Question

Find the equation of the tangent plane to the graph of f(x,y)=2xy 2 +3x 3 y 2 at the point (−3,3). (Use symbolic notation and fractions where needed. Enter your answer using x−y-, z-coordinates.) the equation:

Answer 1

The equation of the tangent plane to the graph of f(x,y)=2xy^2+3x^3y^2 at the point (-3,3) can be found using the gradient vector. The equation of the tangent plane is 162x - 198y + z = -396.

Step-by-step explanation:

The equation of the tangent plane to the graph of f(x,y)=2xy2+3x3y2 at the point (-3,3) can be found using the gradient vector.

The gradient of f(x,y) is given by ∇f = (∂f/∂x, ∂f/∂y).

After finding the partial derivatives and evaluating them at (-3,3), we can use the formula for the equation of a plane to determine the equation of the tangent plane.

The partial derivatives are ∂f/∂x = 2y2+9x2y2 and ∂f/∂y = 4xy+6x3y.

Evaluating these derivatives at (-3,3) gives ∂f/∂x = 162 and ∂f/∂y = -198.

Using the formula for the equation of a plane, the equation of the tangent plane is 162(x+3) - 198(y-3) + z = 0.

Simplifying this equation gives 162x - 198y + z = -396.