Question

Help, parameters for my spacecurve is wrong.
Given 2 surfaces find the curve of intersection.

Plot and work inside.

Answer 1

\[z=3-x^2-y^2\]\[z=x^2+2y^2\]

I solved for the intersection of the surfaces:
\[x=\pm \sqrt{(-y^2+1)}\]\[y= \pm 1\]

Then I performed z1-z2 to recive \[-3x^2-3y^2+3\]
I then divided by 3 to get \[-x^2-y^2+1\]

A parameter for this equation may be represented by\[r(t)=[-\cos(t),-\sin(t)]\] by factoring the negative sign.

When substituting that into my original surfaces i get z parameter = -cos(t)^2+2
finally the parametrization is \[r(t)=[-\cos(t),-\sin(t),2-\cos^2(t)] \] This gives the right curve but raise by 1 so changing the z parameter to 1-cos^2(t) will fix it, what am i doing wrong?

Thanks

Answer 2

For some weir reasons the solution is given to be this, can anyone confirm which is the right? I still think my plot is the right one.

Answer 3

Your parameterization doesn't seem right.

Let's first eliminate \(z\):
\[\begin{cases}3-x^2-y^2=z\\[1ex]x^2+2y^2=z\end{cases}\implies 3-x^2-y^2=x^2+2y^2\]and rewrite this so that it represents the equation of an ellipse:
\[3=2x^2+3y^2\implies 1=\frac{2}{3}x^2+y^2\]If \(x=-\cos t\) and \(y=-\sin t\), the above doesn't hold:
\[\frac{2}{3}(-\cos t)^2+(\sin t)^2=\frac{2}{3}\cos^2t+\sin^2t=1-\frac{1}{3}\cos^2t\neq1\]Your parameterization corresponds to the black path under the orange surface in the attachment. (Red is where you should be)

Answer 4

To get that red curve, you can fix \(x\) and \(y\) quite easily. No minus signs are necessary (they get removed upon squaring anyway).

If \(x=\sqrt{\dfrac{3}{2}}\cos t\) and \(y=\sin t\), then you get
\[\frac{2}{3}\left(\sqrt{\frac{3}{2}}\cos t\right)^2+(\sin t)^2=\frac{2}{3}\left(\frac{3}{2}\cos^2t\right)+\sin^2t=\cos^2t+\sin^2t=1\]so what remains is a parameterization for \(z\).

Well, we already know that \(z=3-x^2-y^2\), so we could just use that and say \(z=3-\dfrac{3}{2}\cos^2t-\sin^2t\), or that \(z=x^2+2y^2\) to choose \(z=\dfrac{3}{2}\cos^2t+2\sin^2t\). (These are both equivalent, and you can justify that by using the Pythagorean identity)

Answer 5

Again there is only you there can help me Holster. Thank you very much as always :) Helpful detailed reply as always also ;)

but you agree the 3rd plot i posted is not a representation of the object? the provoided solution gives me that weired shape.

Answer 6

I will definatly remember to isolate x^2+y^2 in any form and try to force it to equal 1 if i get a similar question on my final tomorrow :)

Another neat little trick from you which i´d apperciate as much as all the help you gave me :)

Answer 7

Yeah that third plot looks more like the intersection of a paraboloid with a parabolic cylinder, not another paraboloid.

As always, happy to help!