Linear approximation of f(x, y) = xy at (2, 3) to estimate f(2.1, 2.9) Step by Step

Step by step

The question

Question: use the tangent plane of f(x, y) = xy at (2, 3) to estimate f(2.1, 2.9).

f(x, y) = xy

1 Put the point in

First the height of the surface at the point: put x = 2 and y = 3 in.

z₀ = (2)(3)

So the surface goes through (2, 3, 6).

z₀ = 6

2 The partial derivatives

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

Term by term.

In xy, y is fixed and waits in front.

x gives 1.

So y.

So ∂f/∂x = y.

∂f∂x = y

Term by term.

In xy, x is fixed and waits in front.

y gives 1.

So x.

So ∂f/∂y = x.

∂f∂y = x

At the point: f_x(2, 3) = 3.

f_x(2, 3) = 3

At the point: f_y(2, 3) = 2.

f_y(2, 3) = 2

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 2 for y.

z − 6 = 3(x − 2) + 2(y − 3)

Careful!

Use the slopes at the point, the numbers 3 and 2.

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

Careful: the slopes are numbers: f_x(2, 3) = 3

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

z − 6 = 3x − 6 + 2y − 6

Add 6 on both sides.

So the tangent plane is z = 3x + 2y − 6.

z = 3x + 2y − 6

4 Estimate

Near the point, the surface is almost the plane.

Put x = 2.1 and y = 2.9 into the plane.

f(2.1, 2.9) ≈ 3(2.1) + 2(2.9) − 6

So f(2.1, 2.9) ≈ 6.1.

f(2.1, 2.9) ≈ 6.1

5 Check

Let's check with a picture: the surface near (2, 3, 6).

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

At the point, the plane gives 6, the height of the surface.

Right!

plane at (2, 3): 3(2) + 2(3) − 6 = 6

Close by, at (2.1, 2.9), the surface is 6.09 and the plane is 6.1.

Almost the same.

Right!

at (2.1, 2.9): surface 6.09, plane 6.1

Let's look back.

First: put the point in.

Then: the partial derivatives.

Then: write the tangent plane.

Then: estimate.

Last: check.

Answer