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Problem 48 Hard Difficulty

Try to sketch by hand the curve of intersection of the parabolic cylinder $ y = x^2 $ and the top half of the ellipsoid $ x^2 + 4y^2 + 4z^2 = 16 $. Then find parametric equations for this curve and use these equations and a computer to graph the curve.

Answer

$x=t, y=t^{2}, z=\frac{1}{2} \sqrt{16-t^{2}-4 t^{4}}, \quad-1.37073 \leq t \leq 1.37073$

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Video Transcript

for this problem we have that y equals X squared and X squared plus four y squared plus four z squared equals 16. So this is the top half of the Ellipse oId, um is going to be the positive square root. So we have is, um if we subtract over this right here and this right here we get that four z squared equals 16 minus x squared, minus four y squared. Um, so then Z is going to equal one half the square root of 16 minus x squared, minus four y squared. Then looking at the two equations, we can see that why is the same thing can be represented by X? So what this is going to suggest is we can come up with a Parametric representation. So if we let x equal t since y equals X squared, that means that why will equal t squared and then Z which equals this whole thing right here. We can now, right as being equal to one half 16 minus T squared, minus four teas to the fourth. So based on that, we see that the value in the square root is greater than or equal to zero. So that means that we're limited two negative 13707 less than or equal to T, which is less than or equal to 1.3707 So we have our X component of the Parametric equation, or white component of the Parametric equation rz component of the Parametric equation, and then we have what T is limited to.