Worked examples
The question
Question: find the tangent plane to z = x² + y² at (1, 2).
z = x2 + y2
1 Put the point in
First the height of the surface at the point: put x = 1 and y = 2 in.
z0 = 12 + 22
Work out each term.
z0 = 1 + 4
So the surface goes through (1, 2, 5).
z0 = 5
2 The partial derivatives
The plane's two slopes are the partial derivatives at the point.
Term by term.
x²: the power 2 comes down.
The power goes down by 1.
So 2x.
y² has no x: it is a fixed number, so it gives 0.
∂f∂x = 2x + 0
Drop the zeros: ∂f/∂x = 2x.
∂f∂x = 2x
Term by term.
x² has no y: it is a fixed number, so it gives 0.
y²: the power 2 comes down.
The power goes down by 1.
So 2y.
∂f∂y = 0 + 2y
Drop the zeros: ∂f/∂y = 2y.
∂f∂y = 2y
At the point: f_x(1, 2) = 2.
fx(1, 2) = 2(1) = 2
At the point: f_y(1, 2) = 4.
fy(1, 2) = 2(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.
z − z0 = fx(x − x0) + fy(y − y0)
Put in the point and the two slopes: 2 for x and 4 for y.
z − 5 = 2(x − 1) + 4(y − 2)
Careful!
Use the slopes at the point, the numbers 2 and 4.
Not the formulas with x and y in them: the plane must be flat.
Careful: the slopes are numbers: fx(1, 2) = 2
What is in front of a bracket multiplies every part inside.
z − 5 = 2x − 2 + 4y − 8
Add 5 on both sides.
So the tangent plane is z = 2x + 4y − 5.
z = 2x + 4y − 5
4 Check
Let's check with a picture: the surface near (1, 2, 5).
The orange plane rests on it and touches it at the point.
At the point, the plane gives 5, the height of the surface.
Right!
plane at (1, 2): 2(1) + 4(2) − 5 = 5
Close by, at (1.1, 2), the surface is 5.21 and the plane is 5.2.
Almost the same.
Right!
at (1.1, 2): surface 5.21, plane 5.2
Let's look back.
First: put the point in.
Then: the partial derivatives.
Then: write the tangent plane.
Last: check.
▶ Watch it solved
The question
Question: find the tangent plane to z = 2x² + y² at (1, −1).
z = 2x2 + y2
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)2
Work out each term.
z0 = 2 + 1
So the surface goes through (1, −1, 3).
z0 = 3
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.
y² has no x: it is a fixed number, so it gives 0.
∂f∂x = 4x + 0
Drop the zeros: ∂f/∂x = 4x.
∂f∂x = 4x
Term by term.
2x² has no y: it is a fixed number, so it gives 0.
y²: the power 2 comes down.
The power goes down by 1.
So 2y.
∂f∂y = 0 + 2y
Drop the zeros: ∂f/∂y = 2y.
∂f∂y = 2y
At the point: f_x(1, −1) = 4.
fx(1, −1) = 4(1) = 4
At the point: f_y(1, −1) = −2.
fy(1, −1) = 2(−1) = −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 = fx(x − x0) + fy(y − y0)
Put in the point and the two slopes: 4 for x and −2 for y.
z − 3 = 4(x − 1) − 2(y + 1)
Careful!
Use the slopes at the point, the numbers 4 and −2.
Not the formulas with x and y in them: the plane must be flat.
Careful: the slopes are numbers: fx(1, −1) = 4
What is in front of a bracket multiplies every part inside.
z − 3 = 4x − 4 − 2y − 2
Add 3 on both sides.
So the tangent plane is z = 4x − 2y − 3.
z = 4x − 2y − 3
4 Check
Let's check with a picture: the surface near (1, −1, 3).
The orange plane rests on it and touches it at the point.
At the point, the plane gives 3, the height of the surface.
Right!
plane at (1, −1): 4(1) − 2(-1) − 3 = 3
Close by, at (1.1, −1), the surface is 3.42 and the plane is 3.4.
Almost the same.
Right!
at (1.1, −1): surface 3.42, plane 3.4
Let's look back.
First: put the point in.
Then: the partial derivatives.
Then: write the tangent plane.
Last: check.
▶ Watch it solved
The question
Question: use the tangent plane of f(x, y) = xy² at (1, 2) to estimate f(1.1, 1.9).
f(x, y) = xy2
1 Put the point in
First the height of the surface at the point: put x = 1 and y = 2 in.
z0 = (1)(2)2
So the surface goes through (1, 2, 4).
z0 = 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.
∂f∂x = y2
Term by term.
In xy², x is fixed and waits in front.
y² gives 2y.
So 2xy.
So ∂f/∂y = 2xy.
∂f∂y = 2xy
At the point: f_x(1, 2) = 4.
fx(1, 2) = 22 = 4
At the point: f_y(1, 2) = 4.
fy(1, 2) = 2(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.
z − z0 = fx(x − x0) + fy(y − y0)
Put in the point and the two slopes: 4 for x and 4 for y.
z − 4 = 4(x − 1) + 4(y − 2)
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.
Careful: the slopes are numbers: fx(1, 2) = 4
What is in front of a bracket multiplies every part inside.
z − 4 = 4x − 4 + 4y − 8
Add 4 on both sides.
So the tangent plane is z = 4x + 4y − 8.
z = 4x + 4y − 8
4 Estimate
Near the point, the surface is almost the plane.
Put x = 1.1 and y = 1.9 into the plane.
f(1.1, 1.9) ≈ 4(1.1) + 4(1.9) − 8
So f(1.1, 1.9) ≈ 4.
f(1.1, 1.9) ≈ 4
5 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!
plane at (1, 2): 4(1) + 4(2) − 8 = 4
Close by, at (1.1, 1.9), the surface is 3.971 and the plane is 4.
Almost the same.
Right!
at (1.1, 1.9): surface 3.971, plane 4
Let's look back.
First: put the point in.
Then: the partial derivatives.
Then: write the tangent plane.
Then: estimate.
Last: check.
▶ Watch it solved
