91Ó°ÊÓ

In calculus, the gradient is a vector that contains the partial derivatives of a function. It points in the direction of the greatest rate of increase for the function. For a function of two variables, like our function \[ g(r, s) = \tan^{-1}(rs) \]its gradient is denoted by \(abla g\) and is calculated as:

• The partial derivative with respect to \( r \): \( \frac{\partial g}{\partial r} \), which tells us how changes in \( r \) affect \( g \).
• The partial derivative with respect to \( s \): \( \frac{\partial g}{\partial s} \), which tells us how changes in \( s \) affect \( g \).

Once calculated, the gradient vector is formed as \( (\frac{s}{1 + (rs)^2}, \frac{r}{1 + (rs)^2}) \).It gives a clear picture of the function's behavior, which is essential for finding the directional derivative.

Partial derivatives represent the rate at which a function changes as one of the variables changes, while the other variable(s) remain constant. To find the partial derivatives of our function, \( g(r, s) = \tan^{-1}(rs) \), we derive it with respect to each variable separately:

• The partial derivative with respect to \( r \), \( \frac{\partial g}{\partial r} = \frac{s}{1 + (rs)^2} \), measures the rate of change of \( g \) as \( r \) changes and \( s \) stays the same.
• The partial derivative with respect to \( s \), \( \frac{\partial g}{\partial s} = \frac{r}{1 + (rs)^2} \), measures the rate of change of \( g \) as \( s \) changes and \( r \) stays the same.

These derivatives form the components of the gradient vector and are crucial in applications where we explore the function's sensitivity to changes in specific directions.

Normalization is a key step in calculating the directional derivative. To normalize a vector means to convert it into a unit vector. A unit vector has a length (or magnitude) of 1 and points in the same direction as the original vector. For the vector \( \mathbf{v} = 5 \mathbf{i} + 10 \mathbf{j} \), normalization involves:1. Calculating its magnitude using the formula: \[ \| \mathbf{v} \| = \sqrt{5^2 + 10^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5} \] 2. Dividing each component of \( \mathbf{v} \) by its magnitude to get the unit vector: \[ \widehat{\mathbf{v}} = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \]This unit vector is then used in the dot product with the gradient to find the directional derivative.

The dot product is a way to multiply two vectors that results in a scalar (a single number). It is essential in finding the directional derivative, as it measures the projection of one vector onto another.For two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), the dot product is calculated as:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2\]In our context, we calculate the directional derivative of the function \( g \) at the point \( (1, 2) \) in the direction of \( \mathbf{v} \) by taking the dot product of the gradient at that point with the normalized direction vector. The calculation:\[D_{\mathbf{v}}g(1, 2) = \left( \frac{2}{5}, \frac{1}{5} \right) \cdot \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) = \frac{4}{5\sqrt{5}}\]This result, \( \frac{4}{5\sqrt{5}} \), represents the rate of change of the function in the direction specified by \( \mathbf{v} \).