• Understand physical interpretation and applications of gradient.

Gradient is a vector with the magnitude and direction of the maximum change of the function in space. Gradient is calculated for scalars and the result of calculation (the gradient itself) is a vector.

$\nabla \phi = \left ( \dfrac{\partial}{\partial x} \mathbf{i} + \dfrac{\partial}{\partial y} \mathbf{\mathbf{j}} + \dfrac{\partial}{\partial z} \mathbf{k} \right ) \phi$

Let us evaluate gradient of a scalar function $\phi$ described below. $\phi = x^3z$ Then, $\dfrac{\partial \phi}{\partial x} = 3x^2z$ $\dfrac{\partial \phi}{\partial y} = 0$ $\dfrac{\partial \phi}{\partial z} = x^3$

$\nabla \phi = \left ( 3x^2z \mathbf{i} + x^3 \mathbf{k} \right )$

If you want to calculate $\nabla \phi$ at certain point, let’s say at (1,2,-1) then substitute these coordinates instead of x, y and z.
$\nabla \phi = \left ( 3(1)^2(-1) \mathbf{i} + (1)^3 \mathbf{k} \right ) = - 3 \mathbf{i} + \mathbf{k}$

The gradient represents the direction of greatest change. The gradient points to the maximum of the function; follow the gradient, and you will reach the local maximum. It is a vector, so it points towards the direction of greatest change.

### Solved Example:

#### 4-9-01

The directional derivative of the scalar function $f (x,y,z) = x^2+ 2y^2+ z$ at the point P = (1, 1, 2) in the direction of the vector a = 3i - 4j is:

Solution:
Direction derivative of a function f along a vector P is given by $a=\nabla f\cdot \dfrac {a}{\left| a\right| }$ $\nabla F=\left( \dfrac {\partial f}{\partial x}\overline {i}+\dfrac {\partial f}{\partial y}\overline {j}+\dfrac {\partial f}{\partial z}\overline {k}\right)$ $f\left( x,y,z\right) =x^{2}+2y^{2}+z$ $a=3\overline {i}-4\overline {j}$ $a=\nabla \left( x^{2}+2y^{2}+z\right) .\dfrac {3\overline {i}-4\overline {j}}{\sqrt {3^{2}+4^{2}}}$ At point P(1, 1, 2) the direction derivative is $=\left[ 2x\overline {i}+4y\overline {j}+\overline {k}\right] .\dfrac {3\overline {i}-4\overline {j}}{\sqrt {25}}$ $a=\dfrac {6\times 1-16\times 1}{5}=-\dfrac {10}{5}=-2$

### Solved Example:

#### 4-9-02

The vector which is normal to the surface $2xz^2-3xy-4x = 7$ at point (1,-1,2) is: (GATE PI 2019)

Solution:
$\mathrm{Let\ } \phi \left( x,y,z\right) =2xz^{2}-3xy-4x=0$ Vector normal to a surface $\phi$ is given by gradient of $\phi$. \begin{align*} \nabla \phi &=\left( \dfrac {\partial }{\partial x}\overline {i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline {k}\right) \left( 2xz^{2}-3xy-4x\right)\\ &= \left( 2z^{2}-3y-4\right) \overline {i} + \left( 0-3x-0\right) \overline {j} + \left( 4xz-0-0\right) \overline {k}\\ &=\left( 2z^{2}-3y-4\right) \overline {i}-3x\overline {j}+4xz\overline {k} \end{align*} At $\left(1,-1,2 \right) \nabla \phi =\left( 8+3-4\right) \overline {i}-3\overline {j}+8\overline {k} =7 \overline{i} - 3\overline {j}+8\overline {k}$

### Solved Example:

#### 4-9-03

A function is defined in the Cartesian coordinate system as $f (x, y) = xe^y$. The value of the directional derivative of the function (in integer) at the point (2,0) along the direction of the straight line segment from point (2,0) to point ($\dfrac{1}{2}$, 2) is: (GATE Civil 2021)

Solution:
$f\left(x, y\right) = xe^{y}$ \begin{align*} \mathrm{Grad}\ f &= \nabla f\\ &=\dfrac{\partial }{\partial x}\overline{i}+\dfrac{\partial }{\partial y}\overline{j}+\dfrac{\partial }{\partial z}\overline{k}\\ &=\left( e^{y}\right) \overline{i}+\left(xe^{y}\right) \overline{j}+\left( 0\right) \overline{k}\\ \mathrm{At}\ \left(2,0\right), \nabla f &=\overline{i}+2\overline{j} \end{align*} Let $P \left(2,0\right) \mathrm{and}\ Q\left(\dfrac{1}{2},2\right)$ \begin{align*} \overline{PQ} &=\left( \dfrac{1}{2}-2\right) \overline{i}+\left( 2-0\right)\overline{j}\\ &=-\dfrac{3}{2}\overline{i}+2\overline{j} \end{align*} Unit vector along $\overline{PQ}$ \begin{align*} \hat{PQ} &= \dfrac{-\dfrac{3}{2}\overline{i}+2\overline{j}}{\sqrt{\left(-\dfrac{3}{2}\right) ^{2}+\left( 2\right) ^{2}}}\\ &= \dfrac{-\dfrac{3}{2}\overline{i}+2\overline{j}}{\left( \dfrac{5}{2}\right) } \end{align*} Directional derivative \begin{align*} D &=\left( \nabla f\right) \cdot \widehat{PQ}\\ &=\left(\overline{i}+2\overline{j}\right) \cdot \dfrac{-\dfrac{3}{2}\overline{i}+2\overline{j}}{\left( \dfrac{5}{2}\right) }\\ &=\dfrac{\dfrac{-3+8}{2}}{\dfrac{5}{2}}\\ &=1 \end{align*}

