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Differential Geometry - Envelopes (FP3)

I need to show that the envelope of the family of circles which passes through the origins and have centres on the circle x2+y22x=0 x^2 + y^2 - 2x = 0 is the cardioid with polar equation r=2(1+cosθ) r=2(1+cos\theta) .

So far, my thoughts are if the centre of the family of circles was (x,y), then the radius squared would be x^2 + y^2 which is also 2x. The centres of these circles will satisfy the equation (x1)2+y2=1 (x-1)^2 + y^2 = 1 . I could introduce a parameter, say x=p x = p , and y=2pp2 y=\sqrt{2p-p^2} , and so: f(x,y,p)=(xp)2+(y2pp2)2=2p f(x, y, p) = (x-p)^2 + (y-\sqrt{2p-p^2})^2 = 2p . But this doesn't look right because when I do fp \frac{\partial f}{\partial p} I don't have p in terms of x's and y's so I can't form simultaneous equations here.

I know that the Cartesian equation of the cardioid is x2+y2=x2+y2+2x x^2 + y^2 = \sqrt{x^2+y^2} + 2x if that's of any help.

Any hints?

EDIT: Actually, fp \frac{\partial f}{\partial p} does work, except it's really ugly, and I still can't see how I could eliminate p...
EDIT 2: Alternative approach: I note that x2+y22x=0 x^2 + y^2 - 2x = 0 is the same as r=2cosθ r = 2cos\theta
(edited 11 years ago)
Original post by Blazy
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I didn't like your parameterisation, since for the circle x2+y22x=0x^2+y^2-2x=0 there are two y values for a given x value, which you don't get.

I used y=sint,x=1costy=\sin t, x=1-\cos t where t is the angle from (1,0) to the centre of the parameterised circle.

This gave me a tant=yx\tan t =-\dfrac{y}{x} from the partial derivative.

My final equation was r=2(cosθ2)r=2(\cos\theta-2), which although different to the given answer, does yield the same curve.

The working was rather horrendous, and I don't know why I've ended up with a different representation of the curve.

Edit: Correct point to (1,0)
(edited 11 years ago)
Reply 2
Original post by ghostwalker
I didn't like your parameterisation, since for the circle x2+y22x=0x^2+y^2-2x=0 there are two y values for a given x value, which you don't get.

I used y=sint,x=1costy=\sin t, x=1-\cos t where t is the angle from (1,0) to the centre of the parameterised circle.

This gave me a tant=yx\tan t =-\dfrac{y}{x} from the partial derivative.

My final equation was r=2(cosθ2)r=2(\cos\theta-2), which although different to the given answer, does yield the same curve.

The working was rather horrendous, and I don't know why I've ended up with a different representation of the curve.

Edit: Correct point to (1,0)


This looks pretty good, I'll have a look at it again later when I'm feeling more awake...
Reply 3
Original post by ghostwalker
x


I have also ended up with the partial derivative: tant=yx tant=-\frac{y}{x} but where should I go from here on? My f(x,y,t) f (x,y,t) is currently x22x+2xcost+y22ysint=0 x^2 - 2x + 2xcost+y^2-2ysint=0 and I'm not sure how I should eiminate the sines and cos...I'm guessing this is where it gets horrendous?
Original post by Blazy
I have also ended up with the partial derivative: tant=yx tant=-\frac{y}{x} but where should I go from here on? My f(x,y,t) f (x,y,t) is currently x22x+2xcost+y22ysint=0 x^2 - 2x + 2xcost+y^2-2ysint=0 and I'm not sure how I should eiminate the sines and cos...I'm guessing this is where it gets horrendous?


Knowing tan t, you can write expressions for sin and cos - use a triangle.

At this point I'd start transforming into polar coordinates; "r" for x2+y2\sqrt{x^2+y^2} as it arises, etc.; reduces some of the horror, but it's still above Boris Karloff level.
(edited 11 years ago)
Reply 5
Original post by ghostwalker
Knowing tan t, you can write expressions for sin and cos - use a triangle.

At this point I'd start transforming into polar coordinates; "r" for x2+y2\sqrt{x^2+y^2} as it arises, etc.; reduces some of the horror, but it's still above Boris Karloff level.


Let sint=yr \displaystyle sint = -\frac{y}{r} and cost=xr \displaystyle cost = \frac{x}{r} , I've ended up with r=2(cosθ1) \displaystyle r = 2(cos\theta -1) which has the same shape as r=2(cosθ+1) r = 2(cos\theta + 1) . I guess this will have to do. Thanks!
Original post by Blazy
Let sint=yr \displaystyle sint = -\frac{y}{r} and cost=xr \displaystyle cost = \frac{x}{r} , I've ended up with r=2(cosθ1) \displaystyle r = 2(cos\theta -1) which has the same shape as r=2(cosθ+1) r = 2(cos\theta + 1) . I guess this will have to do. Thanks!


Yes, I used those values for sint, cost.

It's possible we may get the given solution by assigning the minus sign to cost, rather than to sint.

Haven't thought of a justification for doing one rather than the other though.
Reply 7
Original post by ghostwalker
Yes, I used those values for sint, cost.

It's possible we may get the given solution by assigning the minus sign to cost, rather than to sint.

Haven't thought of a justification for doing one rather than the other though.


Would it work if we say, take a 0 point, go right by x steps, down y steps, and the angle formed would be t. We could go left x steps (so negative x) and go up y steps (positive y) and in turn, create the same angle t and hence preserves the tan t = -y/x? (i.e. Angles opposite to each other are equal).

EDIT: Hang on, surely if tant=-y/x = y/-x then sint = y/r and cost = -x/r because y/r divided by -x/r will give us -y/x?
(edited 11 years ago)
Original post by Blazy

EDIT: Hang on, surely if tant=-y/x = y/-x then sint = y/r and cost = -x/r because y/r divided by -x/r will give us -y/x?


This is the other assignment of the minus sign. But I've not come up with a justification for one or the other being corrrect. And since they both yield the envelope, it's may be that there is no way to differentiate between the two.

I do note that with the given solution r is >= 0, whereas with the one we have it's not.
Reply 9
Original post by ghostwalker
This is the other assignment of the minus sign. But I've not come up with a justification for one or the other being corrrect. And since they both yield the envelope, it's may be that there is no way to differentiate between the two.

I do note that with the given solution r is >= 0, whereas with the one we have it's not.


Fair enough. Thanks a lot for your help!
Original post by Benjy100
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:zomg: We've been Benjy'd.

I hope that was an international competition, and not just the local one for Guadeloupe.

Original post by Blazy
Fair enough. Thanks a lot for your help!


Np. I'll carry on thinking about it, and drop you a pm if any light dawns.

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