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Given a scalar value function

Mathematics · FE Reference Handbook section

Mathematics
2 formulas
10 exam-style examples
~49 min
All Mathematics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • find a vector x*∈Rn such that

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Gradient of a scalar function and divergence of its field — Given a scalar value function

Given the scalar value function φ = 3x² + 2yz + 3z², compute the gradient ∇φ at the point (2, 3, 3), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=3x2+2yz+3z2\phi = 3x^{2} + 2yz + 3z^{2}

  • Point (2, 3, 3)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 6x, ∂φ/∂y = 2z, ∂φ/∂z = 2y + 6z

  3. Substituting

    ∇ϕ=12i+6j+24k∇\phi = 12i + 6j + 24k

  4. Magnitude

    ∣∇ϕ∣=(122+62+242)=27.495|∇\phi| = \sqrt(12^{2} + 6^{2} + 24^{2}) = 27.495

  5. Divergence

    ∇⋅∇ϕ=6+0+6=12∇\cdot∇\phi = 6 + 0 + 6 = 12
Answer:
∇ϕ=(12,6,24),∣∇ϕ∣=27.495,∇⋅∇ϕ=12∇\phi = (12, 6, 24), |∇\phi| = 27.495, ∇\cdot∇\phi = 12

Why the other options are there

  • |∇φ| = 42 (components added)
  • ∇·∇φ = 42 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

Example 2
Gradient of a scalar function and divergence of its field — Given a scalar value function (2)

Given the scalar value function φ = 3x² + 3yz + 2z², compute the gradient ∇φ at the point (1, 2, 2), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=3x2+3yz+2z2\phi = 3x^{2} + 3yz + 2z^{2}

  • Point (1, 2, 2)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 6x, ∂φ/∂y = 3z, ∂φ/∂z = 3y + 4z

  3. Substituting

    ∇ϕ=6i+6j+14k∇\phi = 6i + 6j + 14k

  4. Magnitude

    ∣∇ϕ∣=(62+62+142)=16.371|∇\phi| = \sqrt(6^{2} + 6^{2} + 14^{2}) = 16.371

  5. Divergence

    ∇⋅∇ϕ=6+0+4=10∇\cdot∇\phi = 6 + 0 + 4 = 10
Answer:
∇ϕ=(6,6,14),∣∇ϕ∣=16.371,∇⋅∇ϕ=10∇\phi = (6, 6, 14), |∇\phi| = 16.371, ∇\cdot∇\phi = 10

Why the other options are there

  • |∇φ| = 26 (components added)
  • ∇·∇φ = 26 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

Example 3
Gradient of a scalar function and divergence of its field — Given a scalar value function (3)

Given the scalar value function φ = 5x² + 4yz + 2z², compute the gradient ∇φ at the point (4, 3, 2), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=5x2+4yz+2z2\phi = 5x^{2} + 4yz + 2z^{2}

  • Point (4, 3, 2)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 10x, ∂φ/∂y = 4z, ∂φ/∂z = 4y + 4z

  3. Substituting

    ∇ϕ=40i+8j+20k∇\phi = 40i + 8j + 20k

  4. Magnitude

    ∣∇ϕ∣=(402+82+202)=45.431|∇\phi| = \sqrt(40^{2} + 8^{2} + 20^{2}) = 45.431

  5. Divergence

    ∇⋅∇ϕ=10+0+4=14∇\cdot∇\phi = 10 + 0 + 4 = 14
Answer:
∇ϕ=(40,8,20),∣∇ϕ∣=45.431,∇⋅∇ϕ=14∇\phi = (40, 8, 20), |∇\phi| = 45.431, ∇\cdot∇\phi = 14

Why the other options are there

  • |∇φ| = 68 (components added)
  • ∇·∇φ = 68 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

Example 4
Gradient of a scalar function and divergence of its field — Given a scalar value function (4)

Given the scalar value function φ = 2x² + 6yz + 4z², compute the gradient ∇φ at the point (1, 4, 4), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=2x2+6yz+4z2\phi = 2x^{2} + 6yz + 4z^{2}

  • Point (1, 4, 4)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 4x, ∂φ/∂y = 6z, ∂φ/∂z = 6y + 8z

