JakeB57
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I've been going over and over the following question for hours when it should only take a few minutes, I have been close a few times but not quite getting the end result. Was wondering if anyone could come up with the worked solution.

The volume of a spherical bubble is increasing at a constant rate.
Show that the rate of increase of the radius, r, of the bubble is inversely proportional to r^2.

Many thanks.
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physgradstudent
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So you need to start off by coming up with a differential equation for the rate of change of volume, which you know is constant.

Think about what the volume of a sphere is and how you could get from the rate of change of volume to the rate of change of the radius.
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JakeB57
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(Original post by chapmase)
So you need to start off by coming up with a differential equation for the rate of change of volume, which you know is constant.

Think about what the volume of a sphere is and how you could get from the rate of change of volume to the rate of change of the radius.
From differentiating the volume of the sphere, I came up with 4πr^2.
However, the problem I am having is coming up with the rate of changes equation of which I would need to show the inverse proportionality.
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username3331778
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chain rule on dV/dt
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mathstutor24
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You are told that the volume of the sphere is increasing at a constant rate, so you can write this as \frac{dv}{dt}=c (where c is a constant).

You are asked to show (and therefore find) the rate of increase of the radius, which is \frac{dr}{dt}

To find this, use the chain rule: \frac{dr}{dt}=\frac{dv}{dt} \times \frac{dr}{dv}

You already have \frac{dv}{dt} and so need to work out \frac{dr}{dv}

HINT: \frac{dr}{dv}=\frac{1}{\frac{dv}  {dr}}

Now you should be able to figure out \frac{dr}{dt}

HINT: Remember to gather all the constants into one constant, which you can now label k.

You should get \frac{dr}{dt}=\frac{k}{r^{2}}
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JakeB57
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(Original post by mathstutor24)
You are told that the volume of the sphere is increasing at a constant rate, so you can write this as \frac{dv}{dt}=c (where c is a constant).

You are asked to show (and therefore find) the rate of increase of the radius, which is \frac{dr}{dt}

To find this, use the chain rule: \frac{dr}{dt}=\frac{dv}{dt} \times \frac{dr}{dv}

You already have \frac{dv}{dt} and so need to work out \frac{dr}{dv}

HINT: \frac{dr}{dv}=\frac{1}{\frac{dv}  {dr}}

Now you should be able to figure out \frac{dr}{dt}

HINT: Remember to gather all the constants into one constant, which you can now label k.

You should get \frac{dr}{dt}=\frac{k}{r^{2}}
That's ever so helpful thankyou, that makes the problem look so much easier to go about and solve.
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mathstutor24
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No problem! Happy to help.

Please ignore if I'm asking about something you already know, but just in case - are you happy with how to set up differential equations involving a rate of change and a proportion relation?
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JakeB57
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(Original post by mathstutor24)
No problem! Happy to help.

Please ignore if I'm asking about something you already know, but just in case - are you happy with how to set up differential equations involving a rate of change and a proportion relation?
Yes thanks, I think the issue was the constants and putting all the constants under one constant, k.
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