Question

Consider a tank in the shape of an inverted right circular cone that is leaking water. The dimensions of the conical tank are a height of 12 ft and a radius of 10 ft. How fast does the depth of the water change when the water is 10 ft high if the cone leaks at a rate of 8 cubic feet per minute?

Answer 1

Answer:

$\frac{dh}{dt} = \frac{216}{1875\pi}$

Step-by-step explanation:

Given

Represent radius with r and height with h

$h = 12ft$

$r = 10ft$

$Rate = \frac{8ft^3}{min}$ when $h = 10ft$

Express radius in terms of height

$\frac{r}{h} = \frac{10}{12}$

$\frac{r}{h} = \frac{5}{6}$

$r = \frac{5}{6}h$

First, we need to determine the volume of the cone in terms of height.

$Volume = \frac{1}{3}\pi r^2h$

Substitute $r = \frac{5}{6}h$

$V = \frac{1}{3} * \pi * (\frac{5}{6}h)^2 * h$

$V = \frac{1}{3} * \pi * \frac{25}{36}h^2 * h$

$V = \frac{1}{3} * \pi * \frac{25}{36}h^3$

$V = \frac{25}{108} \pi h^3$

Differentiate both sides w.r.t time (t)

$\frac{dV}{dt} = 3 * \frac{25}{108} \pi h^{3-1} * \frac{dh}{dt}$

$\frac{dV}{dt} = \frac{75}{108} \pi h^{2} \frac{dh}{dt}$

Solve for $\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{dv}{dt} * \frac{108}{75\pi h^2}$

At this point, $h = 10$ and $\frac{dv}{dt} = \frac{8ft^3}{min}$

So, we have:

$\frac{dh}{dt} = 8 * \frac{108}{75\pi 10^2}$

$\frac{dh}{dt} = \frac{8 * 108}{75\pi 10^2}$

$\frac{dh}{dt} = \frac{864}{7500\pi}$

$\frac{dh}{dt} = \frac{216}{1875\pi}$

Hence:

The rate is $\frac{216}{1875\pi} ft/min$