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

Step by step

The question

Question: find the tangent plane to f(x, y) = 2x² + xy at (1, −1).

f(x, y) = 2x² + xy

1 Put the point in

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

z₀ = 2(1)² + (1)(−1)

Work out each term.

z₀ = 2 − 1

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

z₀ = 1

2 The partial derivatives

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

Term by term.

2x²: the power 2 comes down: 2 times 2 is 4.

The power goes down by 1.

So 4x.

In xy, y is fixed and waits in front.

x gives 1.

So y.

So ∂f/∂x = 4x + y.

∂f∂x = 4x + y

Term by term.

2x² has no y: it is a fixed number, so it gives 0.

In xy, x is fixed and waits in front.

y gives 1.

So x.

∂f∂y = 0 + x

Drop the zeros: ∂f/∂y = x.

∂f∂y = x

At the point: f_x(1, −1) = 3.

f_x(1, −1) = 4(1) + (−1) = 3

At the point: f_y(1, −1) = 1.

f_y(1, −1) = 1

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.

z − z₀ = f_x(x − x₀) + f_y(y − y₀)

Put in the point and the two slopes: 3 for x and 1 for y.

z − 1 = 3(x − 1) + (y + 1)

Careful!

Use the slopes at the point, the numbers 3 and 1.

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

Careful: the slopes are numbers: f_x(1, −1) = 3

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

z − 1 = 3x − 3 + y + 1

Add 1 on both sides.

So the tangent plane is z = 3x + y − 1.

z = 3x + y − 1

4 Check

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

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

At the point, the plane gives 1, the height of the surface.

Right!

plane at (1, −1): 3(1) + (-1) − 1 = 1

Close by, at (1.1, −1), the surface is 1.32 and the plane is 1.3.

Almost the same.

Right!

at (1.1, −1): surface 1.32, plane 1.3

Let's look back.

First: put the point in.

Then: the partial derivatives.

Then: write the tangent plane.

Last: check.

Answer