Dated: 30-10-2024

The Triple Scalar or Box Product

The following is called the triple scalar product

\[(\vec{A} \times \vec{B}) \cdot \vec{C} = \lvert \vec{A} \times \vec{B} \rvert \lvert \vec{C} \rvert \lvert \cos(\theta) \rvert\]

\[= \left| \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right| \]

Gradient of a Scalar `Function`¹

\[\nabla = \hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}\]

The \(\nabla\) is called the del operator.
It is also called gradient and operates on a function.¹
It returns us a vector field which is a set² of vectors³ pointing towards the steepest change in the function.¹

Directional Derivative

If \(f(x, y)\) is differentiable at \((x_0, y_0)\), and if \(\vec{u} = (u_1, u_2)\) is a unit vector,³ then the directional derivative of \(f\) at \((x_0, y_0)\) in the direction of \(\vec{u}\) is defined by

\[D_{\vec{u}} f(x_0, y_0) = f_x(x_0, y_0) u_1 + f_y(x_0, y_0) u_2\]

NOTE: There are infinite choices of \(\vec{u}\) at the point \((x_0, y_0)\).

It can also be written as

\[D_{\vec{u}} f(x, y) = \nabla f(x, y) \cdot \hat{u}\]

Gradient of a `Function`¹

If \(f(x, y)\) is a function¹ then the gradient is defined as

\[\nabla f(x, y) = f_x(x, y) \hat{i} + f_y(x, y) \hat{j}\]

Properties of Gradient

\[D_{\vec{u}} f = \nabla f \cdot \hat{u} = \lvert \nabla f \rvert \cos (\theta)\]

\(f\) increases most rapidly in the direction of \(\nabla f\), which means \(\cos(\theta) = 1\) suggesting that direction given by \(\vec{u}\) is in the same direction of \(\nabla f\).

\[D_{\vec{u}} f = \lvert \nabla f \rvert \cos (0) = \lvert \nabla f \rvert\]

It decreases most rapidly in direction of \(- \nabla f\)

\[D_{\vec{u}} f = \lvert \nabla f \rvert \cos (\pi) = - \lvert \nabla f \rvert\]

There is no change if we move in direction perpendicular to \(\nabla f\).

\[D_{\vec{u}} f = \lvert \nabla f \rvert \cos \left(\frac{\pi}{2}\right) = \lvert \nabla f \rvert \cdot 0 = 0\]

Example

Imagine we have a function¹

\[f(x, y) = x^2 + y^2\]

We are interested to know that at \((1, 1)\), in which directions does this function¹ most rapidly

increases
decreases

\[\frac{\partial}{\partial x} f(x, y) = f_x(x, y) = 2x\]

\[\frac{\partial}{\partial y} f(x, y) = f_y(x, y) = 2y\]

\[\nabla f(x, y) = 2x \hat i + 2y \hat j\]

\[\because \nabla f(x, y) \cdot \hat u = \lvert \nabla f(x, y) \rvert \cos(\theta)\]

Also

\[\because \lvert \nabla f(x, y) \rvert = 2 \sqrt{x^2 + y^2}\]

\[\therefore \lvert \nabla f(x, y) \rvert_{(1, 1)} = 2 \sqrt{1^2 + 1^2} = 2 \sqrt 2\]

\[\nabla f(x, y)_{(1, 1)} = 2 \hat i + 2 \hat j\]

Increase

We get maximum increase in the direction of the gradient. Therefore, we need to find the unit vector³ for gradient.

\[\hat u = \frac{2 \hat i + 2 \hat j}{2 \sqrt 2} = \frac{i + j}{\sqrt 2}\]

Decrease

The decrease would be in opposite direction. Therefore,

\[- \frac{i + j}{\sqrt 2}\]