Question

Determine the equation of a plane tangent to a given surface at a point. Find the tangent plane to the equation z=−x2−4y2−3y at the point (−2,−1,−5). z=

Answer 1

To find the equation of a plane tangent to a given surface at a point, we can use partial derivatives. In this example, the equation of the tangent plane to the surface z = -x^2 - 4y^2 - 3y at the point (-2,-1,-5) is 4(x + 2) + 5(y + 1) + (z + 5) = 0.

Step-by-step explanation:

The equation of a plane tangent to a given surface at a point can be found using partial derivatives. First, we need to find the partial derivatives of the given surface with respect to x and y.

Given the equation z = -x^2 - 4y^2 - 3y, we can find the partial derivatives: dz/dx = -2x and dz/dy = -8y - 3.

Next, we substitute the coordinates of the given point (-2,-1,-5) into the partial derivatives to find the values: dz/dx = -2(-2) = 4 and dz/dy = -8(-1) - 3 = 5.

So, the equation of the tangent plane is: 4(x + 2) + 5(y + 1) + (z + 5) = 0.