Question

Air is being pumped into a spherical balloon at a rate of 20ft 3/min. At what rate is the radius changing when the radius is 3 ft?

Answer 1

it asks to find dr/dt it gives as dV/dt = 20; so lets equate volume with radius

V = (4/3) pi r^3 right?

Answer 2

yes

Answer 3

when we derive both sides we get:

\[\frac{dV}{dt} = 4pi\ r^2 \frac{dr}{dt}\]
we need to solve for dr/dt

Answer 4

20 = 4pi (3)^2 * dr/dt

20/(36pi) = dr/dt right?

Answer 5

yep

Answer 6

got it: .176

Answer 7

right?

Answer 8

Another way to look at these is thru the chain rule:
\[\frac{dr}{dt}=\frac{dr}{dV}\frac{dV}{dt}\]
and then you can focus on getting dr/dV

V = (4pi/3) r^3 ; derive with respect to V

dV/dV = (4pi) r^2 dr/dV ; dV/dV = 1

1/(4pi r^2) = dr/dV ; at r=3 we get

1/(36pi) = dr/dV

\[\frac{dr}{dt}=\frac{1}{36pi}\frac{20}{1}\]

Answer 9

.1768... yes