Post a New Question

Calculus

posted by .

Air is being pumped into a spherical balloon so that its volume increases at a rate of 30 cubic centimeters per second. How fast is the surface area of the balloon increasing when its radius is 19 cm?
V= 4/3*pi*r^3
S= 4 pi r^2

  • Calculus -

    from V= 4/3*pi*r^3
    dV/dt = 4pi(r^2)dr/dt
    so when dV/dt=30 and r = 19
    30 = 4pi(361)dr/dt
    dr/dt = 30/[4pi(361)]

    now in
    A = 4pir^2
    dA/dt = 8pi(r)dr/dt
    = 8pi(19)*30/[4pi(361)]
    = you finish it.

  • Calculus -

    Got it down to 3.157 which is correct. Thanks!

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question