A spherical balloon is being inflated so that its volume is increasing at the rate of 5 cubic meter per minute.At what rate is the radius increasing when the radius is 6 meters?

V = (4/3) pi r^3

dV/dt = 4 pi r^2 dr/dt = surface area * dr/dt of course
so
dr/dt = dV/dt /(4 pi r^2)
= 5/(100 pi ) = 1/(20 pi)

thank you so much! :)

uhm, just one more thing. Isn't it 5/144π ? :)

yes, right, forgot 6 and not 5

thank you. :) you really did a big help. Godbless

To find the rate at which the radius is increasing, we can use the formula for the volume of a sphere:

V = (4/3)πr³,

where V is the volume and r is the radius.

We are given that the volume is increasing at a rate of 5 cubic meters per minute, which can be expressed as:

dV/dt = 5.

We want to find the rate at which the radius (r) is increasing, which can be expressed as:

dr/dt = ?

To find the relationship between dV/dt (rate of change of volume with respect to time) and dr/dt (rate of change of radius with respect to time), we can take the derivative of the volume formula with respect to time:

dV/dt = d((4/3)πr³)/dt.
Simplifying, we get:

dV/dt = 4πr²(dr/dt).

Now we can substitute the given values:

5 = 4π(6)²(dr/dt).

Simplifying further, we get:

5 = 144π(dr/dt).

To find dr/dt, we divide both sides by 144π:

dr/dt = 5 / (144π).

Calculating this value, we get:

dr/dt ≈ 0.0109 meters per minute.

Therefore, when the radius is 6 meters, the rate at which the radius is increasing is approximately 0.0109 meters per minute.