Question

Find the equation of the tangent plane to the surface given by y = x² - z² at the point (6, 32, 2).

Answer 1

To find the equation of the tangent plane to the surface y = x² - z² at the point (6, 32, 2), we can use the partial derivatives and the gradient vector of the surface equation. First, calculate the partial derivatives of y = x² - z² with respect to x, y, and z. Then, substitute the values into the partial derivatives to find the gradient vector. Finally, use the gradient vector to write the equation of the tangent plane by substituting the given point. The equation of the tangent plane is 12(x - 6) + (y - 32) - 4(z - 2) = 0.

Step-by-step explanation:

To find the equation of the tangent plane to the surface given by y = x² - z² at the point (6, 32, 2), we need to find the partial derivatives of the surface equation with respect to x, y, and z. Then, we can use the gradient vector of the surface equation to find the equation of the tangent plane. Let's go through the steps:

1. Calculate the partial derivative of y = x² - z² with respect to x: ∂y/∂x = 2x
2. Calculate the partial derivative of y = x² - z² with respect to y: ∂y/∂y = 1
3. Calculate the partial derivative of y = x² - z² with respect to z: ∂y/∂z = -2z
4. At the given point (6, 32, 2), substitute the values into the partial derivatives: ∂y/∂x = 2(6) = 12, ∂y/∂y = 1, ∂y/∂z = -2(2) = -4
5. Now, we have the gradient vector of the surface equation: ∇y = 12i + j - 4k
6. The equation of the tangent plane is given by: ∇y · (x - 6, y - 32, z - 2) = 0
7. Substituting the point (6, 32, 2) into the equation, we get: (12i + j - 4k) · (x - 6, y - 32, z - 2) = 0
8. Simplifying the equation, we get: 12(x - 6) + (y - 32) - 4(z - 2) = 0
9. This is the equation of the tangent plane to the surface y = x² - z² at the point (6, 32, 2).