Answer is: option2
0.354 ft per minSolution:
Rate of Change of Water Level in a Conical Reservoir Volume of a Cone Formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Using Similar Triangles:Since the full cone has \( r = 6 \) ft and \( h = 4 \) ft, the ratio holds:
\[ \frac{r}{h} = \frac{6}{4} = \frac{3}{2} \]
Thus, for any water height \( h \), the radius is:
\[ r = \frac{3}{2} h \]
Substituting \( r \) into the Volume Formula:\[ V = \frac{1}{3} \pi \left(\frac{3}{2} h\right)^2 h \]
\[ V = \frac{1}{3} \pi \frac{9}{4} h^3 \]
\[ V = \frac{3}{4} \pi h^3 \]
Differentiating Both Sides:\[ \frac{dV}{dt} = \frac{3}{4} \pi \cdot 3h^2 \frac{dh}{dt} \]
\[ \frac{dV}{dt} = \frac{9}{4} \pi h^2 \frac{dh}{dt} \]
Solving for \( \frac{dh}{dt} \)Given \( \frac{dV}{dt} = 10 \) and \( h = 2 \):
\[ 10 = \frac{9}{4} \pi (2)^2 \frac{dh}{dt} \]
\[ 10 = \frac{9}{4} \pi (4) \frac{dh}{dt} \]
\[ 10 = \frac{36}{4} \pi \frac{dh}{dt} \]
\[ 10 = 9\pi \frac{dh}{dt} \]
\[ \frac{dh}{dt} = \frac{10}{9\pi} \]
Approximating:\[ \frac{10}{9\pi} \approx \frac{10}{28.27} \approx 0.354 \text{ ft/min} \]
Final Answer: 0.354 ft/min