Rate of change of width of an ellipse as volume changes

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monster22

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Hi, came across a difficult problem that I don't really know how to complete and I can't get my head around it - any help would be appreciated!

If I have the cartesian equation of an ellipse centered at the origin with some width (distance between x ints..) and I'm told that the volume of the ellipse is decreasing (dv/dt = - 50 let's say). I'm asked for the rate of change of width as the volume of the ellipse decreases.

Now I can see that:

dx/dt = dx/dV . dV/dt

I'm given dV/dt in the problem that's no issue - my question is how to find dx/dV.

In the previous part of the question I use an integral to find the volume in terms of x so I feel like the problem wants me to say dV/dx = the derviative of my integral for the volume. ( fundamental thm of calculus ).

Then once I have dV/dx flip it to get dx/dV and away you go.

My question is can I do this? I feel like because the volume of the ellipse is decreasing taking the derivative of my volume integral shouldn't work? Because as the volume of the ellipse decreases so does it's equation?

Any help would be greatly appreciated I can't get my head around this one!

Thanks in advance.
 
monster22 said:
Hi, came across a difficult problem that I don't really know how to complete and I can't get my head around it - any help would be appreciated!

If I have the cartesian equation of an ellipse centered at the origin with some width (distance between x ints..) and I'm told that the volume of the ellipse is decreasing (dv/dt = - 50 let's say). I'm asked for the rate of change of width as the volume of the ellipse decreases.

Now I can see that:

dx/dt = dx/dV . dV/dt

I'm given dV/dt in the problem that's no issue - my question is how to find dx/dV.

In the previous part of the question I use an integral to find the volume in terms of x so I feel like the problem wants me to say dV/dx = the derviative of my integral for the volume. ( fundamental thm of calculus ).

Then once I have dV/dx flip it to get dx/dV and away you go.

My question is can I do this? I feel like because the volume of the ellipse is decreasing taking the derivative of my volume integral shouldn't work? Because as the volume of the ellipse decreases so does it's equation?

Any help would be greatly appreciated I can't get my head around this one!

Thanks in advance.
Click to expand...
A big "problem" with your problem statement is that:

An ellipse is a 2-dimensional geometric figure and

it does NOT have a volume.

However, ellipses have enclosed area.

Please review your problem statement and post corrected statement.

Please show us what you have tried and exactly where you are stuck.​
​
Please follow the rules of posting in this forum, as enunciated at:​
​
​
Please share your work/thoughts about this assignment.​
 
Subhotosh Khan said:
A big "problem" with your problem statement is that:

An ellipse is a 2-dimensional geometric figure and

it does NOT have a volume.

However, ellipses have enclosed area.

Please review your problem statement and post corrected statement.

Please show us what you have tried and exactly where you are stuck.​
​
Please follow the rules of posting in this forum, as enunciated at:​
​
​
Please share your work/thoughts about this assignment.​
Click to expand...

sorry! I meant the 3D ellipsoid say a balloon, that is represented by a 2D ellipse ( with cartesian equation given ) when viewed from the side.

So for the first part: finding the volume for the balloon i use V = pi INT ( y^2) .dx from -x to x .. ( rotating around the x axis for the volume.

it’s whether i can take the derivative of that to get dv/dt when the volume is decreasing i’m
Unsure about.

thanks
 
monster22 said:
sorry! I meant the 3D ellipsoid say a balloon, that is represented by a 2D ellipse ( with cartesian equation given ) when viewed from the side.

So for the first part: finding the volume for the balloon i use V = pi INT ( y^2) .dx from -x to x .. ( rotating around the x axis for the volume.

it’s whether i can take the derivative of that to get dv/dt when the volume is decreasing i’m
Unsure about.

thanks
Click to expand...
You say:

V = pi INT ( y^2) .dx from -x to x

That is not quite correct.

I'll do first few steps.

Let us assume:

2*a = length of the major axis (along x-axis)

2*b = length of the minor axis (along x-axis)

so the equation of the ellipse:

\(\displaystyle y^2 = b^2 * (1 - \frac{x^2}{a^2})\)................................................(1)

Calculate the volume using the equation above.

