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.
(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?
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)
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
yep i got the same answers as you did :) thank you for your help
(a) To find a formula for the volume V of the balloon as a function of time t, we need to start with the given formula V(r) = (4/3)πr^3 and relate the radius r to time t.
Given that the radius of the balloon is increasing at a constant rate of 2 inches per second, we can express the radius as r(t) = 2t.
Substituting this expression for r into the formula for volume, we get:
V(t) = (4/3)π(2t)^3 = (4/3)π(8t^3) = (32/3)πt^3
Therefore, the formula for the volume of the balloon as a function of time is V(t) = (32/3)πt^3.
(b) To determine the volume of the balloon after 5 seconds, we substitute t = 5 into the formula for volume:
V(5) = (32/3)π(5^3) = (32/3)π(125) = (4000/3)π ≈ 4188.79 cubic inches.
Therefore, the volume of the balloon after 5 seconds is approximately 4188.79 cubic inches.
(c) To find the time when the balloon bursts, we set the volume V(t) equal to 10,000 cubic inches and solve for t:
(32/3)πt^3 = 10,000
t^3 = 10,000 * (3/32π)
t = (10,000 * (3/32π))^(1/3)
Using a calculator, we can find the value of t. It is approximately 13.876 seconds.
Therefore, the balloon will burst at approximately 13.876 seconds.
(d) To find a formula for the surface area S of the balloon as a function of time t, we use the given formula for surface area of a sphere, S(r) = 4πr^2, and relate the radius r to time t.
Substituting r = 2t into the formula for surface area, we get:
S(t) = 4π(2t)^2 = 4π(4t^2) = 16πt^2
Therefore, the formula for the surface area of the balloon as a function of time is S(t) = 16πt^2.
(e) To determine the surface area of the balloon after 6 seconds, we substitute t = 6 into the formula for surface area:
S(6) = 16π(6^2) = 16π(36) = 576π ≈ 1809.56 square inches.
Therefore, the surface area of the balloon after 6 seconds is approximately 1809.56 square inches.
(f) To find the surface area of the balloon when it bursts, we substitute the time t = 13.876 into the formula for surface area:
S(13.876) = 16π(13.876^2) = 16π(192.194) ≈ 9,650.88 square inches.
Therefore, the surface area of the balloon when it bursts is approximately 9,650.88 square inches.