Question

A spherical party balloon is being inflated with helium pumped in at a rate of 4 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 2 ft?
The volume of a sphere of radius r is
V =4/3(3.14)r^3

Answer 1

Where are you in relationship to earths core?

Answer 2

does that even matter?

Answer 3

Yes, good question^^.

Answer 4

this is RELATED RATES, calculus

Answer 5

Yes, it does. Without it your lacking the proper centrifugal force.

Answer 6

can someone help me with this question

Answer 7

I just need to do this for practice. Hold on, let me try it on the whiteboard.

Answer 8

dv/dt=4/3*pi*r^2*(dr/dt)
dv/dt=4, plug that in
now solve for dr/dt

Answer 9

and plug in r = 2

Answer 10

i thought u already plugged in 2

Answer 11

sorry dv/dt = 4*pi*r^2*dr/dt

Answer 12

messed up derivative, my bad

Answer 13

you have to plug in the value for r after taking the derivative to that the rate you get (dr/dt) is solved for, right when r = 2

Answer 14

how would i find the derivative of \[4Pir^2\]

Answer 15

16(pi)r?

Answer 16

v=(4/3)*pi*r^3
the only things changing with respect to time in this equation are the radius and the volume. so when you derive this equation you will derive both of these with respect to time.
(4/3)*pi are just constants so d/dt of r^3 = 3*r^2 which leads to
dv/dt=4*pi*r^2*dr/dt

Answer 17

sorry d/dt of r^3 = 3*r^2* (dr/dt),

Answer 18

then find the derivative of that ^^^

Answer 19

6r?