Start with Volume equation for a Cone in terms of base Area.
[tex]V = \frac{1}{3} A h[/tex]
Next relate height in terms of base Area: (Note base = pi*r^2)
[tex]h = 2D = 4r = 4 \sqrt {\frac{A}{\pi}}[/tex]
New volume equation is:
[tex]V = \frac{4}{3 \sqrt{\pi}} A^{3/2}[/tex]
Take derivative with respect to time:
[tex]\frac{dV}{dt} = (\frac{4}{3\sqrt{\pi}})(\frac{3}{2}) \sqrt{A} \frac{dA}{dt} [/tex]
Sub in rate for volume, solve for dA/dt
[tex]\frac{dA}{dt} = \frac{10 \sqrt{\pi}}{\sqrt{A}}[/tex]
Finally we need the Area after 5 min, given the volume after 5 min is 100.
Go back to Volume equation and solve for sqrt(A)
[tex]100 = \frac{4}{3 \sqrt{\pi}} (\sqrt{A})^3 \\ \sqrt{A} = (75 \sqrt{\pi})^{1/3}[/tex]
Final Answer:
[tex]\frac{dA}{dt} = \frac{10 \sqrt{\pi}}{(75 \sqrt{\pi})^{1/3}} = 3.47[/tex]
