24 May 2025

Scalar and Vector Fields, Directional Derivatives, and the Gradient

Fields, directional and normal derivatives, and the geometrical meaning of the gradient.

bsc semester-ii mathematical-physics vector-calculus gradient

A field assigns a physical quantity to every point. A scalar field assigns one number,

\[\phi:\mathbb R^3\to\mathbb R, \qquad (x,y,z)\mapsto\phi(x,y,z),\]

whereas a vector field assigns a vector,

\[\mathbf A=A_x\hat{\mathbf x}+A_y\hat{\mathbf y}+A_z\hat{\mathbf z}.\]

The dimensions belong to the field: if $T$ is temperature, then $[T]=\mathrm{K}$ and $[\partial T/\partial x]=\mathrm{K\,m^{-1}}$.

Change of a scalar field

For $d\mathbf r=dx\,\hat{\mathbf x}+dy\,\hat{\mathbf y}+dz\,\hat{\mathbf z}$, the first-order Taylor expansion is

\[d\phi=\frac{\partial\phi}{\partial x}dx +\frac{\partial\phi}{\partial y}dy +\frac{\partial\phi}{\partial z}dz.\]

Define

\[\nabla=\hat{\mathbf x}\partial_x+\hat{\mathbf y}\partial_y+\hat{\mathbf z}\partial_z, \qquad \nabla\phi=\phi_x\hat{\mathbf x}+\phi_y\hat{\mathbf y}+\phi_z\hat{\mathbf z}.\]

Then

\[\boxed{d\phi=\nabla\phi\cdot d\mathbf r}.\]

Because $d\phi$ is a scalar, $\nabla\phi$ transforms as a vector under rotations of Cartesian axes.

Directional derivative

Move from $\mathbf r_0$ along a unit vector $\hat{\mathbf n}$:

\[\mathbf r(s)=\mathbf r_0+s\hat{\mathbf n}, \qquad \frac{d\mathbf r}{ds}=\hat{\mathbf n}.\]

The chain rule gives

\[\begin{aligned} \frac{d\phi}{ds} &=\phi_x\frac{dx}{ds}+\phi_y\frac{dy}{ds}+\phi_z\frac{dz}{ds}\\ &=\nabla\phi\cdot\hat{\mathbf n}. \end{aligned}\]

Thus

\[\boxed{D_{\hat{\mathbf n}}\phi=\hat{\mathbf n}\cdot\nabla\phi}.\]

The vector must be normalized when a derivative per unit distance is required. For a nonzero displacement direction $\mathbf v$,

\[\hat{\mathbf v}=\frac{\mathbf v}{\lvert\mathbf v\rvert}, \qquad D_{\hat{\mathbf v}}\phi =\nabla\phi\cdot\frac{\mathbf v}{\lvert\mathbf v\rvert}.\]

If $\alpha$ is the angle between $\hat{\mathbf n}$ and $\nabla\phi$,

\[D_{\hat{\mathbf n}}\phi=\lvert\nabla\phi\rvert\cos\alpha.\]

The maximum value $\lvert\nabla\phi\rvert$ occurs along $\nabla\phi$; the minimum $-\lvert\nabla\phi\rvert$ occurs in the opposite direction.

Geometrical interpretation and normal derivative

On a level surface $\phi(x,y,z)=C$, a tangent displacement $d\mathbf r_{\parallel}$ leaves $\phi$ unchanged. Therefore

\[0=d\phi=\nabla\phi\cdot d\mathbf r_{\parallel}.\]

Hence $\nabla\phi$ is normal to the level surface wherever $\nabla\phi\ne\mathbf0$. Choose the unit normal in the direction of increasing $\phi$,

\[\hat{\mathbf n}=\frac{\nabla\phi}{\lvert\nabla\phi\rvert},\]

then the normal derivative is

\[\boxed{\frac{\partial\phi}{\partial n} =\hat{\mathbf n}\cdot\nabla\phi=\lvert\nabla\phi\rvert}.\]

For \(\phi=x^2+y^2\), the level curves are circles and \(\nabla\phi=(2x,2y)\) is radial. The plotted curves and arrows come directly from these equations.

For $\phi=x^2+y^2+3z$,

\[\nabla\phi=2x\hat{\mathbf x}+2y\hat{\mathbf y}+3\hat{\mathbf z}.\]

At $(1,-1,0)$, along $\hat{\mathbf n}=(2,1,2)/3$,

\[D_{\hat{\mathbf n}}\phi=(2,-2,3)\cdot\frac{(2,1,2)}3 =\frac{4-2+6}{3}=\frac83.\]

If the opposite orientation is chosen, the normal derivative is $-\lvert\nabla\phi\rvert$. The units $[\nabla\phi]=[\phi]/\mathrm{length}$ confirm that the gradient is a spatial rate of change.

