▶️ Answer/Explanation
Solution
Step 1: Find Radius
Surface area: \( S = 4\pi r^2 = 100\pi \)
\( r^2 = 25 \)
\( r = 5 \) inches
Step 2: Volume Rate of Change
Volume: \( V = \frac{4}{3}\pi r^3 \)
Derivative: \( \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \)
Given: \( \frac{dr}{dt} = 0.3 \), \( r = 5 \)
\( \frac{dV}{dt} = 4\pi (5)^2 \cdot 0.3 = 4\pi \cdot 25 \cdot 0.3 = 30\pi \)
Answer: E
▶️ Answer/Explanation
Solution
Step 1: Find Radius
Area: \( A = \pi r^2 = 64\pi \)
\( r^2 = 64 \)
\( r = 8 \) meters
Step 2: Radius Rate of Change
Area derivative: \( \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \)
Given: \( \frac{dA}{dt} = 96\pi \), \( r = 8 \)
\( 96\pi = 2\pi (8) \cdot \frac{dr}{dt} \)
\( \frac{dr}{dt} = \frac{96\pi}{16\pi} = 6 \)
Answer: A
▶️ Answer/Explanation
Solution
Step 1: Relate Surface Area Rate
Surface area: \( S = 4\pi r^2 \)
Derivative: \( \frac{dS}{dt} = 8\pi r \cdot \frac{dr}{dt} \)
Given: \( \frac{dS}{dt} = 4 \frac{dr}{dt} \)
\( 8\pi r \cdot \frac{dr}{dt} = 4 \frac{dr}{dt} \)
\( 8\pi r = 4 \)
\( r = \frac{1}{2\pi} \)
Step 2: Calculate Volume
Volume: \( V = \frac{4}{3}\pi r^3 \)
Substitute: \( r = \frac{1}{2\pi} \)
\( V = \frac{4}{3}\pi \left( \frac{1}{2\pi} \right)^3 = \frac{4}{3}\pi \cdot \frac{1}{8\pi^3} = \frac{4}{24\pi^2} = \frac{1}{6\pi^2} \)
Answer: D
Question
A conical funnel has a base diameter of 4 cm and a height of 5 cm. The funnel is initially full, but water is draining at a constant rate of \(2cm^{3}/s\). How fast is the water level falling when the water is 2.5 cm high?
(A) \(\frac{1}{2\pi}\) cm/s
(B) \(-\frac{1}{2\pi}cm/s\)
(C) \(\frac{2}{\pi }cm/s\)
(D)\(-\frac{2}{\pi } cm/s\)
▶️Answer/Explanation
Ans:(D)
\(\frac{2}{5}=\frac{r}{r}\)(Similar triangles)
2h=5r
\(r=\frac{2h}{5}\)
\(V=\frac{1}{3}\pi r^{2}h\)
\(V=\frac{1}{3}\pi \left ( \frac{2h}{5} \right )^{2}h\)
\(V=\frac{1}{3}\pi \left ( \frac{4h^{2}}{25} \right )h\)
\(V=\frac{4}{75}\pi h^{3}\)
\(\frac{\mathrm{d} V}{\mathrm{d} t}=3\left ( \frac{4}{75} \right )\pi h^{2}\frac{\mathrm{d} h}{\mathrm{d} t}\)
\(\frac{\mathrm{d} V}{\mathrm{d} t}=\left ( \frac{4}{25} \right )\pi h^{2}\frac{\mathrm{d} h}{\mathrm{d} t}\)
\(\frac{\mathrm{d} V}{\mathrm{d} t}=-2cm^{3}/s\)
\(-2=\left ( \frac{4}{25} \right )\pi h^{2}\frac{\mathrm{d} h}{\mathrm{d} t}\)
\(-1=\left ( \frac{2}{25} \right )\pi h^{2}\frac{\mathrm{d} h}{\mathrm{d} t}\)
\(\frac{\mathrm{d} h}{\mathrm{d} t}=-\frac{25}{2\pi h^{2}}\)
\(\frac{\mathrm{d} h}{\mathrm{d} t}=-\frac{25}{2\pi (2.5)^{2}}\)
\(\frac{\mathrm{d} h}{\mathrm{d} t}=-\frac{2}{\pi }cm/s\)