Download presentation

Presentation is loading. Please wait.

Published byLauren Axon Modified over 2 years ago

1
The gradient as a normal vector

2
Consider z=f(x,y) and let F(x,y,z) = f(x,y)-z Let P=(x 0,y 0,z 0 ) be a point on the surface of F(x,y,z) Let C be any curve on the surface that passes through P and has the vector equation r(t) =.

3
C P

4
Since C is on S, any point (x(t),y(t),z(t)) must also satisfy F(x(t),y(t),z(t))=0 The chain rule says that:

5
C P r’(t 0 ) In particular, this is true at t 0, so that r(t 0 )=

6
Interpretation The gradient vector at P is perpendicular to the tangent vector to any curve C on S that passes through P This means the tangent plane at P has normal vector

7
We know how to write the equation of that plane! Remark: The same idea holds true for any F(x,y,z)=k, since

8
Consequence: Similarly it holds that for functions of two variables that that the gradient is perpendicular to level curves of the form f(x,y) =k (i.e. contour lines!) Examples….

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google