  3. Substituting

    ∇ϕ=4i+24j+56k∇\phi = 4i + 24j + 56k

  4. Magnitude

    ∣∇ϕ∣=(42+242+562)=61.057|∇\phi| = \sqrt(4^{2} + 24^{2} + 56^{2}) = 61.057

  5. Divergence

    ∇⋅∇ϕ=4+0+8=12∇\cdot∇\phi = 4 + 0 + 8 = 12
Answer:
∇ϕ=(4,24,56),∣∇ϕ∣=61.057,∇⋅∇ϕ=12∇\phi = (4, 24, 56), |∇\phi| = 61.057, ∇\cdot∇\phi = 12

Why the other options are there

  • |∇φ| = 84 (components added)
  • ∇·∇φ = 84 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

Example 5
Gradient of a scalar function and divergence of its field — Given a scalar value function (5)

Given the scalar value function φ = 6x² + 3yz + 3z², compute the gradient ∇φ at the point (2, 3, 3), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=6x2+3yz+3z2\phi = 6x^{2} + 3yz + 3z^{2}

  • Point (2, 3, 3)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 12x, ∂φ/∂y = 3z, ∂φ/∂z = 3y + 6z

  3. Substituting

    ∇ϕ=24i+9j+27k∇\phi = 24i + 9j + 27k

  4. Magnitude

    ∣∇ϕ∣=(242+92+272)=37.229|∇\phi| = \sqrt(24^{2} + 9^{2} + 27^{2}) = 37.229

  5. Divergence

    ∇⋅∇ϕ=12+0+6=18∇\cdot∇\phi = 12 + 0 + 6 = 18
Answer:
∇ϕ=(24,9,27),∣∇ϕ∣=37.229,∇⋅∇ϕ=18∇\phi = (24, 9, 27), |∇\phi| = 37.229, ∇\cdot∇\phi = 18

Why the other options are there

  • |∇φ| = 60 (components added)
  • ∇·∇φ = 60 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

Example 6
Gradient of a scalar function and divergence of its field — Given a scalar value function (6)

Given the scalar value function φ = 4x² + 2yz + 1z², compute the gradient ∇φ at the point (3, 2, 3), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=4x2+2yz+1z2\phi = 4x^{2} + 2yz + 1z^{2}

  • Point (3, 2, 3)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 8x, ∂φ/∂y = 2z, ∂φ/∂z = 2y + 2z

  3. Substituting

    ∇ϕ=24i+6j+10k∇\phi = 24i + 6j + 10k

  4. Magnitude

    ∣∇ϕ∣=(242+62+102)=26.683|∇\phi| = \sqrt(24^{2} + 6^{2} + 10^{2}) = 26.683

  5. Divergence

    ∇⋅∇ϕ=8+0+2=10∇\cdot∇\phi = 8 + 0 + 2 = 10
Answer:
∇ϕ=(24,6,10),∣∇ϕ∣=26.683,∇⋅∇ϕ=10∇\phi = (24, 6, 10), |∇\phi| = 26.683, ∇\cdot∇\phi = 10

Why the other options are there

  • |∇φ| = 40 (components added)
  • ∇·∇φ = 40 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

Example 7
Gradient of a scalar function and divergence of its field — Given a scalar value function (7)

Given the scalar value function φ = 6x² + 5yz + 4z², compute the gradient ∇φ at the point (4, 1, 4), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=6x2+5yz+4z2\phi = 6x^{2} + 5yz + 4z^{2}

  • Point (4, 1, 4)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 12x, ∂φ/∂y = 5z, ∂φ/∂z = 5y + 8z

  3. Substituting

    ∇ϕ=48i+20j+37k∇\phi = 48i + 20j + 37k

  4. Magnitude

    ∣∇ϕ∣=(482+202+372)=63.820|∇\phi| = \sqrt(48^{2} + 20^{2} + 37^{2}) = 63.820

  5. Divergence

    ∇⋅∇ϕ=12+0+8=20∇\cdot∇\phi = 12 + 0 + 8 = 20
Answer:
∇ϕ=(48,20,37),∣∇ϕ∣=63.820,∇⋅∇ϕ=20∇\phi = (48, 20, 37), |∇\phi| = 63.820, ∇\cdot∇\phi = 20

