Expert: Paul Klarreich - 11/21/2005

Question
1) a spherical balloon is to be deflated so that its radius decreases at a constant rate of 15 cm/min. at what rate must air be removed when the radius is 9 cm?
2) Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 6 mi^2/h. how fast is the radius of the spill increasing when the area is 9 mi^2?

Answer
Hi, Amy,

Question:  1) a spherical balloon is to be deflated so that its radius decreases at a constant rate of 15 cm/min. at what rate must air be removed when the radius is 9 cm?

You are really having fun with these.  By now, you know just what to do, but your life experience has never encompassed these large balloons.  No matter.

Variables:
V = volume of the balloon.
r = radius.............

Rates:
dV/dt = rate of deflation, cm^3/min, to be found.
dr/dt = rate of change of radius = -15. (negative because it is decreasing)

Relationship:  Formula for the volume of a sphere.

V = 4pi r^3/3

Differentiate implicitly:
dV   4 pi(3r^2) dr
-- = ---------- --
dt      3       dt

Put in your  dr/dt = -15, your r = 9, and you're on your way.


2) Oil spilled from a ruptured tanker spreads in a circle whose area increases at a constant rate of 6 mi^2/h. how fast is the radius of the spill increasing when the area is 9 mi^2?

By now, you should have the scheme:

Variables:
A = area of spill
r = radius of spill

Rates:
dA/dt = rate of increase of area = 6
dr/dt = ....................radius, to be found.

Relation:
Area of a circle:  A = pi r^2

dA          dr
-- = 2 pi r --
dr          dt

Put in your dA/dt = 6, your A = 9,... oops, you'll have to find the value of r for this value of A, but you have the formula.