Intersection of surface and plane.

[Jimmy5050]

Parameterizing vector function for intersection of cylinder and plane

Homework Statement

Problem asks us to find the vector function of the curve which is created when the plane y= 5/2 intersects the ellptic cyl. (x^2)/4 + (z^2)/6 = 5

Homework Equations

The Attempt at a Solution

I know its going to be an ellipse formed...

I took the given ellptic cyl. equation, and divided by 5 to get (x^2)/20 + (z^2)/30 = 1.

***I parameterized by using x=cos(t) and z=sin(t) and got ((cost)^2)/20 + ((sint)^2)/30 =1.

Now, by looking at examples that are somewhat similar, I could tell the answer by looking at number relationships. However, I am unsure of the true way to go about getting my final solution.

My "made up way" of solving was to set either x or z to zero before parameterizing.

My final answers are x=(sqrt[20])cos(t) y=5/2 z=(sqrt[30])sin(t)

I checked my answer using a graphing program, and it is correct, but I am just unsure about going about the TRUE way of solving once I get to the part labeled *** above.

Thanks.

 

[Jimmy5050]

Any takers? I'm sure I have the answer right, just not sure of the "correct" last couple steps to get to that answer so that I can show work properly.

 

[Mark44]

***I parameterized by using x=cos(t) and z=sin(t) and got ((cost)^2)/20 + ((sint)^2)/30 =1.

This step seems flaky to me.

The correct parametrization is, I believe, x = a*cos(t), z = b*sin(t). Then you have
$$\frac{a^2 cos^2(t)}{20} + \frac{b^2 sin^2(t)}{30} = 1$$

For this equation to be identically true for all t, it must be that a² = 20 and b² = 30, which makes the parametric form of the ellipse (in the x-z plane)
x = $\sqrt{20}$ cos(t)
z = $\sqrt{30}$ sin(t)

I think this is the right way to go about it.