Question

A spherical balloon is expanding. Given that the radius is increasing at the rate of 2 inches per minute, at what rate is the volume increasing when the radius is 5 inches?

Answer 1

ok... so you need

\[\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}\]

if

\[V = \frac{4}{3} \pi r^3\]

what is

\[\frac{dV}{dr} = ?\]

Answer 2

and you are given \[\frac{dr}{dt} = 2\]

so \[\frac{dV}{dt} = \frac{dV}{dr} \times 2\]

just substitute r =5 for find the rate at which the volume is changing with respect to time

Answer 3

SO IS DV/DR=4pir^2?

Answer 4

thats correct

so

\[\frac{dV}{dt} = 4\pi r^2 \times 2\]

now substitute r = 5

and the units for the rate are cubic inches per minute

Answer 5

where do i substitute the 5....

Answer 6

replace r with 5

Answer 7

so you are looking at

\[\frac{dV}{dt} = 4 \pi (5)^2 \times 2\]

Answer 8

4pi50

Answer 9

do i multiply 4*50?

Answer 10

well \[\frac{dV}{dt} = 4 \times \pi \times 25 \times 2 = 200\pi\]

Answer 11

okay! so is the answer left as 200pi?

Answer 12

cubic inches per minute?

Answer 13

yep... thats my best guess

Answer 14

thank you!