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.
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)