A spherical balloon is being inflated at a rate of 10 cubic inches per second. How fast is the radius of the balloon increasing when the surface area of the balloon is 2pi square inches? Enter your answer correct to 3 decimal places.

V = (4/3)πr^3

dV/dt = 4πr^2 dr/dt (#1)

when 4πr^2 = 2π
r^2 = 1/2
r = 1/√2

in #1:
10 = 4π(1/√2)^2 dr/dt
10= 2π dr/dt
dr/dt = 5/π or 1.592 inches/sec

check my arithmetic

To find how fast the radius of the balloon is increasing, we need to use related rates and differentiate the equation for the surface area of a sphere.

The formula for the surface area of a sphere is given by:

A = 4πr^2

Where A is the surface area and r is the radius of the sphere.

We are given that the surface area of the balloon is 2π square inches, so we can set up the equation:

2π = 4πr^2

Dividing both sides by 4π, we get:

r^2 = 0.5

Now, we can differentiate both sides of the equation with respect to time (t) using the chain rule:

d/dt(r^2) = d/dt(0.5)

2r * dr/dt = 0

Since the balloon is being inflated at a rate of 10 cubic inches per second, the rate of change of the volume, dV/dt, is equal to 10. And since the volume of a sphere is given by:

V = (4/3)πr^3

We can differentiate the volume equation with respect to time:

dV/dt = 4πr^2 * dr/dt

Substituting the given rate of change of volume, dV/dt = 10, and rearranging the equation:

10 = 4πr^2 * dr/dt

We already know that r^2 = 0.5, so we can substitute it into the equation:

10 = 4π(0.5) * dr/dt

Simplifying, we get:

10 = 2π * dr/dt

Now, we can solve for dr/dt:

dr/dt = 10 / (2π)

Evaluating this expression, we find:

dr/dt ≈ 1.591

Therefore, the radius of the balloon is increasing at a rate of approximately 1.591 inches per second when the surface area of the balloon is 2π square inches.