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NikolaiAcosta
Hi! How can I solve this problem? - Find the points of the cylinder x=4-z^2 so that the tangent plane is paralell to the plane yz.
Here you can find similar problems: http://www.math.psu.edu/adler/courses/math230/HWs/15_6_hw_soln.pdf
Just perfect! Thank you very much!
you welcome, hope it hepls!!
First of all, this "cylinder" is not a circular cylinder like we're used to from pre-calc, but rather, a similar shape formed from a parabola projected along the y-axis. That is, it doesn't matter what y is; the shape will be the same for all y (or, at any y it will be the same as at any other y-value).
This means that, in effect, the shape (or rather, the surface) is formed by lines parallel to the y-axis|dw:1350320098224:dw|
This means we get (for free) a bunch of lines inside planes parallel to the y-axis. Even better, every plane tangent to this surface will automatically be parallel to the y-axis. So, we just need to find which of these planes is/are parallel to the z-axis. (Note that saying the yz-plane is the same as saying the plane perpendicular to the x-axis.) This reduces it to a pre-calculus question: Find the point in the x-z plane (perpendicular to the y-axis) where the slope of the parabola is the same as the slope of the z-axis, ie is zero. Since this parabola is lined up with the x- and z-axes (ie, is not diagonal), this is the same as saying, "find the vertex", which also happens to be centered on the x-axis in this case.
I agree with annag's answer... I'm not sure if my thought process is correct, but here I go anyways... The gradient of the function is <1,0,2z>. At any point of the form <a,0,0> (i.e., any value for x; and z set to 0) the gradient would be <1,0,0>. Thus, the gradient vector is along the x-axis which is perpendicular to y-z plane. This in turn implies that the tangent plane described by this normal vector would turn out to be parallel to the y-z plane. The answer is basically z = 0!!