Question

Find the equation of the tangent plane to the surface z=9y²−0x²at the point (−3,−2,36).
z=___

Answer 1

The equation of the tangent plane to the surface at the point (-3, -2, 36) is z = -36y + 36.

Step-by-step explanation:

To find the equation of the tangent plane to the surface z = 9y² at the point (-3, -2, 36), we first need to find the partial derivatives of z with respect to x and y.

Given the surface equation z = 9y² - 0x², we find:

• The partial derivative with respect to x is ∂z/∂x = 0, since z does not depend on x.
• The partial derivative with respect to y is ∂z/∂y = 18y.

At the point (-3, -2, 36), we have:

• ∂z/∂x (at (-3, -2)) = 0
• ∂z/∂y (at (-3, -2)) = 18(-2) = -36

The equation of the tangent plane at a point (x0, y0, z0) on the surface z = f(x, y) is given by:

z - z0 = (∂f/∂x (x0, y0))*(x - x0) + (∂f/∂y (x0, y0))*(y - y0)

Thus, the equation of the tangent plane to the given surface at (-3, -2, 36) is:

z - 36 = 0*(x + 3) - 36*(y + 2)

Simplifying, we obtain the equation of the tangent plane: z = -36y - 36*2 + 36, which simplifies to z = -36y + 36.