Gradient Engineering Maths, Btech first year

Gradient

Let's clear some basics first:

Scalar Point Function : $f(x,y,z)=5x+2y+10z$

Vector Point Function : $f(x,y,z)=x\hat{i}+y\hat{j}+z\hat{k}$

Position Vector : $\overrightarrow{R}=x\hat{i}+y\hat{j}+z\hat{k}$ where $\hat{i},\hat{j},\hat{k}$ are unit vectors of x-axis, y-axis and z-axis respectively

Magnitude : $|\overrightarrow{R}|=r=\sqrt{x^2+y^2+z^2}$

Example: $\overrightarrow{A}=2\hat{i}+3\hat{j}-4\hat{k}$
$|\overrightarrow{A}|=\sqrt{2^2+3^2+(-4{)}^2}=\sqrt{29}$

Direction : $\hat{A}=\frac{\overrightarrow{A}}{|\overrightarrow{A}|}$
$\hat{A}=\frac{2\hat{i}+3\hat{j}-4\hat{k}}{\sqrt{29}}=\frac{2}{\sqrt{29}}\hat{i}+\frac{3}{\sqrt{29}}\hat{j}-\frac{4}{\sqrt{29}}\hat{k}$

Now let's jump on Gradient.

Gradient of a Scalar

$\mathrm{\nabla }$ or $\overrightarrow{\mathrm{\nabla }}⟶\text{Del Operator/ Vector Differential Operator/ Vector Operator }$

$$\boxed{{\displaystyle \mathrm{\nabla }=\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z}}}$$

Gradient : If $f$ be a scalar point function, then $\mathrm{\nabla }f$ is defined as gradient of $f$ $$\mathrm{\nabla }f=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})f$$

$$\boxed{{\displaystyle grad(f)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}}}$$

Note : $f$ is a scalar, but its gradient is a vector.

Gometrical Interpretation of Gradient :

Gradient of a scalar point function represents the Normal Vector to that surface

Sample Problem :

If $\varphi =3x^{2}y-y^3{z}^2$ , find grad($\varphi$) at the point (1,-2,-1).

Solution :

To find gradient we simply partially differentiate the function w.r.t. $x,y,z$ and multiply it with the respective unit vectors i.e. $\hat{i},\hat{j},\hat{k}$ as you can observe from the formula $$\varphi =3x^{2}y-y^3{z}^2$$ $$grad(\varphi )=\mathrm{\nabla }\varphi =(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})(3x^{2}y-y^3{z}^2)$$ $$=\hat{i}\frac{\partial }{\partial x}(3x^{2}y-y^3{z}^2)+\hat{j}\frac{\partial }{\partial y}(3x^{2}y-y^3{z}^2)+\hat{k}\frac{\partial }{\partial z}(3x^{2}y-y^3{z}^2)$$ $$\Rightarrow grad\varphi =\hat{i}(6xy)+\hat{j}(3x^2-3y^2{z}^2)+\hat{k}(-2y^{3}z)$$ At point (1,-2,-1) $$grad(\varphi )=6(1)(-2)\hat{i}+[3(1{)}^2-3(-2{)}^2(-1{)}^2]\hat{j}-2(-2{)}^3(-1)\hat{k}$$ $$grad(\varphi )=-12\hat{i}-9\hat{j}-16\hat{k}$$ It is the normal vector to the surface $\varphi$

Sample Problem 2 :

Find a unit vector normal to the surface $x^3+y^3+3xyz=3$ at point (1,2,-1)

Solution

Here we have to find two things: (a)Normal vector(i.e. gradient) (b) unit vector of that normal at given point.
Equation of the given surface is - $$x^3+y^3+3xyz=3$$ $$x^3+y^3+3xyz-3=0$$ $$\text{Let}\varphi =x^3+y^3+3xyz-3$$ Normal vector = grad($\varphi$) $$=(\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\hat{k}\frac{\partial }{\partial z})(x^3+y^3+3xyz-3)$$ $$\Rightarrow grad(\varphi )=(3x^2+3yz)\hat{i}+(3y^2+3xz)\hat{j}+3xy\hat{k}$$ At point (1,2,-1) $$\overrightarrow{N}=[3(1{)}^2+3(2)(-1)]\hat{i}+[3(2{)}^2+3(1)(-1)]\hat{j}+3(1)(2)\hat{k}$$ $$\overrightarrow{N}=-3\hat{i}+9\hat{j}+6\hat{k}$$ Now we will find its unit vector $$\hat{N}=\frac{\overrightarrow{N}}{|\overrightarrow{N}|}$$ $$=\frac{-3\hat{i}+9\hat{j}+6\hat{k}}{\sqrt{(-3{)}^2+(9{)}^2+(6{)}^2}}$$ $$\Rightarrow \hat{N}=\frac{-3\hat{i}+9\hat{j}+6\hat{k}}{\sqrt{126}}$$