### Solved Example:

#### 4-9-04

The magnitude of the gradient for the function $(𝑥, 𝑦, 𝑧) = 𝑥^2 + 3𝑦^2 + 𝑧^3$ at the point (1, 1, 1) is: (GATE EC 2014 Shift IV)

Solution:
$\phi = 𝑥^2 + 3𝑦^2 + 𝑧^3$ $\nabla \phi = 2x \bar{i} + 6y \bar{j} + 3z^2 \bar{k}$ At (1,1,1), $\nabla \phi = 2 \bar{i} + 6 \bar{j} + 3 \bar{k}$ Magnitude can be calculated as: $|\nabla \phi| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{49} = 7$

### Solved Example:

#### 4-9-01

The directional derivative of the scalar function $f (x,y,z) = x^2+ 2y^2+ z$ at the point P = (1, 1, 2) in the direction of the vector a = 3i - 4j is:

Solution:
Direction derivative of a function f along a vector P is given by $a=\nabla f\cdot \dfrac {a}{\left| a\right| }$ $\nabla F=\left( \dfrac {\partial f}{\partial x}\overline {i}+\dfrac {\partial f}{\partial y}\overline {j}+\dfrac {\partial f}{\partial z}\overline {k}\right)$ $f\left( x,y,z\right) =x^{2}+2y^{2}+z$ $a=3\overline {i}-4\overline {j}$ $a=\nabla \left( x^{2}+2y^{2}+z\right) .\dfrac {3\overline {i}-4\overline {j}}{\sqrt {3^{2}+4^{2}}}$ At point P(1, 1, 2) the direction derivative is $=\left[ 2x\overline {i}+4y\overline {j}+\overline {k}\right] .\dfrac {3\overline {i}-4\overline {j}}{\sqrt {25}}$ $a=\dfrac {6\times 1-16\times 1}{5}=-\dfrac {10}{5}=-2$

### Solved Example:

#### 4-9-02

The vector which is normal to the surface $2xz^2-3xy-4x = 7$ at point (1,-1,2) is: (GATE PI 2019)

Solution:
$\mathrm{Let\ } \phi \left( x,y,z\right) =2xz^{2}-3xy-4x=0$ Vector normal to a surface $\phi$ is given by gradient of $\phi$. \begin{align*} \nabla \phi &=\left( \dfrac {\partial }{\partial x}\overline {i}+\dfrac {\partial }{\partial y}\overline {j}+\dfrac {\partial }{\partial z}\overline {k}\right) \left( 2xz^{2}-3xy-4x\right)\\ &= \left( 2z^{2}-3y-4\right) \overline {i} + \left( 0-3x-0\right) \overline {j} + \left( 4xz-0-0\right) \overline {k}\\ &=\left( 2z^{2}-3y-4\right) \overline {i}-3x\overline {j}+4xz\overline {k} \end{align*} At $\left(1,-1,2 \right) \nabla \phi =\left( 8+3-4\right) \overline {i}-3\overline {j}+8\overline {k} =7 \overline{i} - 3\overline {j}+8\overline {k}$

### Solved Example:

#### 4-9-03

A function is defined in the Cartesian coordinate system as $f (x, y) = xe^y$. The value of the directional derivative of the function (in integer) at the point (2,0) along the direction of the straight line segment from point (2,0) to point ($\dfrac{1}{2}$, 2) is: (GATE Civil 2021)

Solution:
$f\left(x, y\right) = xe^{y}$ \begin{align*} \mathrm{Grad}\ f &= \nabla f\\ &=\dfrac{\partial }{\partial x}\overline{i}+\dfrac{\partial }{\partial y}\overline{j}+\dfrac{\partial }{\partial z}\overline{k}\\ &=\left( e^{y}\right) \overline{i}+\left(xe^{y}\right) \overline{j}+\left( 0\right) \overline{k}\\ \mathrm{At}\ \left(2,0\right), \nabla f &=\overline{i}+2\overline{j} \end{align*} Let $P \left(2,0\right) \mathrm{and}\ Q\left(\dfrac{1}{2},2\right)$ \begin{align*} \overline{PQ} &=\left( \dfrac{1}{2}-2\right) \overline{i}+\left( 2-0\right)\overline{j}\\ &=-\dfrac{3}{2}\overline{i}+2\overline{j} \end{align*} Unit vector along $\overline{PQ}$ \begin{align*} \hat{PQ} &= \dfrac{-\dfrac{3}{2}\overline{i}+2\overline{j}}{\sqrt{\left(-\dfrac{3}{2}\right) ^{2}+\left( 2\right) ^{2}}}\\ &= \dfrac{-\dfrac{3}{2}\overline{i}+2\overline{j}}{\left( \dfrac{5}{2}\right) } \end{align*} Directional derivative \begin{align*} D &=\left( \nabla f\right) \cdot \widehat{PQ}\\ &=\left(\overline{i}+2\overline{j}\right) \cdot \dfrac{-\dfrac{3}{2}\overline{i}+2\overline{j}}{\left( \dfrac{5}{2}\right) }\\ &=\dfrac{\dfrac{-3+8}{2}}{\dfrac{5}{2}}\\ &=1 \end{align*}

The magnitude of the gradient for the function $(𝑥, 𝑦, 𝑧) = 𝑥^2 + 3𝑦^2 + 𝑧^3$ at the point (1, 1, 1) is: (GATE EC 2014 Shift IV)
$\phi = 𝑥^2 + 3𝑦^2 + 𝑧^3$ $\nabla \phi = 2x \bar{i} + 6y \bar{j} + 3z^2 \bar{k}$ At (1,1,1), $\nabla \phi = 2 \bar{i} + 6 \bar{j} + 3 \bar{k}$ Magnitude can be calculated as: $|\nabla \phi| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{49} = 7$