A spherical party balloon is being inflated with helium pumped in at a rate of 12 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 1 ft?

V = (4/3)πr^3

dV/dt = 4π r^2 dr/dt
given : dV/dt = 12 , and r = 1

12 = 4π (1) dr/dt
dr/dt = 3/π ft/min

Thank you! I initially thought the given rate of 12 cubic feet per minute was dr/dt.

The units are usually your give-away.

it says 12 cubic feet/min which is volume/time , so dV/dt

This will work in most cases.

To find the rate at which the radius is growing, we need to use the relationship between the volume and the radius of a sphere.

The volume of a sphere is given by the formula V = (4/3)πr^3, where V is the volume and r is the radius.

Since we know that the balloon is being inflated at a rate of 12 cubic feet per minute, we can set up the following equation:

dV/dt = 12,

where dV/dt represents the rate at which the volume is changing with respect to time.

To find the rate at which the radius is changing, we need to differentiate both sides of the equation with respect to time. Recall that the derivative of V with respect to t (dV/dt) represents the rate of change of V with respect to t. Similarly, the derivative of (4/3)πr^3 with respect to t will represent the rate of change of the volume with respect to time.

Differentiating the equation (4/3)πr^3 = V with respect to t using implicit differentiation, we get:

(d/dt) [(4/3)πr^3] = dV/dt.

Differentiating on the left side using the chain rule, we have:

(4/3)π * 3r^2 * (dr/dt) = dV/dt.

Simplifying, we have:

(4/3)π * 3r^2 * (dr/dt) = 12.

Now, we can plug in the value of r when it reaches 1 ft, which means r = 1.

(4/3)π * 3(1^2) * (dr/dt) = 12.

(4/3)π * 3 * (dr/dt) = 12.

(4/3)π * 3 * (dr/dt) = 12.

4π * (dr/dt) = 12.

Now, we can solve for (dr/dt) to find the rate at which the radius is growing:

(dr/dt) = 12 / 4π.

(dr/dt) = 3 / π.

Therefore, the rate at which the radius is growing at the instant when the radius has reached 1 ft is 3/π feet per minute.

12