Scalar fields and the gradient
Before you start: you'll need partial differentiation.
A scalar field gives a single number at every point in space: temperature in a room, height on a map, pressure in a pipe. We write it as \(\phi(x, y, z)\).
The del operator collects the three partial derivatives into one vector operator:
\[\nabla = \mathbf{i}\,\frac{\partial}{\partial x} + \mathbf{j}\,\frac{\partial}{\partial y} + \mathbf{k}\,\frac{\partial}{\partial z}\]
\[\operatorname{grad}\phi = \nabla\phi = \frac{\partial\phi}{\partial x}\,\mathbf{i} + \frac{\partial\phi}{\partial y}\,\mathbf{j} + \frac{\partial\phi}{\partial z}\,\mathbf{k}\]
The gradient turns a scalar field into a vector field.
What the gradient tells you
- Direction: \(\nabla\phi\) points in the direction in which \(\phi\) increases fastest.
- Size: \(|\nabla\phi|\) is that greatest rate of increase.
- Normal to surfaces: \(\nabla\phi\) is perpendicular to the level surface \(\phi = c\) through the point. So the unit normal to a surface \(\phi(x, y, z) = c\) is \[\hat n = \frac{\nabla\phi}{|\nabla\phi|}\]
On a hill, the gradient of height points straight up the steepest slope, and it is at right angles to the contour lines on the map.
Useful results
Let \(\vec r = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) be the position vector and \(r = |\vec r| = \sqrt{x^2 + y^2 + z^2}\).
| Result | Note |
|---|---|
| \(\nabla(\phi\psi) = \phi\,\nabla\psi + \psi\,\nabla\phi\) | product rule |
| \(\nabla f(r) = f'(r)\,\dfrac{\vec r}{r}\) | for any function of \(r\) alone |
| \(\nabla r = \dfrac{\vec r}{r}\) | a unit vector pointing away from the origin |
| \(\nabla r^n = n\,r^{n-2}\,\vec r\) | so \(\nabla\dfrac{1}{r} = -\dfrac{\vec r}{r^3}\) |
The first entry follows from \(\dfrac{\partial r}{\partial x} = \dfrac{x}{r}\), which you get by differentiating \(r^2 = x^2 + y^2 + z^2\).
Solved examples
Example 1: Gradient at a point
EasyFind \(\nabla\phi\) and \(|\nabla\phi|\) at \((1, 2, 1)\) for \(\phi = x^2y + yz^2\).
- \(\phi_x = 2xy\), \(\phi_y = x^2 + z^2\), \(\phi_z = 2yz\).
- At \((1, 2, 1)\): \(\nabla\phi = 4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\).
- \(|\nabla\phi| = \sqrt{16 + 4 + 16} = 6\).
Answer\(\nabla\phi = 4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\), \(\;|\nabla\phi| = 6\)
Example 2: Gradient of a power of r
MediumShow that \(\nabla r^n = n\,r^{n-2}\,\vec r\).
- \(\dfrac{\partial}{\partial x}r^n = n\,r^{n-1}\dfrac{\partial r}{\partial x} = n\,r^{n-1}\dfrac{x}{r} = n\,r^{n-2}\,x\). Similarly for \(y\) and \(z\).
- So \(\nabla r^n = n\,r^{n-2}(x\mathbf{i} + y\mathbf{j} + z\mathbf{k})\).
Answer\(\nabla r^n = n\,r^{n-2}\,\vec r\). With \(n = -1\): \(\nabla\dfrac{1}{r} = -\dfrac{\vec r}{r^3}\).
Example 3: Unit normal to a sphere
EasyFind the unit normal to the surface \(x^2 + y^2 + z^2 = 9\) at \((1, 2, 2)\).
- \(\phi = x^2 + y^2 + z^2\), so \(\nabla\phi = (2x, 2y, 2z) = (2, 4, 4)\) at the point.
- \(|\nabla\phi| = \sqrt{4 + 16 + 16} = 6\).
Answer\(\hat n = \dfrac{1}{3}(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\), which points straight out from the centre, as expected.
Example 4: Unit normal to another surface
MediumFind the unit normal to \(xy + yz + zx = 3\) at \((1, 1, 1)\).
- \(\nabla\phi = (y + z,\; x + z,\; x + y) = (2, 2, 2)\) at the point.
- \(|\nabla\phi| = 2\sqrt 3\).
Answer\(\hat n = \dfrac{1}{\sqrt 3}(\mathbf{i} + \mathbf{j} + \mathbf{k})\)
Example 5: Angle between two surfaces
Exam levelFind the angle between the surfaces \(x^2 + y^2 + z^2 = 9\) and \(z = x^2 + y^2 - 3\) at the point \((2, -1, 2)\).
