Finding Tangent Plane and Normal Line Equations for a Given Surface

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SUMMARY

The discussion focuses on finding the equations of the tangent plane and normal line for the surface defined by the equation x² - 2y² + z² + yz = 7 at the point (5, 3, -3). The correct equation for the tangent plane is derived using the partial derivatives fx, fy, and fz at the specified point. The normal line equations are confirmed as x = 10t + 5, y = -15 + 3, and z = -3 - 3, following the correct application of linear approximation.

PREREQUISITES
  • Understanding of partial derivatives (fx, fy, fz)
  • Knowledge of tangent plane equations in multivariable calculus
  • Familiarity with normal line equations in three-dimensional space
  • Experience with linear approximation techniques
NEXT STEPS
  • Study the derivation of tangent plane equations for multivariable functions
  • Learn how to compute partial derivatives for complex surfaces
  • Explore the concept of linear approximation in calculus
  • Review examples of normal lines in three-dimensional geometry
USEFUL FOR

Students in calculus courses, particularly those studying multivariable calculus, as well as educators teaching concepts related to tangent planes and normal lines in three-dimensional surfaces.

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Homework Statement


Find equations of the following.

x2-2y2+z2+yz=7, (5,3,-3)
(a) the tangent plane
(b) the normal line to the given surface at the point

Homework Equations



I know it involves fx, fy, fz

The Attempt at a Solution


I got 10x-15y-3z=7. Is this correct? Because its not true at the point (5,3,-3).

I got it by f(5,3,-3)+fx(5,3,-3)(x-5)+fy(5,3,-3)(y-3)+fz(5,3,-3)(z+3)

As for the normal line I know the answers are
x=10t+5
y=-15+3
z=-3-3
 
Last edited:
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ktobrien said:

Homework Statement


Find equations of the following.

x2-2y2+z2+yz=7, (5,3,-3)
(a) the tangent plane
(b) the normal line to the given surface at the point

Homework Equations



I know it involves fx, fy, fz

The Attempt at a Solution


I got 10x-15y-3z=7. Is this correct? Because its not true at the point (5,3,-3).
No. The point (5, 3 -3) has to satisfy both the equation of the plane and the equation of the surface.
ktobrien said:
I got it by f(5,3,-3)+fx(5,3,-3)(x-5)+fy(5,3,-3)(y-3)+fz(5,3,-3)(z+3)
For one thing, this is not an equation, so there's no way to get an equation out of it. For another thing, if my memory is correct, the equation of the tangent plane is fx(5, 3, -3)(x - 5) + fy(5, 3, -3)(y - 3) + fz(5, 3, -3)(z - (-3)) = 0.

You didn't show the partial derivatives that you calculated, so it might also be that you have an error in one or more of them.

ktobrien said:
As for the normal line I know the answers are
x=10t+5
y=-15+3
z=-3-3
 
Ive got it now. I used the linear approximation and set it = to 0. I just had to take out the f(5,3,-3). I figured it was something stupid like that. Thanks for the help though.
 

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