SOLUTION: Very confused on how to solve this related rates problem: Suppose a spherical balloon grows in such a way that after t seconds, its volume V is given by V = 4 &#8730;(t) cm^3.

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Question 994513: Very confused on how to solve this related rates problem:
Suppose a spherical balloon grows in such a way that after t seconds, its volume V is given by V = 4 √(t) cm^3.

After 64 seconds the radius of the balloon is r =
How fast is the radius changing after 64 seconds?
dr/dt =

Thank you!
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
Volume(V) = 4 * square root(t)
and the volume of a sphere is (4/3)*pi*r^3
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after 64 seconds V = 4 * square root(64) = 32
now we use the second equation
32 = (4/3)*pi*r^3
multiply both sides of = by 3
96 = 4*pi*r^3
divide both sides of = by 4
pi*r^3 = 24
divide both sides of = by pi
r^3 = 24 / pi = 7.639437267
*********************************************************
r = (7.639437267)^(1/3) = 1.969490044 cm
*********************************************************
(4/3)*pi*r^3 = 4 * square root(t)
divide both sides of = by 4
(1/3)*pi*r^3 = square root(t)
multiply both sides of = by 3
pi*r^3 = 3*square root(t)
divide both sides of = by pi
r^3 = (3*t^(1/2)) / pi
r = 0.954929659 * t^(1/6) approx t^(1/6)
dr/dt = 1 / t^(5/6)

SOLUTION: Very confused on how to solve this related rates problem: Suppose a spherical balloon grows in such a way that after t seconds, its volume V is given by V = 4 &#8730;(t) cm^3.

SOLUTION: Very confused on how to solve this related rates problem: Suppose a spherical balloon grows in such a way that after t seconds, its volume V is given by V = 4 √(t) cm^3.