anonymous
  • anonymous
Find the tangent plane to the surface x^2+y^2+xyz=4+z^3(y-2x) at the point (1, 1, 1). So the equation of the tangent plane is z-z0=f'x(x-x0)+f'y(y-y0). I find the partials using implicit differentiation: f'x=(2x)/(3yz^2-6z^2-y) @ (1,1,1) = -1/2 f'y=(2y)/(3z^2-6xz^2-x) @ (1,1,1) = -1/2 Then the equation of the plane (ie plugging into the above equation) is z+(1/2)x+(1/2)y=2 But the answer is 5x+2y+4z=11 I have two question from here: 1, Where am I going wrong in the implicit differentiation? Wolfram gives me a different result. 2, I know using this method is incorrect. Why is that? Why must I use the gradient vector to the level surface instead of implicitly deriving?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I'm not sure what you mean by that, the way I would solve this would be to define F(x,y,z)=x^2+y^2+xyz-z^3(y-2x) solve for gradF and dot gradF with
anonymous
  • anonymous
This is calc III by the way
anonymous
  • anonymous
I'm not so confused about the problem itself but rather why these two methods are not equivalent

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anonymous
  • anonymous
And also, why wolfram gives me a completely different partial derivatives when solved implicitly. I have no idea where I am going wrong
anonymous
  • anonymous
So where \(f(x, y, z) = (x^2+y^2+xyz) - (4+z^3(y-2x))\) The surface you are considering is \(f^{-1}(0,0,0)\) So the method you discussed earlier is the only reasonable one I can think of. That is where you consider the vectors orthogonal to the gradient at a given point on the level set.
anonymous
  • anonymous
In our textbook, they taught us how to find the equation of a tangent plane for z=f(x,y) using the formula z-z0=f'x(x-x0)+f'y(y-y0). What I'm trying to do is define my equation above as a implicit equation with z(x,y). So I use the process of implicit partial differentiation to find f'x and f'y and plug into the equation for the tangent plane. I don't know why that isn't equivalent to what you just wrote down
anonymous
  • anonymous
That is my real question. I understand the method you outlined but I don't get why that wouldn't be equivalent to the method I just described above
anonymous
  • anonymous
The way I define the equation would be: x^2+y^2+xyz(z,y)=4+(z(x,y))^3 (y-2x) So z is implicitly a function of (x,y) And wondering now why my solution for the first partials of this equation are wrong according to wolfram.
anonymous
  • anonymous
The methods are indeed equivalent, my partial derivatives were incorrect. I found the correct partial derivatives and arrived at the right answer. Thanks for the help

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