Tangent plane to f(x, y) = xy^2 at (-1, 2)

Step by step

The question

Question: find the tangent plane to f(x, y) = xy² at (−1, 2).

1 Put the point in

First the height of the surface at the point: put x = −1 and y = 2 in.

So the surface goes through (−1, 2, −4).

2 The partial derivatives

The plane's two slopes are the partial derivatives at the point.

Term by term.

In xy², y² is fixed and waits in front.

x gives 1.

So y².

So ∂f/∂x = y².

Term by term.

In xy², x is fixed and waits in front.

y² gives 2y.

So 2xy.

So ∂f/∂y = 2xy.

At the point: f_x(−1, 2) = 4.

At the point: f_y(−1, 2) = −4.

3 Write the tangent plane

The rule: z minus z₀ equals the x slope times x minus x₀, plus the y slope times y minus y₀.

Put in the point and the two slopes: 4 for x and −4 for y.

Careful!

Use the slopes at the point, the numbers 4 and −4.

Not the formulas with x and y in them: the plane must be flat.

What is in front of a bracket multiplies every part inside.

Take away 4 on both sides.

So the tangent plane is z = 4x − 4y + 8.

4 Check

Let's check with a picture: the surface near (−1, 2, −4).

The orange plane rests on it and touches it at the point.

At the point, the plane gives −4, the height of the surface.

Right!

Close by, at (−0.9, 2), the surface is −3.6 and the plane is −3.6.

Almost the same.

Right!

Let's look back.

First: put the point in.

Then: the partial derivatives.

Then: write the tangent plane.

Last: check.