anonymous
  • anonymous
Find an equation of the tangent plane at the given point. f(x,y) = x^2y + xy^3 at point (2,1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Recall our gradient is normal to our tangent plane at a point. Can you determine the gradient of our function at \((2,1)\)?
anonymous
  • anonymous
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anonymous
  • anonymous
Correct, \(\nabla f=(5,10)\). But we wish to describe our surface using an equation of a function in three variables \(F(x,y,z)=0\). Let \(F(x,y,z)=f(x,y)-z\). Can you find its gradient? at (2,1)?

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anonymous
  • anonymous
(2,1,6)*
anonymous
  • anonymous
Wait where did you get the 6 from? Sorry
anonymous
  • anonymous
$$\nabla F=(2xy+y^2,x^3+3xy^2,-1)=(5,10,-1)$$
anonymous
  • anonymous
The \(6\) is just \(f(2,1)\).
anonymous
  • anonymous
Ok
anonymous
  • anonymous
Okay, now we know our gradient is normal to our tangent plane at said point. If it's normal, we know it's orthogonal to every vector in said plane: $$(\mathbf{x}-\mathbf{x}_0)\cdot\nabla F=0\\(x-2,y-1,z-6)\cdot(5,10,-1)=0\\5(x-2)+10(y-1)-(z-6)=0\\5x+10y-z-10-10+6=0\\5x+10y-z=14$$
anonymous
  • anonymous
LOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOL That's awesome
anonymous
  • anonymous
Can you come up with a problem for me to practice? And I'll try it all by myself
anonymous
  • anonymous
Heck, make your own question and tag me in it and ill answer it! :))))))))))))
anonymous
  • anonymous
I see why I was stuck, the book I look at just tells me to use the linearization rule :P @oldrin.bataku
anonymous
  • anonymous
The linearization rule works just the same. Unfortunately I'm going to bed soon as it's rather late here.

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