Worked examples
The question
Question: find the gradient of f(x, y) = x²y at (1, 2).
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 = 1 and y = 2 in: ∇f(1, 2) = ⟨4, 1⟩.
∇f(1, 2) = ⟨4, 1⟩
Careful!
The answer has two parts: one for each letter.
It is an arrow, not one number.
Careful: the gradient is a vector, not one number
3 Check
Let's look at it on the xy grid.
The blue curve is a level curve: f is 2 all along it.
The gradient points straight uphill, at a square corner to the level curve.
A tiny step of 0.01 along the arrow: f grows from 2 to about 2.041.
Uphill.
Right!
f(1.0097, 2.0024) ≈ 2.041 > f(1, 2) = 2
Let's look back.
First: the partial derivatives.
Then: find the gradient.
Last: check.
▶ Watch it solved
The question
Question: find the gradient of f(x, y) = x³ − xy at (2, 1).
f(x, y) = x3 − xy
1 The partial derivatives
Term by term.
x³: the power 3 comes down.
The power goes down by 1.
So 3x².
In xy, y is fixed and waits in front.
x gives 1.
So y.
So ∂f/∂x = 3x² − y.
∂f∂x = 3x2 − y
Term by term.
x³ has no y: it is a fixed number, so it gives 0.
In xy, x is fixed and waits in front.
y gives 1.
So x.
∂f∂y = 0 − x
Drop the zeros: ∂f/∂y = −x.
∂f∂y = −x
2 Find the gradient
The gradient ∇f puts the partial derivatives together as one vector.
∇f = ⟨∂f∂x, ∂f∂y⟩
So ∇f = ⟨3x² − y, −x⟩.
∇f = ⟨3x2 − y, −x⟩
At the point, put x = 2 and y = 1 in: ∇f(2, 1) = ⟨11, −2⟩.
∇f(2, 1) = ⟨11, −2⟩
Careful!
The answer has two parts: one for each letter.
It is an arrow, not one number.
Careful: the gradient is a vector, not one number
3 Check
Let's look at it on the xy grid.
The blue curve is a level curve: f is 6 all along it.
The gradient points straight uphill, at a square corner to the level curve.
A tiny step of 0.01 along the arrow: f grows from 6 to about 6.112.
Uphill.
Right!
f(2.0098, 0.9982) ≈ 6.112 > f(2, 1) = 6
Let's look back.
First: the partial derivatives.
Then: find the gradient.
Last: check.
Answer ∇f(2, 1) = ⟨11, −2⟩
▶ Watch it solved
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.
▶ Watch it solved
