Answer is: option2
0.28 ft/minSolution:
The sand pile forms a cone where the radius r is always three times the height h.
Given: \( r = 3h \)
The volume of a cone is:
\( V = \frac{1}{3} \pi r^2 h \)
Substituting \( h = \frac{r}{3} \) into the volume formula:
\( V = \frac{1}{3} \pi r^2 \left(\frac{r}{3}\right) \)
\( V = \frac{1}{9} \pi r^3 \)
Differentiating both sides with respect to t:
\( \frac{dV}{dt} = \frac{1}{9} \pi (3r^2) \frac{dr}{dt} \)
\( \frac{dV}{dt} = \frac{1}{3} \pi r^2 \frac{dr}{dt} \)
Solving for \( \frac{dr}{dt} \):
\( \frac{dr}{dt} = \frac{3}{\pi r^2} \frac{dV}{dt} \)
From the graph we have:
- \( V(6) = 4\pi \)
- \( \frac{dV}{dt} = \pi \)
Using \( V = \frac{1}{9} \pi r^3 \), we substitute \( V(6) = 4\pi \):
\( 4\pi = \frac{1}{9} \pi r^3 \)
Cancel \( \pi \):
\( 4 = \frac{1}{9} r^3 \)
\( r^3 = 36 \)
\( r = \sqrt{36} \)
Approximating:
\( r \approx 3.3 \)
Cancel \( \pi \):
\( \frac{dr}{dt} = \frac{3}{\pi r^2} \cdot \pi \)
\( \frac{dr}{dt} = \frac{3}{r^2} \)
Substituting \( r \approx 3.3 \):
\( \frac{dr}{dt} = \frac{3}{(3.3)^2} \)
Approximating:
\( (3.3)^2 = 10.89 \)
\( \frac{dr}{dt} \approx \frac{3}{10.89} \)
\( \frac{dr}{dt} \approx 0.2756 \)
Final Answer: \( \frac{dr}{dt} \approx 0.28 \) ft/min