Find the tangent plane given at the stationary point.

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SUMMARY

The discussion focuses on finding the tangent plane at a stationary point for a surface defined by the equation \( z = f(x, y) \). A participant initially calculated the tangent plane incorrectly, arriving at -1 instead of the correct value of 6. The conversation highlights the importance of accurately substituting values into the x-partial derivative and emphasizes the relationship between the normal vector \((a, b, c)\) and the tangent plane's equation \((a, b, c) \cdot (x-x_0,y-y_0,z-z_0) = 0\).

PREREQUISITES
  • Understanding of partial derivatives and their application in multivariable calculus
  • Familiarity with the equation of a plane in three-dimensional space
  • Knowledge of how to compute normal vectors to surfaces
  • Experience with evaluating functions at specific points
NEXT STEPS
  • Study the derivation of the tangent plane formula in multivariable calculus
  • Learn how to compute partial derivatives using tools like Wolfram Alpha or MATLAB
  • Explore examples of finding normal vectors to surfaces defined by \( z = f(x, y) \)
  • Practice solving problems involving stationary points and their tangent planes
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Students studying multivariable calculus, educators teaching surface geometry, and anyone interested in understanding the application of tangent planes in mathematical analysis.

Jozefina Gramatikova
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Homework Statement


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Homework Equations


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The Attempt at a Solution


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I got -1, but the answer says "6". Could you help me, please?
 

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I see a mistake in the x-partial. You have not substituted in correctly.
 
verty said:
I see a mistake in the x-partial. You have not substituted in correctly.
Oh, thanks! But then 3-3=0 and the whole thing is 0
 
So what do you think the tangent plane looks like? And what is its formula?

PS. This is all the help I can give, sorry.
 
Jozefina Gramatikova said:
Oh, thanks! But then 3-3=0 and the whole thing is 0
In the relevant equations, you have correctly given the equation of a plane. You can rewrite this as
##(a, b, c) \cdot (x-x_0,y-y_0,z-z_0) = 0##

This says that ##(a, b, c)## is normal to all of the vectors in the plane.

If this plane is tangent to the surface described by ##(x, y,f(x,y))## at ##(x_0,y_0,z_0)##, then ##(a, b, c)## would also have to be normal to that surface at that point. How would you find the vector ##(a, b, c)## that is normal to the surface?
 

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