The volume V of a sphere of radius r is given by the formula V (r) = (4/3)πr^3. A balloon in the shape of a sphere is being inflated with gas. Assume that the radius of the balloon is increasing at the constant rate of 2 inches per second, and is zero whent = 0.

Question

(a) Find a formula for the volume V of the balloon as a function of time t.
(b) Determine the volume of the balloon after 5 seconds.
(c) Starting with an empty balloon, suppose that the balloon will burst when
its volume is 10, 000 cubic inches. At what time will the balloon burst?
(d) Find a formula for the surface area S of the balloon as a function of time
t; recall the surface area formula for a sphere of radius r is S(r) = 4πr^2.
(e) Determine the surface area of the balloon after 6 seconds.
(f) What will be the surface area of the balloon when it bursts?

Answer 1

V = 4/3 πr^3
(a) dV = 4πr^2 dr
V = ∫[0,t] 4πr^2 dr
(b) now plug in t=5
(c) find when (a) = 10000
(d) dA = 8πr dr
work it as in (a)
(e) work as in (b)
(f) plug in t as in (c)

Answer 2

since r = 2t, let's replace r in the equation

V = (4/3)π r^3
= (4/3)π (2t)^3
= (32/3)π t^3

b) replace t with 5 in my formula.
c) (32/3)π t^3 = 10000
solve for r

d) S(r) = 4πr^2 , let do the same thing: r = 2t
= 4π(2t)^2 = 16πt^2

e) let t = 6 and evaluate using the new formula from d)

f) sub in the value of t you got in c) into the new S(r) formula

let me know what your answers are so I can check them

Answer 3

yep i got the same answers as you did :) thank you for your help