Rate of change of area in sphereical baloon, .

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Homework Help Overview

The discussion revolves around a problem involving the rate of change of area of a spherical balloon being inflated at a constant volume increase rate. The original poster is tasked with finding the rate of change of the surface area when the radius is 6 inches, given the volume increases at a rate of 10 cubic inches per second.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between volume and radius, applying the chain rule to relate the rates of change of volume and area. Questions arise about how to isolate variables and the proper application of derivatives in this context.

Discussion Status

Some participants have provided guidance on applying the chain rule correctly and have pointed out potential errors in the original poster's approach. There is an ongoing exploration of how to derive the necessary rates of change from the given information, with no explicit consensus reached.

Contextual Notes

There is some confusion regarding the values used in the calculations, particularly the rate of volume increase and the radius. Participants are encouraged to clarify their assumptions and ensure consistency in their variables.

christian0710
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Homework Statement


I have a homework assignment that I find challengingA spherical balloon is being inflated at a rate of 10 cu.in/sec (i assume it's cubic inches per second)

Find the rate of change of area when the balloon has radius=6

Homework Equations



So far I know that

V=(4/3)*pi*r^3
A=4*pi*r^2

dV/dt = 10 in^3/secSo the question is, how do I find dA/dt when radius=6 and the volume increases at a rate of 10 in^3/sec ?

I'm realyl stuck here. Can someone help?

The Attempt at a Solution



I've just been wild guessing here. I know that dV/dt= 4*pi*r*dr/dt and that dV/dt=10, but if i substitute it into
10= 4*pi*r*dr/dt and isolate dr/dt, I don't really get anywhere.

The answer is supposed to be 10/3 in^2 sec.
 
Last edited:
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Apply the chain rule more carefully. You have some errors.

I find it helpful to slow down and take more careful (smaller steps):

dV/dt = dV/dr * dr/dt

dV/dr = stuff
...
 
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So dV/dr = 4*pi*r^2
dr/dt = How do i find this?
 
christian0710 said:
So dV/dr = 4*pi*r^2
dr/dt = How do i find this?

It's the only unknown left in the dV/dr equation.

Solve for it as a number, then plug that number into your equation for dA/dt = dA/dr * dr/dt.
 
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You are given a rate of inflation, this is dV/dt. First you say it is 6 but in your work you refer to 10. Please be more clear.
You have the stuff you need already.
dV/dt = dV/dr * dr/dt
dA/dt = dA/dr * dr/dt

You can simply use the given information to find dr/dt, then plug that into your equation for change in area.

[edit] I was too late.
 
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Thank you so much, it works!

just a last question: So how do you just know that you must apply the 2 chain rules beneath?

dV/dt = dV/dr * dr/dt
dA/dt = dA/dr * dr/dt.

Is it beacuse you know that the change in V and A depends on time and on radius r?
 
It is because the change in r depends on t.
If G(a) = G(a(t)) then dG/dt = dG/da*da/dt. That is what the chain rule says.
 
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I knew what to do from years of teaching Calculus and for the reasons RUber described.

Students may not always see what to do, so I encourage them:

When you do not know what to do, take some steps you do know how to execute properly.

Then reassess and see if you are any closer to finding the variable you are trying to solve for.
 
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Thank you very much! I really appreciate the help !
 
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