Why the other options are there

  • |∇φ| = 105 (components added)
  • ∇·∇φ = 105 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

Example 8
Gradient of a scalar function and divergence of its field — Given a scalar value function (8)

Given the scalar value function φ = 3x² + 5yz + 3z², compute the gradient ∇φ at the point (1, 4, 2), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=3x2+5yz+3z2\phi = 3x^{2} + 5yz + 3z^{2}

  • Point (1, 4, 2)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 6x, ∂φ/∂y = 5z, ∂φ/∂z = 5y + 6z

  3. Substituting

    ∇ϕ=6i+10j+32k∇\phi = 6i + 10j + 32k

  4. Magnitude

    ∣∇ϕ∣=(62+102+322)=34.059|∇\phi| = \sqrt(6^{2} + 10^{2} + 32^{2}) = 34.059

  5. Divergence

    ∇⋅∇ϕ=6+0+6=12∇\cdot∇\phi = 6 + 0 + 6 = 12
Answer:
∇ϕ=(6,10,32),∣∇ϕ∣=34.059,∇⋅∇ϕ=12∇\phi = (6, 10, 32), |∇\phi| = 34.059, ∇\cdot∇\phi = 12

Why the other options are there

  • |∇φ| = 48 (components added)
  • ∇·∇φ = 48 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

Example 9
Gradient of a scalar function and divergence of its field — Given a scalar value function (9)

Given the scalar value function φ = 2x² + 6yz + 2z², compute the gradient ∇φ at the point (4, 2, 2), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=2x2+6yz+2z2\phi = 2x^{2} + 6yz + 2z^{2}

  • Point (4, 2, 2)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 4x, ∂φ/∂y = 6z, ∂φ/∂z = 6y + 4z

  3. Substituting

    ∇ϕ=16i+12j+20k∇\phi = 16i + 12j + 20k

  4. Magnitude

    ∣∇ϕ∣=(162+122+202)=28.284|∇\phi| = \sqrt(16^{2} + 12^{2} + 20^{2}) = 28.284

  5. Divergence

    ∇⋅∇ϕ=4+0+4=8∇\cdot∇\phi = 4 + 0 + 4 = 8
Answer:
∇ϕ=(16,12,20),∣∇ϕ∣=28.284,∇⋅∇ϕ=8∇\phi = (16, 12, 20), |∇\phi| = 28.284, ∇\cdot∇\phi = 8

Why the other options are there

  • |∇φ| = 48 (components added)
  • ∇·∇φ = 48 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

Example 10
Gradient of a scalar function and divergence of its field — Given a scalar value function (10)

Given the scalar value function φ = 5x² + 2yz + 5z², compute the gradient ∇φ at the point (3, 2, 1), its magnitude, and the divergence ∇·(∇φ).

Given

  • ϕ=5x2+2yz+5z2\phi = 5x^{2} + 2yz + 5z^{2}

  • Point (3, 2, 1)

Find

∇φ, |∇φ| and ∇·∇φ

Start with the thinking

  • The gradient collects the partial derivatives; it points toward the steepest increase of the scalar function.
  • Divergence of a gradient is the Laplacian — a scalar, not a vector.

Step-by-step solution

  1. Formula — ∇φ = (∂φ/∂x)i + (∂φ/∂y)j + (∂φ/∂z)k

  2. Partials — ∂φ/∂x = 10x, ∂φ/∂y = 2z, ∂φ/∂z = 2y + 10z

  3. Substituting

    ∇ϕ=30i+2j+14k∇\phi = 30i + 2j + 14k

  4. Magnitude

    ∣∇ϕ∣=(302+22+142)=33.166|∇\phi| = \sqrt(30^{2} + 2^{2} + 14^{2}) = 33.166

  5. Divergence

    ∇⋅∇ϕ=10+0+10=20∇\cdot∇\phi = 10 + 0 + 10 = 20
Answer:
∇ϕ=(30,2,14),∣∇ϕ∣=33.166,∇⋅∇ϕ=20∇\phi = (30, 2, 14), |∇\phi| = 33.166, ∇\cdot∇\phi = 20

Why the other options are there

  • |∇φ| = 46 (components added)
  • ∇·∇φ = 46 (used the gradient values)

Reference: FE Reference Handbook — Mathematics → Given a scalar value function

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