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The unit normal vector of the cone of revolution 22 – 6(x² + y²) at the point P:(2,0,1), is o 12 0, 145 V 145 12 V145, 0, 145 12 »0 V145 145 12 ,0, 145 145

Answer :

Final answer:

The unit normal vector at the point P:(2,0,1) on the cone of revolution [tex]22 – 6(x² + y²) is (-24/sqrt(577), 0/sqrt(577), 1/sqrt(577)).[/tex]

Explanation:

To find the unit normal vector at the point P:(2,0,1) on the cone of revolution [tex]22 – 6(x² + y²)[/tex], we need to evaluate the partial derivatives of the equation with respect to x and y, and then normalize the resulting vector.

Let's start by finding the partial derivative with respect to x:

[tex]∂z/∂x = -12x[/tex]

Next, let's find the partial derivative with respect to y:

[tex]∂z/∂y = -12y[/tex]

Now, we can calculate the unit normal vector by evaluating these partial derivatives at the point P:(2,0,1) and normalizing the resulting vector:

[tex]N = (-12(2), -12(0), 1) = (-24, 0, 1)[/tex]

To normalize the vector, we divide each component by the magnitude of the vector:

[tex]|N| = sqrt((-24)² + 0² + 1²) = sqrt(576 + 1) = sqrt(577)[/tex]

Therefore, the unit normal vector at the point P:(2,0,1) on the cone of revolution 22 – 6(x² + y²) is:

[tex]N = (-24/sqrt(577), 0/sqrt(577), 1/sqrt(577))[/tex]

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