- The angle between surfaces is the angle between their normals. Check the point lies on both: \(4 + 1 + 4 = 9\) ✓ and \(4 + 1 - 3 = 2\) ✓.
- Surface 1: \(\nabla\phi_1 = (2x, 2y, 2z) = (4, -2, 4)\), with length 6.
- Surface 2: write it as \(x^2 + y^2 - z = 3\). \(\nabla\phi_2 = (2x, 2y, -1) = (4, -2, -1)\), with length \(\sqrt{21}\).
- \(\cos\theta = \dfrac{16 + 4 - 4}{6\sqrt{21}} = \dfrac{8}{3\sqrt{21}}\).
Answer\(\theta = \cos^{-1}\dfrac{8}{3\sqrt{21}} \approx 54.4^\circ\)
Example 6: Greatest rate of increase
MediumIn which direction does \(\phi = xyz^2\) increase fastest at \((1, 0, 3)\), and what is that rate?
- \(\nabla\phi = (yz^2,\; xz^2,\; 2xyz) = (0, 9, 0)\) at the point.
AnswerFastest increase along \(\mathbf{j}\), at the rate \(|\nabla\phi| = 9\).
For a normal to a surface, first move everything to one side so the surface reads \(\phi(x, y, z) = \text{constant}\). Then the normal is simply \(\nabla\phi\). Forgetting to rearrange (as in Example 5) gives the wrong vector.
Common mistakes
1. Giving \(\nabla\phi\) as a number
The gradient of a scalar field is a vector. Its size \(|\nabla\phi|\) is a number.
2. Forgetting to normalise
A unit normal must be divided by \(|\nabla\phi|\).
3. Not rearranging the surface first
For \(z = x^2 + y^2 - 3\), use \(\phi = x^2 + y^2 - z\), not \(x^2 + y^2 - 3\).
4. Using an obtuse angle between surfaces
The angle between two surfaces is usually taken as the acute angle, so use \(|\cos\theta|\) if the dot product comes out negative.
How it's asked in exams
- Find \(\nabla\phi\) at a point, and its magnitude.
- Unit normal to a surface at a point.
- Angle between two surfaces at a common point.
- Prove a result such as \(\nabla r^n = n r^{n-2}\vec r\) or \(\nabla f(r) = f'(r)\dfrac{\vec r}{r}\).
- Find \(\phi\) given \(\nabla\phi\) (see scalar potential).
Practice questions
Try each one on paper before you open the answer.
Q1Find \(\nabla\phi\) for \(\phi = x^3 + y^3 + z^3 - 3xyz\).
\(3(x^2 - yz)\mathbf{i} + 3(y^2 - zx)\mathbf{j} + 3(z^2 - xy)\mathbf{k}\).
Q2Show that \(\nabla(\log r) = \dfrac{\vec r}{r^2}\).
\(\nabla f(r) = f'(r)\dfrac{\vec r}{r}\) with \(f = \log r\), \(f' = \dfrac{1}{r}\).
Q3Find the unit normal to \(x^2y + 2xz = 4\) at \((2, -2, 3)\).
\(\nabla\phi = (2xy + 2z,\; x^2,\; 2x) = (-2, 4, 4)\), with length 6. So \(\hat n = \dfrac{1}{3}(-\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})\).
Q4Find the greatest rate of increase of \(\phi = x^2 + y^2 + z^2\) at \((1, 1, 1)\).
\(\nabla\phi = (2, 2, 2)\), so the rate is \(2\sqrt 3\), in the direction \(\dfrac{1}{\sqrt 3}(\mathbf{i} + \mathbf{j} + \mathbf{k})\).
Q5If \(\nabla\phi = 2xyz^3\,\mathbf{i} + x^2z^3\,\mathbf{j} + 3x^2yz^2\,\mathbf{k}\), find \(\phi\).
Integrate the first component with respect to \(x\): \(x^2yz^3\). It also matches the other two components. So \(\phi = x^2yz^3 + c\).
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Frequently asked questions
What is the gradient of a scalar field?
The vector \(\nabla\phi = \phi_x\mathbf{i} + \phi_y\mathbf{j} + \phi_z\mathbf{k}\). It points in the direction of fastest increase, and its size is that rate of increase.
Why is the gradient normal to a level surface?
Along the surface \(\phi\) doesn't change, so the rate of change in any tangent direction is 0. That rate is \(\nabla\phi\cdot\hat t = 0\), so \(\nabla\phi\) is perpendicular to every tangent direction.
What is the difference between gradient, divergence and curl?
Gradient acts on a scalar field and gives a vector. Divergence acts on a vector field and gives a scalar. Curl acts on a vector field and gives a vector.
Where the gradient leads next
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.