Question

related rates problem. Air is being pumped into a spherical balloon at a rate of 5cm^3/min. Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20cm.

Answer 1

do you have the formula for the area of a sphere given the radius?

Answer 2

yep. V= 4/3pi(r^3)

Answer 3

what next?

Answer 4

ok then take the derivative with respect to time and you get
\[V'=4\pi r^2r'\]

Answer 5

you are told that
\[V'=5\] so solve for
\[r'\]

Answer 6

where did the 3 from the 4/3 go?

Answer 7

\[\frac{d}{dr}r^3=3r^2\] so the 3 canceled

Answer 8

so the 3 on 3r\[^{2}\] was canceled?

Answer 9

chain rule says
\[\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}\]

Answer 10

you know
\[\frac{dV}{dt}=5\] because that is what you were told in the problem

Answer 11

\[V=\frac{4\pi}{3}r^3\] so \[\frac{dV}{dr}=4\pi r^2\]

Answer 12

ok. that sort of makes sense.

Answer 13

and therefore you get
\[\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}\] so
\[5=4\pi r^2 \frac{dr}{dt}\]

Answer 14

you want to find r' when r=10

Answer 15

you want
\[\frac{dr}{dt}\] so replace r by 10 and solve for it

Answer 16

ok. one sec..

Answer 17

you get
\[5=40\pi \frac{dr}{dt}\] etc

Answer 18

?

Answer 19

oops

Answer 20

wait.. isn't r^2 100?

Answer 21

maybe it should be
\[5=400\pi r'\]

Answer 22

\[5=400 \pi r'\]

Answer 23

that makes more sense. :)

Answer 24

what myininaya said. good thing she is here to catch my mistakes....

Answer 25

thats what i'm good for

Answer 26

(and vice versa!)

Answer 27

yes

Answer 28

now back to work for me!

Answer 29

so now that I have 5= 400pi(dr/dt), what next?

Answer 30

lol solve for r'

Answer 31

divide both sides by 400pi

Answer 32

so.. divide both sides by 400\[\pi\] to get: 5/400pi = dr/dt

Answer 33

5/(400pi)
you can reduced the fraction