Maximum rate of change of f(x, y) = 3x + 4y at (1, 1) Step by Step

Step by step

The question

Question: find the fastest rate of change of f(x, y) = 3x + 4y at (1, 1), and its direction.

f(x, y) = 3x + 4y

1 The partial derivatives

Term by term.

3x gives 3.

4y has no x: it is a fixed number, so it gives 0.

∂f∂x = 3 + 0

Drop the zeros: ∂f/∂x = 3.

∂f∂x = 3

Term by term.

3x has no y: it is a fixed number, so it gives 0.

4y gives 4.

∂f∂y = 0 + 4

Drop the zeros: ∂f/∂y = 4.

∂f∂y = 4

2 Find the gradient

The gradient ∇f puts the partial derivatives together as one vector.

∇f = ⟨∂f∂x, ∂f∂y⟩

So ∇f = ⟨3, 4⟩.

∇f = ⟨3, 4⟩

At the point, put x = 1 and y = 1 in: ∇f(1, 1) = ⟨3, 4⟩.

∇f(1, 1) = ⟨3, 4⟩

3 Find the length

The fastest growth is the length of the gradient.

Its direction is the gradient itself.

The length ‖∇f‖ is Pythagoras with two parts.

Square each part, add, then take the square root.

‖∇f‖ = √3² + 4²

3 squared is 9.

4 squared is 16.

‖∇f‖ = √9 + 16

Add: 9 plus 16 is 25.

‖∇f‖ = √25

5 times 5 is 25.

5 × 5 = 25

So ‖∇f‖ = 5.

‖∇f‖ = 5

4 Check

Let's look at it on the xy grid.

The blue curve is a level curve: f is 7 all along it.

The gradient points straight uphill, at a square corner to the level curve.

A tiny step of 0.001 along the arrow: f changes about 5 per step.

That is 5.

Right!

along ∇f: f(…) − f(1, 1)0.001 ≈ 5

Let's look back.

First: the partial derivatives.

Then: find the gradient.

Then: find the length.

Last: check.

Answer