What is the Tangent Plane at a Given Point on a Level Surface?

  • Thread starter Thread starter makyol
  • Start date Start date
makyol
Messages
17
Reaction score
0
f(x, y, z) = 0 f(x, y, z) = ...

Homework Statement



[PLAIN]http://www.netbookolik.com/wp-content/uploads/2010/07/q12.png

Homework Equations





The Attempt at a Solution



Sorry have no idea:(
 
Last edited by a moderator:
Physics news on Phys.org


If a surface is given by f(x,y,z)= 0 (or any constant) we can think of it as a "level surface" so \nabla f= f_x\vec{i}+ f_y\vec{j}+ f_z\vec{k} is normal to the surface. The tangent plane at (x_0,y_0,z_0) is of the form f_x(x_0,y_0,z_0)(x- x_0)+f_y(x_0,y_0,z_0)(y-y_0)+ f_z(x_0,y_0, z_0)(z- z_0)= 0.

For what (x_0, y_0, z_0) does (0, 0, 0) lie on that tangent plane?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top