Question

2. [0/0.57 Points] PREVIOUS ANSWERS Find an equation of the tangent plane to the surface at the given point. x2 + y2 422 126, (-7,-9, 1) x DETAILS LARCALCET7 13.7.015.

Answer 1

To find the equation of the tangent plane to the surface x^2 + y^2 = 422 at the point (-7, -9, 1), we can find the gradient vector at that point and use it to write the equation of the tangent plane.

Step-by-step explanation:

To find the equation of the tangent plane to the surface, we need to find the gradient vector and the point of tangency.

Given the surface equation x^2 + y^2 = 422, we can take the partial derivatives with respect to x and y to find the gradient vector:

∇f = (∂f/∂x, ∂f/∂y) = (2x, 2y)

Plug in the coordinates of the given point (-7, -9, 1) to find the gradient vector at that point:

∇f(-7, -9) = (2(-7), 2(-9)) = (-14, -18)

So the equation of the tangent plane is:

-14(x+7) - 18(y+9) + z - 1 = 0

Answer 2

To find the equation of the tangent plane to the surface at the given point, we can use partial derivatives and the equation of a plane.

Step-by-step explanation:

An equation of the tangent plane to the surface at the given point can be found using partial derivatives. We can start by finding the partial derivatives of the given function with respect to x and y.

Given function: f(x, y) = x² + y²

Partial derivative with respect to x: f_x = 2x

Partial derivative with respect to y: f_y = 2y

Next, we can use the point (-7, -9, 1) and the values of the partial derivatives to find the equation of the tangent plane. The equation of a plane can be written as:

A(x - x_0) + B(y - y_0) + C(z - z_0) = 0

Where (x_0, y_0, z_0) is the given point and A, B, and C are the coefficients of the equation.

Plugging in the values, we get:

2x(x - (-7)) + 2y(y - (-9)) + C(z - 1) = 0

Simplifying further, we have:

2x(x + 7) + 2y(y + 9) + C(z - 1) = 0

Therefore, the equation of the tangent plane to the surface at the given point is 2x(x + 7) + 2y(y + 9) + C(z - 1) = 0.