx$ (squaring the difference instead of the difference of squares) — this loses the form point.

Question 6 9 points No calculatorDifficulty 4/5

Consider the curve defined by the equation $$x^{2}-xy+y^{2}=12.$$ (a) Show that $\dfrac{dy}{dx}=\dfrac{2x-y}{x-2y}.$ (b) Find the coordinates of each point on the curve at which the tangent line is horizontal. (c) Find the coordinates of each point on the curve at which the tangent line is vertical. (d) A particle moves along the curve. At the instant the particle is at the point $(2,-2)$, its $x$-coordinate is increasing at a rate of $\dfrac{dx}{dt}=5$ units per second. Find the rate of change of its $y$-coordinate, $\dfrac{dy}{dt}$, at that instant.

Show worked solution

Part (a)2 PTS

Derive $dy/dx$ by implicit differentiation.

Differentiate both sides of $x^{2}-xy+y^{2}=12$ with respect to $x$, treating $y$ as a function of $x$ (product rule on $xy$, chain rule on $y^{2}$): $$2x-\left(y+x\frac{dy}{dx}\right)+2y\frac{dy}{dx}=0.$$ Group the $\dfrac{dy}{dx}$ terms: $$2x-y+(-x+2y)\frac{dy}{dx}=0\quad\Longrightarrow\quad (-x+2y)\frac{dy}{dx}=-(2x-y).$$ Solve for the derivative, dividing numerator and denominator by $-1$: $\dfrac{dy}{dx}=\dfrac{-(2x-y)}{-x+2y}=\dfrac{2x-y}{x-2y}.$ Therefore $$\frac{dy}{dx}=\frac{2x-y}{x-2y}.$$

<span class="math-block">\[2x-\big(y+x\,y'\big)+2y\,y'=0\]</span>

<span class="math-block">\[(-x+2y)\,y'=-(2x-y)\]</span>

<span class="math-block">\[y'=\frac{2x-y}{x-2y}\]</span>

2 points: 1 for correct implicit differentiation including the product rule on $xy$ (i.e. $\frac{d}{dx}(xy)=y+xy'$) and chain rule on $y^2$; 1 for correctly solving to the stated form. Treating $y$ as a constant (dropping $y'$ terms) loses both points.

Part (b)3 PTS

Find points with a horizontal tangent.

A horizontal tangent requires $\dfrac{dy}{dx}=0$, i.e. the numerator $2x-y=0$ (with denominator nonzero), so $y=2x$. Substitute into the curve: $$x^{2}-x(2x)+(2x)^{2}=12\ \Rightarrow\ x^{2}-2x^{2}+4x^{2}=12\ \Rightarrow\ 3x^{2}=12\ \Rightarrow\ x=\pm 2.$$ Then $y=2x$ gives the points $(2,4)$ and $(-2,-4)$. At each, the denominator $x-2y=x-4x=-3x\neq 0$, so these are genuine horizontal tangents. **Horizontal tangents at $(2,4)$ and $(-2,-4)$.**

<span class="math-block">\[\frac{dy}{dx}=0 \Rightarrow 2x-y=0 \Rightarrow y=2x\]</span>

<span class="math-block">\[x^{2}-2x^{2}+4x^{2}=3x^{2}=12 \Rightarrow x=\pm 2\]</span>

<span class="math-block">\[(2,4)\ \text{and}\ (-2,-4)\]</span>

3 points: 1 for setting the numerator $2x-y=0$; 1 for substituting $y=2x$ and solving $3x^2=12$; 1 for both points $(2,4),(-2,-4)$ (and checking the denominator $\neq0$). Missing one of the two points, or setting the denominator to zero instead, loses a point.

Part (c)2 PTS

Find points with a vertical tangent.

A vertical tangent requires the denominator $x-2y=0$ (with numerator nonzero), so $x=2y$. Substitute into the curve: $$(2y)^{2}-(2y)y+y^{2}=12\ \Rightarrow\ 4y^{2}-2y^{2}+y^{2}=12\ \Rightarrow\ 3y^{2}=12\ \Rightarrow\ y=\pm 2.$$ Then $x=2y$ gives the points $(4,2)$ and $(-4,-2)$. At each, the numerator $2x-y=4y-y=3y\neq 0$, so the tangents are vertical. **Vertical tangents at $(4,2)$ and $(-4,-2)$.**

<span class="math-block">\[x-2y=0 \Rightarrow x=2y\]</span>

<span class="math-block">\[4y^{2}-2y^{2}+y^{2}=3y^{2}=12 \Rightarrow y=\pm 2\]</span>

<span class="math-block">\[(4,2)\ \text{and}\ (-4,-2)\]</span>

2 points: 1 for setting the denominator $x-2y=0$ and substituting $x=2y$; 1 for both points $(4,2),(-4,-2)$. Solving the numerator instead (that gives horizontal tangents) earns no credit here.

Part (d)2 PTS

Related rates: find $dy/dt$ at $(2,-2)$ given $dx/dt=5$.

First confirm $(2,-2)$ is on the curve: $2^{2}-(2)(-2)+(-2)^{2}=4+4+4=12.$ ✓ Differentiate the curve with respect to $t$ (every term, using the chain rule): $$2x\frac{dx}{dt}-\left(\frac{dx}{dt}\,y+x\frac{dy}{dt}\right)+2y\frac{dy}{dt}=0.$$ Substitute $x=2,\ y=-2,\ \dfrac{dx}{dt}=5$: $$2(2)(5)-\big(5(-2)+2\tfrac{dy}{dt}\big)+2(-2)\tfrac{dy}{dt}=0$$ $$20-(-10)-2\tfrac{dy}{dt}-4\tfrac{dy}{dt}=0\ \Rightarrow\ 30-6\tfrac{dy}{dt}=0\ \Rightarrow\ \frac{dy}{dt}=5.$$ (Equivalently, $\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}=\dfrac{2(2)-(-2)}{2-2(-2)}\cdot 5=\dfrac{6}{6}\cdot5=5.$) So $\dfrac{dy}{dt}=5$ units per second.

<span class="math-block">\[2x\,\dot x-(\dot x\,y+x\,\dot y)+2y\,\dot y=0\]</span>

<span class="math-block">\[20-(-10)-2\dot y-4\dot y=0 \Rightarrow 30=6\dot y\]</span>

<span class="math-block">\[\frac{dy}{dt}=5\]</span>

2 points: 1 for differentiating the relation with respect to $t$ (chain/product rules, keeping both $\dot x$ and $\dot y$ terms) and substituting the given values; 1 for $\frac{dy}{dt}=5$. Substituting numbers before differentiating, or dropping the $\dot x\,y$ product term, loses credit.

Originality statement

This practice exam is original instructional content created by Tian2 AP, aligned to the concepts and skills in College Board’s Course and Exam Description. It is not a reproduction of, nor affiliated with, College Board’s official materials.

AB Practice Exam 1 — Free Response.