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NikolaiAcosta
 3 years ago
Hi! How can I solve this problem?
 Find the points of the cylinder x=4z^2 so that the tangent plane is paralell to the plane yz.
NikolaiAcosta
 3 years ago
Hi! How can I solve this problem?  Find the points of the cylinder x=4z^2 so that the tangent plane is paralell to the plane yz.

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Deni_Morales28
 3 years ago
Best ResponseYou've already chosen the best response.1Here you can find similar problems: http://www.math.psu.edu/adler/courses/math230/HWs/15_6_hw_soln.pdf

NikolaiAcosta
 3 years ago
Best ResponseYou've already chosen the best response.0Just perfect! Thank you very much!

Deni_Morales28
 3 years ago
Best ResponseYou've already chosen the best response.1you welcome, hope it hepls!!

annag
 3 years ago
Best ResponseYou've already chosen the best response.0First of all, this "cylinder" is not a circular cylinder like we're used to from precalc, but rather, a similar shape formed from a parabola projected along the yaxis. 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 yvalue).

annag
 3 years ago
Best ResponseYou've already chosen the best response.0This means that, in effect, the shape (or rather, the surface) is formed by lines parallel to the yaxisdw:1350320098224:dw

annag
 3 years ago
Best ResponseYou've already chosen the best response.0This means we get (for free) a bunch of lines inside planes parallel to the yaxis. Even better, every plane tangent to this surface will automatically be parallel to the yaxis. So, we just need to find which of these planes is/are parallel to the zaxis. (Note that saying the yzplane is the same as saying the plane perpendicular to the xaxis.) This reduces it to a precalculus question: Find the point in the xz plane (perpendicular to the yaxis) where the slope of the parabola is the same as the slope of the zaxis, ie is zero. Since this parabola is lined up with the x and zaxes (ie, is not diagonal), this is the same as saying, "find the vertex", which also happens to be centered on the xaxis in this case.

snadig
 3 years ago
Best ResponseYou've already chosen the best response.0I 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 xaxis which is perpendicular to yz plane. This in turn implies that the tangent plane described by this normal vector would turn out to be parallel to the yz plane. The answer is basically z = 0!!
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