The parabolic cylinder y=x^2 and the top half of the ellipsoid x^2 + 4y^2 + 4z^2 = 16. Find parametric equations for this curve.

Answer 1

We will need to find the intersection of the two surfaces.

Substituting y = x^2 into the equation of the ellipsoid, we get:

x^2 + 4(x^2)^2 + 4z^2 = 16

Simplifying and solving for z, we get:

z = ±sqrt(4 - 5x^2 - 4x^4)/2

Since we only want the curve in the top half of the ellipsoid, we will only use the positive square root.

So our parametric equations are:

x = t

y = t^2

z = sqrt(4 - 5t^2 - 4t^4)/2

for -sqrt(2/5) ≤ t ≤ sqrt(2/5) (this is the range of t that ensures that the expression under the square root is non-negative so that z is real).

We can also write the equations in vector form as:

r(t) = <t, t^2, sqrt(4 - 5t^2 - 4t^4)/2>

for -sqrt(2/5) ≤ t ≤ sqrt(2/5)