Ask Question
14 August, 16:59

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?

+1
Answers (1)
  1. 14 August, 20:48
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers