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 : \( \vec{R} = x\hat{i} + y\hat{j} + z\hat{k} \)
where \(\hat{i},\hat{j}, \hat{k} \) are unit vectors of x-axis, y-axis and z-axis respectively
Magnitude : \(|\vec{R}| = r = \sqrt{x^2 + y^2 + z^2} \)
Example: \(\vec{A} = 2\hat{i} + 3\hat{j} - 4\hat{k} \)
\( |\vec{A}| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{29} \)
Direction : \( \hat{A} =\Large \frac{\vec{A}}{|\vec{A}|} \)
\(\hat{A} = \Large \frac{2\hat{i} + 3\hat{j} - 4\hat{k}}{\sqrt{29}} = \frac{2}{\sqrt{29}}\hat{i} + \frac{3}{\sqrt{29}}\hat{j} - \frac{4}{\sqrt{29}}\hat{k} \)
Now let's jump on Gradient.
Gradient of a Scalar
\(\nabla \) or \(\vec{\nabla} \longrightarrow \text{Del Operator/ Vector Differential Operator/ Vector Operator } \)
\[\large\boxed{\nabla = \hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} } \]
Gradient : If \(f\) be a scalar point function, then \(\nabla f \) is defined as gradient of \(f\) \[ \nabla f = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right)f \]
\[ \large\boxed{grad(f) = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j} + \frac{\partial f}{\partial z}\hat{k}} \]
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 \(\phi = 3x^2y - y^3z^2\) , find grad(\(\phi\)) 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 \[\phi = 3x^2y - y^3z^2 \] \[grad(\phi) = \nabla \phi = \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} + \hat{k}\frac{\partial}{\partial z} \right)(3x^2y - y^3z^2) \] \[ = \hat{i}\frac{\partial}{\partial x}(3x^2y - y^3z^2) + \hat{j}\frac{\partial}{\partial y}(3x^2y - y^3z^2) + \hat{k}\frac{\partial}{\partial z}(3x^2y - y^3z^2) \] \[ \Rightarrow grad{\phi} = \hat{i}(6xy) + \hat{j}(3x^2 - 3y^2z^2) + \hat{k}(-2y^3z) \] At point (1,-2,-1) \[ grad(\phi) = 6(1)(-2)\hat{i} + \left[ 3(1)^2 -3(-2)^2(-1)^2 \right]\hat{j} -2(-2)^3(-1)\hat{k} \] \[ grad(\phi) = -12\hat{i} - 9\hat{j} - 16\hat{k} \] It is the normal vector to the surface \(\phi\)
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 -
\[ x^3 + y^3 + 3xyz = 3 \]
\[ x^3 + y^3 + 3xyz - 3 = 0 \]
\[ \text{Let} \hspace{5pt} \phi = x^3 + y^3 + 3xyz - 3 \]
Normal vector = grad(\(\phi\))
\[= \left(\hat{i}\frac{\partial}{\partial x} + \hat{j}\frac{\partial}{\partial y} +
\hat{k}\frac{\partial}{\partial z} \right)\left(x^3 + y^3 + 3xyz - 3 \right) \]
\[\Rightarrow grad(\phi) = (3x^2 + 3yz)\hat{i} + (3y^2 + 3xz)\hat{j} + 3xy\hat{k} \]
At point (1,2,-1)
\[ \overrightarrow{N} = \left[3(1)^2 + 3(2)(-1) \right]\hat{i} + \left[3(2)^2 + 3(1)(-1) \right]\hat{j} + 3(1)(2)\hat{k} \]
\[ \vec{N} = -3\hat{i} + 9\hat{j} + 6\hat{k} \]
Now we will find its unit vector
\[ \hat{N} = \large\frac{\vec{N}}{|\vec{N}|} \]
\[ = \frac{-3\hat{i} + 9\hat{j} + 6\hat{k}}{\sqrt{(-3)^2 + (9)^2 + (6)^2}} \]
\[\Rightarrow \hat{N} = \large\frac{-3\hat{i} + 9\hat{j} + 6\hat{k}}{\sqrt{126}} \]
