AP Calculus AB FRQ Practice: Mixed Free-Response Set

Solution

1. The midpoints of the subintervals are $t=3$, $t=9$ and $t=14$, all in the table; each width multiplies the value of $W$ at the midpoint.
2. ${\int }_0^{16}W(t)dt\approx W(3)(6)+$$W(9)(12-6)+$$W(14)(16-12)=$$66(6)+$$40(6)+$$39(4)=$$792$.
3. $W(t)$ is a rate in cubic meters per hour, so its integral over $0\le t\le 16$ is an amount in cubic meters.
4. ${\int }_0^{16}W(t)dt$ is the total volume of water, in cubic meters, that the creek carries into the pond from time $t=0$ to time $t=16$ hours.

Solution

1. The amount that arrives over a time interval is the integral of the rate over that interval: ${\int }_9^{26}V(t)dt$, with $V(t)=32+25(t-9)e^{-0.2(t-9)}$ there.
2. On the calculator, ${\int }_9^{26}V(t)dt=1077.224$ cubic meters.

Solution

1. The table gives $W(9)=40$ and $W(12)=34$; on the calculator, $V(9)=32.000$ and $V(12)=73.161$.
2. So $D(9)=W(9)-V(9)=40-32.000=8.000>0$ and $D(12)=W(12)-V(12)=34-73.161=-39.161<0$: $D$ changes sign on $[9,12]$.
3. Yes. Because $D$ is differentiable, $D$ is continuous on $9\le t\le 12$, and $D$ changes sign there, so by the Intermediate Value Theorem there is a time $t$, with $9<t<12$, at which $D(t)=0$: the creek and the ditch carry water into the pond at the same rate.

Solution

1. At $x$ the base of the rectangle runs across $R$, with length $f(x)-g(x)$; its area is $(5-x)(f(x)-g(x))$.
2. Volume $={\int }_{-1.861}^{2.115}(5-x)(f(x)-$$g(x))dx=$$51.892$.

Solution

1. The line $y=5$ lies above $R$: the outer radius is $5-g(x)$ and the inner radius is $5-f(x)$.
2. Volume $=\pi {\int }_{-1.861}^{2.115}((5-g(x){)}^2-(5-f(x){)}^2)dx=$$\pi {\int }_{-1.861}^{2.115}({(7-x^2)}^2-{(5-\sqrt{x+4})}^2)dx$.

Solution

1. $f^{\mathrm{\prime }}(x)={\displaystyle \frac{1}{2\sqrt{x+4}}}$ and $g^{\mathrm{\prime }}(x)=2x$. The tangent lines are parallel where $f^{\mathrm{\prime }}(x)=g^{\mathrm{\prime }}(x)$:
   ${\displaystyle \frac{1}{2\sqrt{x+4}}}=2x$.
2. Solving on $0<x<1$ (calculator): $x=0.123$.

Solution

1. Because $D(t)>0$, ${\displaystyle \frac{dD}{dt}}={\displaystyle \frac{9-t^2}{2D}}$ equals 0 only where its numerator is 0: $9-t^2=0$ gives $t=3$ (the value $t=-3$ is outside the interval).
2. For $0<t<3$, ${\displaystyle \frac{dD}{dt}}>0$, and for $3<t<5$, ${\displaystyle \frac{dD}{dt}}<0$, since the numerator is positive before $t=3$ and negative after it while the denominator stays positive.
3. So ${\displaystyle \frac{dD}{dt}}$ changes sign from positive to negative at $t=3$, and $D$ has a relative maximum at $t=3$.

Solution

1. Separate the variables:
   $DdD=({\displaystyle \frac{9}{2}}-{\displaystyle \frac{1}{2}}{t}^2)dt$.
2. Integrate:
   ${\displaystyle \frac{D^2}{2}}={\displaystyle \frac{9}{2}}t-{\displaystyle \frac{1}{6}}{t}^3+C$.
3. With $D(0)=2$:
   ${\displaystyle \frac{4}{2}}=C$, so $C=2$ and $D^2=4+9t-{\displaystyle \frac{1}{3}}{t}^3$.
4. Because $D(t)>0$, take the positive square root:
   $D(t)=\sqrt{4+9t-{\displaystyle \frac{1}{3}}{t}^3}$.

Solution

1. The graph of $f$ changes concavity where $f^{\mathrm{\prime }\mathrm{\prime }}$ changes sign, that is, where $f^{\mathrm{\prime }}$ changes from increasing to decreasing or from decreasing to increasing.
2. From the graph, $f^{\mathrm{\prime }}$ is increasing on $(0,3)$ and decreasing on $(-2,0)\cup (3,5)$.
3. The graph of $f$ has a point of inflection at $x=0$ and $x=3$ because $f^{\mathrm{\prime }}$ changes from increasing to decreasing at $x=3$ and from decreasing to increasing at $x=0$.

