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The unit normal vector to the surface X2 + Y2 + Z2 - 48 = 0 at the point (4, 4, 4) is

Question

The unit normal vector to the surface X2+Y2+Z248=0X^2 + Y^2 + Z^2 - 48 = 0 at the point (4,4,4)(4, 4, 4) is

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Solution

To find the unit normal vector to the surface X2 + Y2 + Z2 - 48 = 0 at the point (4, 4, 4), we can follow these steps:

Step 1: Rewrite the equation of the surface in the form of a function f(x, y, z) = 0. In this case, the equation is X2 + Y2 + Z2 - 48 = 0.

Step 2: Calculate the gradient of the function f(x, y, z) at the given point (4, 4, 4). The gradient is a vector that points in the direction of the steepest increase of the function at a given point. In this case, the gradient vector is given by ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z).

Step 3: Evaluate the partial derivatives of the function f(x, y, z) with respect to x, y, and z. In this case, we have ∂f/∂x = 2X, ∂f/∂y = 2Y, and ∂f/∂z = 2Z.

Step 4: Substitute the coordinates of the given point (4, 4, 4) into the partial derivatives. We get ∂f/∂x = 2(4) = 8, ∂f/∂y = 2(4) = 8, and ∂f/∂z = 2(4) = 8.

Step 5: Combine the partial derivatives to form the gradient vector ∇f = (8, 8, 8).

Step 6: Normalize the gradient vector to obtain the unit normal vector. To normalize a vector, we divide each component of the vector by its magnitude. In this case, the magnitude of the gradient vector is |∇f| = √(8^2 + 8^2 + 8^2) = √(192) = 8√3. Therefore, the unit normal vector is (∇f)/|∇f| = (8/8√3, 8/8√3, 8/8√3) = (1/√3, 1/√3, 1/√3).

So, the unit normal vector to the surface X2 + Y2 + Z2 - 48 = 0 at the point (4, 4, 4) is (1/√3, 1/√3, 1/√3).

This problem has been solved

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