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You are blowing air into a spherical balloon at a rate of 33 cubic inches per second. Given that the radius of the balloon is 22 inches when t=4t=4 seconds answer the following questions:

a. How fast is the radius of the balloon growing at t=4 seconds.
b. What is the rate of change of the surface area at t =4 seconds?

Answer :

Answer:

a) the radius of the balloon increases at a rate of 5.42 in/s

b) the surface area of the balloon increases at a rate of 3 in²/s

Step-by-step explanation:

a) since the volume of a sphere V is

V= 4/3*π*R³

where R= radius , then the rate of change of the volume is

V' = dV/dR= 4*π*R²

using the chain rule

dV/dt = dV/dR*dR/dt

thus

k = 4*π*R² * dR/dt

dR/dt = k/(4*π*R²)

replacing values

dR/dt = k/(4*π*R²) = (33 in³/s) /(4*π*(22 in)²] = 5.42 in/s

then the radius of the balloon increases at a rate of 5.42 in/s

b) since the surface area is

S=4*π*R²

then

S' = dS/dR= 8*π*R

and

dS/dt = dS/dR*dR/dt = 8*π*R * k/(4*π*R²) = 2*k/R

replacing values

dS/dt = 2*k/R = 2*(33 in³/s)/( 22 in) = 3 in²/s

then the surface area of the balloon increases at a rate of 3 in²/s

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