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Calc III - finding tangent plane

  • Thread starter 1MileCrash
  • Start date
  • #1
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Homework Statement



attached.

Homework Equations





The Attempt at a Solution



I thought the problem was easy, but my answer is wrong (aparently; I still disagree.)

First I defined x = (y^3)(z^3) to be a surface of function F

So
F(x,y,z) = (y^3)(z^3) - x = 0

Then, the gradient of F:

Partial wrt x = -1
Partial wrt y = 3y^2(z^3)
Partial wrt z = 3z^2(y^3)

Gradient F = -1i + 3y^2(z^3)j + 3z^2(y^3)k

Gradient F(1,-1,-1) = -1i + 3(-1)^2((-1)^3)j + 3(-1)^2((-1)^3)k
= -i - 3j - 3k

Then the tangent plane formula is

- x - 3y - 3z = 5

or

x + 3y + 3z = -5

Where am I going wrong with this?
 

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Answers and Replies

  • #2
I'm 'inclined' to agree with you
Pic related - graphs of the surface and the tangent plane to surface at 1,-1,-1

I am pretty tired atm so we could both be falling into the same trap though..
 

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  • #3
HallsofIvy
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That's the same result I get.
 
  • #4
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Darn you wiley plus!

I'm not about to start guessing through their wrong answers to see which one is "correct." Guess I'll email my professor.
 

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