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

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)

Scroll to Top