It will be a function of 'a' and 'b'. In this case 'a' and 'b' will be functions of 't'. Remember when you are blowing a balloon, the size of the balloon changes by changing the "width" and the "length".

You differentiate the "volume" above (eqn. 1) and calculate the rate of change of "a" and "b". This will be partial derivative.

I suspect that you have NOT posted the complete problem. There might be important information not reported in your problem statement.
 
Right, (thanks)

So am I correct in saying the suggested solution doesn't work? Which was that we can take the derivative of V(t) = Pi* (integral expression) to give us dv/dx. then use this with our related rates to find dx/dt. (Partial derivatives haven't been covered here so can't be used)

Logically it doesn't seem to work - because the balloon is deflating it's shape will change which means I can't take the derivative of the volume expression (the integral) as the equation in the integral represents the initial shape of the balloon.

Thanks again.
 
monster22 said:
Right, (thanks)
Which was that we can take the derivative of V(t) = Pi* (integral expression) to give us dv/dx. then use this with our related rates to find dx/dt. (Partial derivatives haven't been covered here so can't be used)

Logically it doesn't seem to work - because the balloon is deflating it's shape will change which means I can't take the derivative of the volume expression (the integral) as the equation in the integral represents the initial shape of the balloon.

Thanks again.
Click to expand...
You say:

So am I correct in saying the suggested solution doesn't work?​

I would say - No - we have not seen the exact problem statement

I suspect that you have NOT posted the complete problem. There might be important information not reported in your problem statement. Can you post a picture of the problem?
 
monster22 said:
So am I correct in saying the suggested solution doesn't work? Which was that we can take the derivative of V(t) = Pi* (integral expression) to give us dv/dx. then use this with our related rates to find dx/dt. (Partial derivatives haven't been covered here so can't be used)

Logically it doesn't seem to work - because the balloon is deflating it's shape will change which means I can't take the derivative of the volume expression (the integral) as the equation in the integral represents the initial shape of the balloon.
Click to expand...

One thing that has to be stated is what is remaining the same as the balloon deflates. Does it keep the same shape and just change size, or do two axes remain the same while the third changes, or does one axis stay the same while the other two remain equal, or does it keep the same surface area (that would be a very hard problem), or what?
 
Hi,

The problem isn't clear on this (I don't think). It just states the rate at which the volume of the balloon decreases and it tells us that the width is defined as the maximum horizontal distance when looking at the 2D cross section (the ellipse) then asks us to find the rate of change of width of the balloon when the balloon is a certain width wide.

I would imagine we are to assume it keeps the same shape (but it's size decreases in the same ratio?) That seems the most likely with no clear definition of a different scenario.

As I said - no problems finding the volume of the balloon with it's initial size - and no problem solving the related rates problem in the end - it's whether I can take the derivative of the V(x) equation (that I get from the volume integral in previous part) to get dV/dx for the related rates. It just doesn't seem right that I can do that when the equation for the 2D ellipse cross section would be changing to satisfy the change in volume.
 
If it doesn't say what remains fixed, then it is a bad problem, and you have no obligation to try to solve it. If you could show us an image of the actual problem, we might see more in it than you do, though. PLEASE DO SO, AS YOU HAVE BEEN ASKED BEFORE!

You do know the formula for the volume of an ellipsoid, right? [MATH]V = \frac{4}{3}\pi a b c[/MATH]. You don't need to work this out as an integral if the problem is about related rates.

If you want to assume it has a fixed shape (that is, a constant ratio of dimensions), then do so.

Let [MATH]a = x[/MATH], [MATH]b = sx[/MATH], [MATH]c = tx[/MATH], so that the semiaxes are in the constant ratio 1:s:t, and go for it.

On the other hand, since you've mentioned knowing the equation of a side view, it sounds like two semiaxes are fixed, and only the third is variable. If you think so, then try it that way.
 
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