AP Calculus BC: 4.3 Rates of Change in Applied Contexts Other  Than Motion – Exam Style questions with Answer- FRQ

Question

A conical funnel has a base diameter of 6 cm and a height of 5 cm. The funnel sits over a cylindrical can with an open top. The can has a diameter of 4 cm and a height of 5 cm. The funnel is initially full, but water is draining from the funnel bottom into the can at a constant rate of \(2cm^{3}/s\). Answer the following questions. (Show your work.)
(A) How fast is the water level in the funnel falling when the water is 2.5 cm high?
(B) How fast is the water level in the can rising?
(C) Will the can overflow? If not, how high will the final level of water be in the can?

Answer/Explanation

(A)

\(\frac{3}{5}=\frac{r}{h}\)
3h=5r
\(r=\frac{3h}{5}\)
\(V=\frac{1}{3}\pi r^{2}h\)
\(V=\frac{1}{3}\pi \left ( \frac{3h}{5} \right )^{2}h\)
\(V=\frac{1}{3}\pi \left ( \frac{9h^{2}}{25} \right )h\)
\(V=\frac{3}{25}\pi h^{3}\)
\(\frac{\mathrm{d} V}{\mathrm{d} t}=3\left ( \frac{3}{25} \right )\pi h^{2}\frac{\mathrm{d} h}{\mathrm{d} t}\)
\(\frac{\mathrm{d} V}{\mathrm{d} t}=\left ( \frac{9}{25} \right )\pi h^{2}\frac{\mathrm{d} h}{\mathrm{d} t}\)
\(\frac{\mathrm{d} V}{\mathrm{d} t}=-2cm^{3}/s\)
\(-2=\left ( \frac{9}{25} \right )\pi h^{2}\frac{\mathrm{d} h}{\mathrm{d} t}\)
\(\frac{\mathrm{d} h}{\mathrm{d} t}=-\frac{50}{9\pi h^{2}}\)
\(\frac{\mathrm{d} h}{\mathrm{d} t}=-\frac{50}{9\pi (2.5)^{2}}\)
\(\frac{\mathrm{d} h}{\mathrm{d} t}=-\frac{50}{9\pi \left ( \frac{25}{4} \right )}=-\frac{8}{9\pi }=0.28cm/s\)

(B) 

\(V=\pi r^{2}h\)
\(\frac{\mathrm{d} V}{\mathrm{d} t}=\pi r^{2}\frac{\mathrm{d} h}{\mathrm{d} t}\) (r is constant)
\(\pi r^{2}\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{\mathrm{d} V}{\mathrm{d} t}\)
\(\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{1}{\pi r^{2}}\left ( \frac{\mathrm{d} V}{\mathrm{d} t} \right )\)
\(\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{1}{\pi (2cm)^{2}}(2cm^{3}/min)\)
\(\frac{\mathrm{d} h}{\mathrm{d} t}=\left ( \frac{1cm/s}{2\pi } \right )=0.16cm/s\)

(C)

Question

You are to make a cylindrical tin can with closed top to hold \(360 cm^3\).
(A) Make a drawing and label what is given.
(B) What are its dimensions if the amount of tin used is to be the minimum?
(C) What is the surface area?

Answer/Explanation

(A)

(B)\(V=\pi r^{2}h\)
\(360=\pi r^{2}h\)
\(h=\frac{360}{\pi r^{2}}\)
\(S=2\pi rh+2\pi r^{2}\)
\(S=2\pi r\left ( \frac{360}{\pi r^{2}} \right )+2\pi r^{2}\)
\(S=2\left ( \frac{360}{r} \right )+2\pi r^{2}\)
\(S=720r^{-1}+2\pi r^{2}\)
\(\frac{\mathrm{d} S}{\mathrm{d} r}=-720r^{-2}+4\pi r\)
\(0=\frac{-720}{r^{2}}+4\pi r\)
\(4\pi r=\frac{720}{r^{2}}\)
\(4\pi r^{3}=720\)
\(r^{3}=\frac{720}{4\pi }\)
\(r^{3}=\frac{180}{\pi }\)
\(r=\sqrt[3]{\frac{180}{\pi }}\)
r=3.86cm
\(h=\frac{360}{\pi r^{2}}=\frac{360}{\pi (3.86cm)^{2}}=7.72cm\)

(C) \(S=2\pi rh+2\pi r^{2}\)
\(S=2\pi (3.86cm)(7.71cm)+2\pi (3.86cm)^{2}\)
\(S=187cm^{2}+94cm^{2}\)
\(S=281cm^{2}\)

Question

The rate of change in temperature of a greenhouse from 7 p.m. to 7 a.m. is given by the function: \(f(t)=-3\sin \left ( \frac{t}{3} \right )\) where temperature is measured in degrees Fahrenheit and t is the  number of hours after 7 p.m.
(A) If at 7 p.m. the temperature is 1058F, find the temperature of the greenhouse at 2 a.m.
(B) Write an integral expression to represent the temperature of the greenhouse at time t, where t is between 7 p.m. and 7 a.m.
(C) Find the average change in temperature of the greenhouse between 7 p.m. and 7 a.m. to the nearest 10th of a degree.
(D) Is there a point during the night when the rate of change it temperature of the greenhouse is the average change in temperature from part (c)? If yes, state the time. Either way, show all work to justify your answer.Ans:

Answer/Explanation

(A)F(0)=105,f(t) represents the rate of change of the temperature, and 2 a.m. is 7h after 7p.m.
\(F(7)=105+\int_{0}^{7}(-3\sin (t/3))dt\Rightarrow F(7)=105+9\cos (t/3)|^{7}_{0}\Rightarrow F(7)\approx 89.78\)
(B)\(F(t)=105+\int_{0}^{t}(-3\sin (x/3))dx\)
(C) Average value=\(\frac{1}{12-0}\int_{0}^{12}(-3\sin (t/3))dt\approx -1.24^{\circ}F/hour\)
(D) Since f(t) is continuous on [0, 12], by the Mean Value Theorem for integrals there
exists a number c in [0, 12] such that \(\int_{0}^{12}f(t)dt=f(c)(b-a)\)
You can use our solution from part (d) to show that f(c)=-1.24. Now solve for c.
\(f(c)=-1.24\Rightarrow -3\sin (t/3)=-1.24\Rightarrow \sin (t/3)=0.413\Rightarrow t/3=\sin ^{-1}(0.413)\Rightarrow t\approx 1.28\)
Approximately 1.28 h after 7 p.m., the rate of change of the temperature in the greenhouse equals the average rate of change of the temperature for the night. Now convert 0.28 to minutes to find the time.
\((0.28h)\times (60)=16.8min\approx 17 min\)

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