Question

The height of a cone is increasing at a rate of 10 cm/sec and its radius is decreasing so that its volume remains constant. How fast is the radius changing when the radius is 4 cm and the height is 10 cm?

Answer 1

Answer:

dr/dt = -2 cm/s.

Explanation:

The volume of a cone is given by:

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

• r is the radius
• h is the height

Let's take the derivative with respect to time in each side of (1).

$\frac{dV}{dt}=\frac{1}{3} \pi \frac{d}{dt}(r^{2}h)=\frac{1}{3} \pi \left(2r\frac{dr}{dt}h+r^{2}\frac{dh}{dt} \right)$ (2)

We know that:

• dh/dt = 10 cm / s (rate increasing of height)
• dV/dt = 0 (constant volume means no variation with respect of time)
• r = 4 cm
• h = 10 cm

We can calculate how fast is the radius changing using the above information.

$0=\frac{1}{3} \pi \left( 2\cdot 4\cdot \frac{dr}{dt} \cdot 10 + 4^{2}\cdot 10)\right$  

Therefore dr/dt will be:

$\frac{dr}{dt}=-\frac{160}{80}=-2 cm/s$

The minus signs means that r is decreasing.

I hope it helps you!