Step by step
The question
Question: find the tangent plane to z = x³ + y² at (1, 2).
1 Put the point in
First the height of the surface at the point: put x = 1 and y = 2 in.
Work out each term.
So the surface goes through (1, 2, 5).
2 The partial derivatives
The plane's two slopes are the partial derivatives at the point.
Term by term.
x³: the power 3 comes down.
The power goes down by 1.
So 3x².
y² has no x: it is a fixed number, so it gives 0.
Drop the zeros: ∂f/∂x = 3x².
Term by term.
x³ 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.
Drop the zeros: ∂f/∂y = 2y.
At the point: f_x(1, 2) = 3.
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: 3 for x and 4 for y.
Careful!
Use the slopes at the point, the numbers 3 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.
Add 5 on both sides.
So the tangent plane is z = 3x + 4y − 6.
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!
Close by, at (1.1, 2), the surface is 5.331 and the plane is 5.3.
Almost the same.
Right!
Let's look back.
First: put the point in.
Then: the partial derivatives.
Then: write the tangent plane.
Last: check.