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

Step by step

The question

Question: find the fastest rate of change of f(x, y) = x²y at (2, 1), and its direction.

f(x, y) = x²y

1 The partial derivatives

Term by term.

In x²y, y is fixed and waits in front.

x² gives 2x.

So 2xy.

So ∂f/∂x = 2xy.

∂f∂x = 2xy

Term by term.

In x²y, x² is fixed and waits in front.

y gives 1.

So x².

So ∂f/∂y = x².

∂f∂y = x²

2 Find the gradient

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

∇f = ⟨∂f∂x, ∂f∂y⟩

So ∇f = ⟨2xy, x²⟩.

∇f = ⟨2xy, x²⟩

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

∇f(2, 1) = ⟨4, 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‖ = √4² + 4²

4 squared is 16.

‖∇f‖ = √16 + 16

Add: 16 plus 16 is 32.

‖∇f‖ = √32

But 32 has a square number in it: 16.

32 is 16 times 2.

‖∇f‖ = √16 × 2

A root of a times splits into two roots.

‖∇f‖ = √16 × √2

√16 is 4.

4 × 4 = 16

So ‖∇f‖ = 4√2.

That is the exact answer, made simple.

‖∇f‖ = 4√2

With a calculator, that is about 5.7, to the nearest tenth.

‖∇f‖ ≈ 5.7

4 Check

Let's look at it on the xy grid.

The blue curve is a level curve: f is 4 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 5.66 per step.

That is about 5.7.

Right!

along ∇f: f(…) − f(2, 1)0.001 ≈ 5.659

Let's look back.

First: the partial derivatives.

Then: find the gradient.

Then: find the length.

Last: check.

Answer