Answer to Question #155194 in Calculus for Phyroe

Question

A conical cistern is 10 ft across the top and 12 ft deep. If water is poured into the cistern at the rate of 1 cubic foot per second, how fast is the surface rising when the water is 8 ft deep?

Answer 1

let the water level have a radius of r feet

let the height of the water be h feet

by ratio:

$\large\frac{r}{h} = \large\frac{5}{12}$

$12r = 5h \to r = \large\frac{5h}{12}$

$V = \large\frac{1}{3}\pi r^2h = \large\frac{1}{3}\pi (\large\frac{25h^2}{144})(h)$

$\large\frac{dV}{dt} = \large\frac{25}{144}\pi h^2 \large\frac{dh}{dt}$

plug in our given stuff 8 feet deep

$1 = \large\frac{25}{144}\pi(64)$ $\large\frac{dh}{dt}$

$\large\frac{dh}{dt} = \large\frac{144}{25*64} = \frac{9}{100\pi}$