A spherical balloon is being inflated so that its volume is increasing at the rate of 200 cm3/min. At what rate is the radius increasing when the radius is 15 cm.
Volume increase rate = surface area*(radius increase rate)
200 cm^3/min = (4 pi r*2)*dr/dt
Solve for dr/dt
To find the rate at which the radius is increasing, we can use the relationship between the volume and radius of a sphere.
The volume of a sphere is given by the formula:
V = (4/3) * π * r^3
Taking the derivative of both sides with respect to time (t), we get:
dV/dt = (4/3) * π * 3r^2 * (dr/dt)
where dV/dt represents the rate at which the volume is changing with respect to time, and dr/dt represents the rate at which the radius is changing with respect to time.
Given that dV/dt = 200 cm^3/min, we need to find dr/dt when r = 15 cm.
Plug in the values into the equation:
200 = (4/3) * π * 3(15)^2 * (dr/dt)
Simplify:
200 = 60π * (dr/dt) * 225
Divide both sides by 60π * 225:
(dr/dt) = 200 / (60π * 225)
Simplify further:
(dr/dt) = 1 / (9π)
So, when the radius is 15 cm, the rate at which the radius is increasing is approximately 1 / (9π) cm/min.