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Ive never seen this type of problem before but i got it for homework today.

 

A spherical ballon is being inflated in such a way that its radius is increasing at the constant rate of 3cm/min. If the volume of the balloon is 0 at time 0, at what rate is the Volume increasing after 5 minutes?

 

I think the equation is sorta like V(r) = 4r3/3 but that is for feet and I need cm.

 Sep 7, 2016
 #1
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A spherical ballon is being inflated in such a way that its radius is increasing at the constant rate of 3cm/min.drdt=3cm/min

The relation between the volume and radius of the spherical ballon is v(r)=4r3/3

The dervative of volume v  respect to variable v is dvdr=4r2

At time 5 minutes dvdr=452=100cm^2

Applying chain rule,dvdt=dvdrdrdt

dvdt=100cm23cm/min=300cm3/min

This is a kind of problem relative rate of change,application of differvative.

laugh

 Sep 7, 2016
 #2
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Thank you so much

IronDome  Sep 7, 2016
 #3
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Volume of Sphere=4/3. r^3. Pi

 Sep 7, 2016
 #4
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That's a derivative problem..... Not a usual volume problem.

MaxWong  Sep 7, 2016

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