- #1
JosephTraverso2
- 8
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A balloon's volume is increasing at a rate of dV/dt. Express the rate of change of the circumference with respect to time (dc/dt) in terms of the volume and radius.
Vsphere = (4/3)(π)(r^3)
C = (2)(π)(r)
My strategy was to come up with two separate equations of dr/dt and substitute leading to an equation relating dv/dt and dc/dt
dV/dt = (4)(π)(r^2) (dr/dt)
dc/dt = (2)(π)(dr/dt)
(dc/dt) (1/2π) = dr/dt
Substitute...
(4)(π)(r^2) (dc/dt) (1/2π) = dv/dt
Canceling like terms
(2r^2)(dc/dt) = dv/dt
(1/2r^2) (dv/dt) = (dc/dt)
Now I was curious and wanted to graph the function just for fun. I know dv/dt is some number so I assumed it is 3. This yields
3/2r^2 = dc/dt
If I were to graph this function on a graphing utility would I be getting the circumference on the y-axis and time on the x? I ask this because the function is decreasing but the circumference of the balloon is increasing. Please help.
Homework Equations
Vsphere = (4/3)(π)(r^3)
C = (2)(π)(r)
The Attempt at a Solution
[/B]My strategy was to come up with two separate equations of dr/dt and substitute leading to an equation relating dv/dt and dc/dt
dV/dt = (4)(π)(r^2) (dr/dt)
dc/dt = (2)(π)(dr/dt)
(dc/dt) (1/2π) = dr/dt
Substitute...
(4)(π)(r^2) (dc/dt) (1/2π) = dv/dt
Canceling like terms
(2r^2)(dc/dt) = dv/dt
(1/2r^2) (dv/dt) = (dc/dt)
Now I was curious and wanted to graph the function just for fun. I know dv/dt is some number so I assumed it is 3. This yields
3/2r^2 = dc/dt
If I were to graph this function on a graphing utility would I be getting the circumference on the y-axis and time on the x? I ask this because the function is decreasing but the circumference of the balloon is increasing. Please help.