Maximum rate of change of f(x, y) = x^2 + 3y^2 at (2, 1) Step by Step

Step by step

The question

Question: find the fastest rate of change of f(x, y) = x² + 3y² at (2, 1), and its direction.

f(x, y) = x² + 3y²

1 The partial derivatives

Term by term.

x²: the power 2 comes down.

The power goes down by 1.

So 2x.

3y² 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.

3y²: the power 2 comes down: 3 times 2 is 6.

The power goes down by 1.

So 6y.

∂f∂y = 0 + 6y

Drop the zeros: ∂f/∂y = 6y.

∂f∂y = 6y

2 Find the gradient

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

∇f = ⟨∂f∂x, ∂f∂y⟩

So ∇f = ⟨2x, 6y⟩.

∇f = ⟨2x, 6y⟩

At the point, put x = 2 and y = 1 in: ∇f(2, 1) = ⟨4, 6⟩.

∇f(2, 1) = ⟨4, 6⟩

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‖ = √4² + 6²

4 squared is 16.

6 squared is 36.

‖∇f‖ = √16 + 36

Add: 16 plus 36 is 52.

‖∇f‖ = √52

But 52 has a square number in it: 4.

52 is 4 times 13.

‖∇f‖ = √4 × 13

A root of a times splits into two roots.

‖∇f‖ = √4 × √13

√4 is 2.

2 × 2 = 4

So ‖∇f‖ = 2√13.

That is the exact answer, made simple.

‖∇f‖ = 2√13

With a calculator, that is about 7.2, to the nearest tenth.

‖∇f‖ ≈ 7.2

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 7.21 per step.

That is about 7.2.

Right!

along ∇f: f(…) − f(2, 1)0.001 ≈ 7.213

Let's look back.

First: the partial derivatives.

Then: find the gradient.

Then: find the length.

Last: check.

Answer