Question

Air Is being pumped into a sphere balloon so that its volume increase a rate of 20 cm^3/s. How fast is radius of the balloon increasing when the diameter is 10 cm?
V= 3/4 pi*r^2*h

Answer 1

This is a related rates problem... you first need to decide what the "snapshot" variables are. That means, what are the starting measurements. THEN, you need to determine the "dynamic" variables. This means, what quantities are changing over time. Then, you need to differentiate the Volume equation (keeping in mind that you must differentiate EACH variable with respect to time).

Try that and see if it helps you... if not, let me know.

Answer 2

so d/dt= 3/4 2r (r) h
v=20 cm^3
diameter is 10=5
like this ??

Answer 3

dV/dt=20

Answer 4

dr/dt=?

Answer 5

forget to put V'

Answer 6

dr/dt=5?

Answer 7

dr/dt = the rate at which your radius is changing

Answer 8

which is what the question is asking

Answer 9

then i need find the derivate of r?

Answer 10

no you need to first find a proportion that relates height and radius

Answer 11

found out that i wrote the wrong formula for the volume of the sphere

Answer 12

4/3 r^3 pi

Answer 13

correct alright so now that you have that you need to use implicit differentiation

Answer 14

differentiate in respects to t. what do you get?

Answer 15

\[\frac{dV}{dt}=\frac{4\pi\frac{d}{dt}[r^3]}{3}\]

Answer 16

2-= 4pi (5)^2*r' or dr/dt

Answer 17

don't plug in anything before you differentiate because the rat is always changign

Answer 18

k

Answer 19

The volume expression for a spherical balloon is:\[V=\frac{4 \pi r^3}{3} \]The Total Derivative of the above is:\[dV=4 \pi r^2 dr \]Plug in the given values for dV and dr and then solve for dr.

Answer 20

that wat i did @robtobey

Answer 21

\[20=4pi(5)^2(\frac{dr}{dt})\]=
\[\frac{20}{4\pi(5)^2}=\frac{dr}{dt}\]

Answer 22

dr/dt=1/5 pi

Answer 23

thank for the help