Tangent plane, directional derivatives

Click For Summary
SUMMARY

The discussion focuses on finding the equation of the tangent plane for the surface defined by the equation \( yz = \ln(x + z) \) at the point (0, 0, 1). The correct approach involves determining the gradient vector of the function \( F(x, y, z) = \ln(x + z) - yz \). The partial derivatives calculated at the point yield the normal vector (1, -1, 1), confirming that the tangent plane equation is \( x - y + z - 1 = 0 \). The solution is validated, with a suggestion to refine the calculation of the partial derivatives for general cases.

PREREQUISITES
  • Understanding of multivariable calculus concepts, particularly tangent planes.
  • Familiarity with partial derivatives and gradient vectors.
  • Knowledge of implicit functions and their representations.
  • Proficiency in evaluating logarithmic functions in multiple dimensions.
NEXT STEPS
  • Study the derivation of tangent plane equations in multivariable calculus.
  • Learn how to compute partial derivatives for implicit functions.
  • Explore the application of gradient vectors in determining surface normals.
  • Investigate examples of tangent planes for various surfaces beyond logarithmic functions.
USEFUL FOR

Students and educators in multivariable calculus, mathematicians working with implicit functions, and anyone seeking to deepen their understanding of tangent planes and directional derivatives.

Laura1321412
Messages
23
Reaction score
0

Homework Statement



find the equation on the tangent plane of yz=ln(x+z) at point (0, 0, 1 )


Homework Equations



Tangent plane equation...

The Attempt at a Solution



I wasn't sure how to determine the partials on this equation. My attempt was to rearange as ln(x+z)-yz=0 so Fx = 1/(x+z) Fy= -1 Fz = 1/(x+z) at the point (001) Fx= 1 Fy= -1 Fz= 1

into the plane equations and i get

x-y+z-1=0

... I am not sure if the answer is right, i think the issue is when i rearanged the equation, how can i determine the partials on this correctly?

Thank you!
 
Physics news on Phys.org
Hello? :(
 
You are finding the gradient vector to the surface F(x,y,z)=ln(x+z)-yz=0. Some of your components aren't quite correct for a general x,y,z but they are correct for (x,y,z)=(0,0,1). So yes, the normal vector is (1,-1,1) and the plane has to pass through (0,0,1) so I think your answer is correct. Can you correct some of the 'partials'?
 

Similar threads

Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
7
Views
2K