Solution

1. $g^{\mathrm{\prime }}(x)=f^{\mathrm{\prime }}(x)+1$, and $g$ is decreasing where $g^{\mathrm{\prime }}(x)<0$, that is, where $f^{\mathrm{\prime }}(x)<-1$.
2. Compare the graph of $f^{\mathrm{\prime }}$ with the horizontal line $y=-1$: it lies below the line on $(-2,1)$.
3. So $g$ is decreasing on $(-2,1)$.

Solution

1. $g^{\mathrm{\prime }}(x)=f^{\mathrm{\prime }}(x)+1$ is $0$ where the graph of $f^{\mathrm{\prime }}$ meets the line $y=-1$: at $x=1$. The candidates are these points and the endpoints $x=-2$ and $x=5$.
2. From $f(-1)=-4$ and the signed areas under the graph of $f^{\mathrm{\prime }}$:
   $f(-2)=$$-{\displaystyle \frac{11}{4}},$ $f(1)=$$-{\displaystyle \frac{29}{4}},$ $f(5)=$${\displaystyle \frac{\pi }{2}}-$${\displaystyle \frac{33}{4}}$.
3. $g(-2)=$$f(-2)+$$(-2)=$$-{\displaystyle \frac{11}{4}}-$$2=$$-{\displaystyle \frac{19}{4}},$ $g(1)=$$f(1)+$$(1)=$$-{\displaystyle \frac{29}{4}}+$$1=$$-{\displaystyle \frac{25}{4}},$ $g(5)=$$f(5)+$$(5)=$${\displaystyle \frac{\pi }{2}}-$${\displaystyle \frac{33}{4}}+$$5=$${\displaystyle \frac{\pi }{2}}-$${\displaystyle \frac{13}{4}}$.
4. The absolute maximum value of $g$ on $-2\le x\le 5$ is $g(5)={\displaystyle \frac{\pi }{2}}-{\displaystyle \frac{13}{4}}$, the largest of these values.

Solution

1. $v(2)=4$, which is positive, and from part (a), $a(2)=-6$, which is negative.
2. The speed increases when velocity and acceleration have the same sign and decreases when they have opposite signs, so the cart is slowing down.

Solution

1. On $0\le t\le 3$, $v(t)=2t{(3-t)}^2\ge 0$, so the cart never moves backward and the distance is ${\int }_0^{3}v(t)dt$.
2. An antiderivative of $2t^3-12t^2+18t$ is ${\displaystyle \frac{t^4}{2}}-4t^3+9t^2$.
3. The distance is ${[{\displaystyle \frac{t^4}{2}}-4t^3+9t^2]}_0^3={\displaystyle \frac{27}{2}}$ centimeters.

Solution

1. The average velocity on $5\le t\le 9$ is the average value of $v$ there:
   ${\displaystyle \frac{1}{9-5}}{\int }_5^{9}v(t)dt$, where $v(t)=-5\sin \text{\hspace{0.17em}⁣}\left({\displaystyle \frac{\pi (t-5)}{8}}\right)$.
2. By substitution (with $u={\displaystyle \frac{\pi (t-5)}{8}}$), an antiderivative of $-5\sin \text{\hspace{0.17em}⁣}\left({\displaystyle \frac{\pi (t-5)}{8}}\right)$ is ${\displaystyle \frac{40}{\pi }}\cos \text{\hspace{0.17em}⁣}\left({\displaystyle \frac{\pi (t-5)}{8}}\right)$.
3. So the average velocity is ${\displaystyle \frac{1}{4}}\text{\hspace{0.17em}⁣}(0-{\displaystyle \frac{40}{\pi }})=-{\displaystyle \frac{10}{\pi }}$ centimeters per second.

Solution

1. By the chain rule, $w^{\mathrm{\prime }}(x)=h^{\mathrm{\prime }}\left(f(x)\right)\cdot f^{\mathrm{\prime }}(x)$.
2. $f\left({\displaystyle \frac{\pi }{4}}\right)=5$, and $h^{\mathrm{\prime }}(5)=2$ is the slope of the segment of the graph of $h$ through $x=5$.
3. $w^{\mathrm{\prime }}\left({\displaystyle \frac{\pi }{4}}\right)=(2)(2)=4$, using $f^{\mathrm{\prime }}\left({\displaystyle \frac{\pi }{4}}\right)=2$ from part (a).

Solution

1. By the product rule and the chain rule, $p^{\mathrm{\prime }}(x)=-g^{\mathrm{\prime }}(-x)h(x)+g(-x)h^{\mathrm{\prime }}(x)$.
2. From the table, $g(-2)=12$ and $g^{\mathrm{\prime }}(-2)=2$. From the graph, $h(2)=-1$ and $h^{\mathrm{\prime }}(2)=-1$, the slope of the segment through $x=2$.
3. $p^{\mathrm{\prime }}(2)=-(2)(-1)+(12)(-1)=-10$.

Solution

1. $g$ is differentiable everywhere, so the hypotheses of the Mean Value Theorem hold on $[-4,-3]$.
2. From the table, ${\displaystyle \frac{g(-3)-g(-4)}{-3-(-4)}}={\displaystyle \frac{10-9}{1}}=1$.
3. So some $c$ between $-4$ and $-3$ has $g^{\mathrm{\prime }}(c)=1$.