1) A spherical weather balloon is being inflated. The radius of the balloon is i

Question

1) A spherical weather balloon is being inflated. The radius of the balloon is increasing at the rate of 4 cm/s. Express the surface area of the balloon as a function of time t (in seconds). (Let S(0) = 0.)

S(t)= ________cm^2

2) A spherical balloon is being inflated. The radius of the balloon is increasing at the rate of 9 cm/s. (Find Part C)

(a) Find a function f that models the radius as a function of time t, in seconds.
f(t) = 9t

(b) Find a function g that models the volume as a function of the radius r, in cm.
g(r) = (4r^3/3)

(c) Find g f.
g f =

Explanation / Answer

1) surface area of sphere = 4pir^2

dr/dt = 4 cm/s

integrating dr/st

r = 4t

plugging this r into surface area equation

S(t) = 4pi ( 4t)^2

S(t) = 64 pi*t^2 cm^2

2) f(t) = 9t

g(r) = ( 4pir^3 / 3 )

c) gof = ( 4pi (9t)^3 / 3 )

= 972 pi t^3

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