Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
NikolaiAcosta
Group Title
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.
 2 years ago
 2 years ago
NikolaiAcosta Group Title
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.
 2 years ago
 2 years ago

This Question is Open

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

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

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

annag Group TitleBest ResponseYou've already chosen the best response.0
First 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).
 2 years ago

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

annag Group TitleBest ResponseYou've already chosen the best response.0
This 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.
 2 years ago

snadig Group TitleBest ResponseYou've already chosen the best response.0
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 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!!
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.