At a regular point $\mathbf r_0$ of the level surface, where $\nabla\phi(\mathbf r_0)\ne\mathbf0$, its tangent plane is

\[\boxed{\nabla\phi(\mathbf r_0)\cdot(\mathbf r-\mathbf r_0)=0}.\]

If $\nabla\phi(\mathbf r_0)=\mathbf0$, this first-order equation does not determine a unique normal. For a time-dependent field sampled by a moving observer, $\phi=\phi(\mathbf r(t),t)$, the same chain rule gives

\[\boxed{\frac{d\phi}{dt}=\frac{\partial\phi}{\partial t} +\frac{d\mathbf r}{dt}\cdot\nabla\phi}.\]

The first term is local temporal change; the second is change caused by motion through the spatial gradient.

Solved Problems

1. A temperature field is $T=300+2x^2-yz$ K, with $x,y,z$ measured in metres and the numerical coefficients carrying the corresponding SI units. At $P=(1,2,-1)$, find the directional derivative towards $Q=(3,1,1)$, the direction and magnitude of steepest increase, and the tangent plane to the isotherm through $P$.

   The displacement and its unit vector are

   \[\mathbf v=Q-P=(2,-1,2),\qquad \lvert\mathbf v\rvert=3,\qquad \hat{\mathbf v}=\frac{(2,-1,2)}3.\]

   Since

   \[\nabla T=(4x,-z,-y)\ \mathrm{K\,m^{-1}},\]

   at $P$ one has $\nabla T=(4,1,-2)\ \mathrm{K\,m^{-1}}$. Therefore

   \[D_{\hat{\mathbf v}}T =(4,1,-2)\cdot\frac{(2,-1,2)}3 =1\ \mathrm{K\,m^{-1}}.\]

   The steepest-increase direction and rate are

   \[\hat{\mathbf n}_{\max}=\frac{(4,1,-2)}{\sqrt{21}}, \qquad \lvert\nabla T\rvert=\sqrt{21}\ \mathrm{K\,m^{-1}}.\]

   Finally, the tangent plane is

   \[(4,1,-2)\cdot(x-1,y-2,z+1)=0, \qquad \boxed{4x+y-2z=8}.\]

   The positive derivative towards $Q$ means that the temperature initially rises along that direction. The unit-vector normalization supplies the required kelvin-per-metre units.

2. Let $\phi=x^2y+z$ and let a particle follow $\mathbf r(t)=(t,t^2,1-t)$. Find $d\phi/dt$ at $t=1$ in two independent ways.

   Direct substitution gives

   \[\phi(\mathbf r(t))=t^2(t^2)+(1-t)=t^4+1-t,\]

   and hence

   \[\left.\frac{d\phi}{dt}\right\rvert_{t=1} =\left.(4t^3-1)\right\rvert_{t=1}=3.\]

   Alternatively,

   \[\nabla\phi=(2xy,x^2,1),\qquad \frac{d\mathbf r}{dt}=(1,2t,-1).\]

   At $t=1$, $\mathbf r=(1,1,0)$, so

   \[\nabla\phi\cdot\frac{d\mathbf r}{dt} =(2,1,1)\cdot(1,2,-1)=3.\]

   The agreement is the chain-rule check: the field value is increasing at three units of $\phi$ per unit of the path parameter.

Descriptive Questions

1. Starting from the first-order Taylor expansion, derive the gradient and the directional derivative along an arbitrary unit vector.
2. Prove that $\nabla\phi$ is perpendicular to a regular level surface, and obtain the equation of its tangent plane.
3. Distinguish scalar and vector fields using physical examples, dimensions, and their transformation under a rotation of Cartesian axes.
4. Use the Cauchy–Schwarz inequality to establish the maximum and minimum directional derivatives and explain the sign of an oriented normal derivative.

Numerical Problems

1. The temperature is $T=20+x^2+3y^2$ °C, with distances in metres. Find its directional derivative at $(2,-1)$ along $(3,4)$.

   Answer: $-12/5\ \mathrm{^\circ C\,m^{-1}}$.

2. For $\phi=xe^y+z^2$, find the unit direction and magnitude of maximum increase at $(1,0,-1)$.

   Answer: direction $(1,1,-2)/\sqrt6$; maximum rate $\sqrt6$.

3. Find the tangent plane to $x^2+2y^2+3z^2=6$ at $(1,1,1)$.

   Answer: $x+2y+3z=6$.

4. For $\phi=x^2y-yz$, calculate the directional derivative at $(1,2,3)$ along $(2,-2,1)$.

   Answer: $10/3$.

5. A particle follows $\mathbf r(s)=(\cos s,\sin s,s)$ through $\phi=x^2+y^2+z^2$. Find $d\phi/ds$ at $s=\pi$.

   Answer: $2\pi$.

6. On the plane $x+y+z=3$, take the normal outward from $x+y+z\leq3$. Find the normal derivative of $\phi=x^2+y^2+z^2$ at $(1,1,1)$.

   Answer: $2\sqrt3$.

All symbolic reductions and numerical answers are checked in the Unit I Maxima worksheet; every printed residual is zero.

References

1. Gradient — Wikipedia
2. OpenStax, Calculus Volume 3, §4.6: Directional Derivatives and the Gradient
3. George B. Arfken, Hans J. Weber, and Frank E. Harris, Mathematical Methods for Physicists, 7th ed., Chapter 3, §§3.1–3.6.