If a snowball melts so that the surface area S=4*pi*r^2 decreases at a rate of 3 cm^2/min, find the rate at which the radius decreases when the radius is 5cm?

Question

If a snowball melts so that the surface area S=4*pi*r^2  decreases at a rate of 3 cm^2/min, find the rate at which the radius decreases when the radius is 5cm?

ash2326 · Answer

We have
$$S= 4\pi r^2$$
It's given that the snow ball decreases at a rate of 3 cm^2/min
so
$$\frac{ds}{dt}=-3 cm^2/min$$
We have to find the rate of decrease of radius $\frac{dr}{dt}$ when radius= 5 cm
Let's find $\frac{ds}{dt}$
$$\frac{ds}{dt}= 8 \pi r \frac{dr}{dt}$$
we have $\frac{ds}{dt}=-3$  and radius = 5cm
$$ -3 = 8\pi \times 5 \frac{dr}{dt}$$
we'll get
$$\frac{dr}{dt}=-\frac{3}{40\pi}$$
We get
$$\frac{dr}{dt}=-0.02388 cm/minute$$

anonymous · Answer

Thank you

ash2326 · Answer

welcome , did you understand it ?

anonymous · Answer

yeah i got down to the part of 8pi *40 but i wasnt sure that the 3 needed to be negative and i wasnt understanding how it was on top but i understand now why its that way

ash2326 · Answer

great, since it's decreasing it's negative