Respuesta :
Answer:
Therefore (0,1,3) lies on the given surface.
Therefore a vector normal to the surface at (0,1,3) is
[tex]\hat{n}=2\hat{i}+9\hat {j}+3\hat{k}[/tex]
Step-by-step explanation:
Given equation of the surface is
[tex]y= \frac{9}{2x+3z}[/tex]
First we arrange the equation of the surface in the form of f(x,y,z) =0.
[tex]y(2x+3z)=9[/tex]
[tex]\Rightarrow y(2x+3z)-9=0[/tex]
Therefore [tex]f(x,y,z)=y(2x+3z)-9=0[/tex]
If the point (0,1,3) lies on the given surface. Then it will be satisfy the equation.
putting x=0, y= 1 and z=3
f(0,1,3) = 1(2.0+3.3)-9 = 9-9=0
Therefore (0,1,3) lies on the given surface.
In order to find the normal at any point in vector space we use the gradient operator:
[tex]\triangledown f(x,y,z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}[/tex]
Here, [tex]\frac{\partial }{\partial x}[y(2x+3z)] = 2y[/tex]
[tex]\frac{\partial }{\partial y}[y(2x+3z)] = 2x+3z[/tex]
[tex]\frac{\partial }{\partial z}[y(2x+3z)] = 3y[/tex]
[tex]\triangledown f(x,y,z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}[/tex]
[tex]=2y \hat{i}+(2x+3z)\hat{j}+3y\hat{k}[/tex]
The normal vector of the surface at (0,1,3) is
[tex]\triangledown f(0,1,3)=2.1 \hat{i} +(2.0+3.3)\hat{j}+3.1\hat{k}[/tex]
[tex]=2\hat{i}+9\hat {j}+3\hat{k}[/tex]
Therefore a vector normal to the surface at (0,1,3) is
[tex]\hat{n}=2\hat{i}+9\hat {j}+3\hat{k}[/tex]
