Question

Suppose that both the radius r and height h of a circular cone are increasing at a rate of 7 cm/s. How fast is the volume of the cone increasing when r=15 cm and h=10 cm?

Answer 1

Answer:

The volume of the cone is increasing at the rate oof 3848.45 cm³/s when r=15 cm and h=10 cm.

Step-by-step explanation:

The volume of the cone is given by the following formula:

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

In which V is measured in cm³ while r and h are measured in cm.

Suppose that both the radius r and height h of a circular cone are increasing at a rate of 7 cm/s.

This means that $\frac{dh}{dt} = \frac{dr}{dt} = 7$

How fast is the volume of the cone increasing when r=15 cm and h=10 cm?

This is $\frac{dV}{dt}$ when $r = 15, h = 10$.

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

Applying implicit differentiation:

We have three variables, V, r and h. So

$\frac{dV}{dt} = \frac{2r\frac{dr}{dt}h \pi + r^{2}\pi \frac{dh}{dt}}{3}$

$\frac{dV}{dt} = \frac{2*15*7*10\pi + 225*\pi*7}{3}$

$\frac{dV}{dt} = 3848.45$

The volume of the cone is increasing at the rate oof 3848.45 cm³/s when r=15 cm and h=10 cm.