Question

How to find the tangent plane?

Answer 1

\[f(x,y) = \frac{ x }{ \sqrt{y} }\]
At the point (4,4), I need to find the tangent plane.

Answer 2

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:$$(\mathbf{x}-\mathbf{x}_0)\cdot\nabla F=0\\(\mathbf{x}-(4,4,2))\cdot\left(-\frac12,-\frac14,1\right)=0\\-\frac12(x-4)-\frac14(y-4)+(z-2)=0$$

Answer 3

Well said.

Answer 4

How did you get that the gradient at (4,4,2) is normal?

Answer 5

How did you get the Z coordinate to equal 2 anyway?

Answer 6

you have z=x/sqrt(y) with x=2, y=2

Answer 7

$$f(4,4)=\frac4{\sqrt4}=\sqrt4=2$$