anonymous
  • anonymous
How to find the tangent plane?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[f(x,y) = \frac{ x }{ \sqrt{y} }\] At the point (4,4), I need to find the tangent plane.
anonymous
  • anonymous
Consider that \(f(x,y)=\dfrac{x}{\sqrt{y}}\) describes the same surface as \(F(x,y,z)=0\) for \(F(x,y,z)=z-\dfrac{x}{\sqrt{y}}\). Now, it's intuitive that the gradient at \((4,4,2)\) is normal to our surface at that point, so to determine a surface normal to our tangent plane we merely compute our gradient at the point:$$\nabla F=\frac{\partial F}{\partial x}\mathbf{i}+\frac{\partial F}{\partial y}\mathbf{j}+\frac{\partial F}{\partial z}\mathbf{k}=-\frac1{\sqrt{y}}\mathbf{i}+-\frac{x}{2\sqrt{y^3}}\mathbf{j}+\mathbf{k}\\\nabla F(4,4,2)=\left(-\frac12,-\frac14,1\right)$$Now, recall that for a plane all points \((x,y,z)\) on the plane are orthogonal to our normal:$$(\mathbf{x}-\mathbf{x}_0)\cdot\nabla F=0\\(\mathbf{x}-(4,4,2))\cdot\left(-\frac12,-\frac14,1\right)=0\\-\frac12(x-4)-\frac14(y-4)+(z-2)=0$$
anonymous
  • anonymous
Well said.

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anonymous
  • anonymous
How did you get that the gradient at (4,4,2) is normal?
anonymous
  • anonymous
How did you get the Z coordinate to equal 2 anyway?
sirm3d
  • sirm3d
you have z=x/sqrt(y) with x=2, y=2
anonymous
  • anonymous
$$f(4,4)=\frac4{\sqrt4}=\sqrt4=2$$

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