Question

The volume of the spherical ball is increasing at the rate of 4π cc/sec. Find the rate at which the radius and the surface area are changing when the volume is 288 π cc.

Answer 1

Let r be the radius, s be the surface area and V be the volume of the spherical ball.

Then, `(dV)/dt` = 4π cc/sec, V = 288π cc   ...[Given]

V = `4/3 pir^3`     ...(i)

Differentiating w.r.t. t, we get

`(dV)/dt = 4/3 pi(3r^2)*(dr)/dt`

∴ 4π = `4pi"r"^2("dr"/"dt")`

∴ `(dr)/dt = 1/r^2`   ...(ii)

Now, V = 288π

∴ `4/3 pir^3` = 288π    ...[From (i)]

∴ r³ = 216

⇒ r = 6

From (ii), we get

`(dr)/dt = 1/6^2 = 1/36`

Thus, the radius of the spherical ball is increasing at the rate of `1/36` cm/sec.

Now, s = 4πr²

Differentiating w.r.t. t, we get

`((ds)/dt) = 4pi(2r)*(dr)/dt`

= `8pir*(dr)/dt`

When r = 6 cm,

`((ds)/dt)_(r = 6) = 8pi(6)*1/36`

= `(4pi)/3`

Thus, the surface area of the spherical ball is increasing at the rate of `(4pi)/3` cm²/sec.