Problem 14 - Calculus Early Transcendentals

Question

\(11-17\) Find the directional derivative of the function at the given point in the direction of the vector \(\mathbf{v}\) . $$g(r, s)=\tan ^{-1}(r s), \quad(1,2), \quad \mathbf{v}=5 \mathbf{i}+10 \mathbf{j}$$

Step-by-Step Solution

Answer

Directional derivative is \( \frac{4}{5\sqrt{5}} \).

Step 1: Compute the Gradient of g(r, s)

The gradient of the function \( g(r, s) = \tan^{-1}(r s) \) is the vector of partial derivatives. First, find the partial derivative with respect to \( r \):\[\frac{\partial g}{\partial r} = \frac{s}{1 + (r s)^2}\]Now, find the partial derivative with respect to \( s \):\[\frac{\partial g}{\partial s} = \frac{r}{1 + (r s)^2}\]Thus, the gradient is:\[abla g(r, s) = \left( \frac{s}{1 + (r s)^2}, \frac{r}{1 + (r s)^2} \right)\]

Step 2: Evaluate the Gradient at the Given Point

Evaluate the gradient at the point \((1, 2)\):\[abla g(1, 2) = \left( \frac{2}{1 + (1 \times 2)^2}, \frac{1}{1 + (1 \times 2)^2} \right)\]Simplify the components:\[abla g(1, 2) = \left( \frac{2}{1 + 4}, \frac{1}{1 + 4} \right) = \left( \frac{2}{5}, \frac{1}{5} \right)\]

Step 3: Normalize the Direction Vector

Normalize the direction vector \( \mathbf{v} = 5\mathbf{i} + 10\mathbf{j} \) by dividing it by its magnitude:\[\|\mathbf{v}\| = \sqrt{5^2 + 10^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5}\]Thus, the unit vector in the direction \( \mathbf{v} \) is:\[\widehat{\mathbf{v}} = \left( \frac{5}{5\sqrt{5}}, \frac{10}{5\sqrt{5}} \right) = \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right)\]

Step 4: Calculate the Directional Derivative

The directional derivative of \( g \) at \( (1, 2) \) in the direction of the vector \( \mathbf{v} \) is given by the dot product of the gradient at \( (1, 2) \) and the unit direction vector:\[D_{\mathbf{v}}g(1, 2) = abla g(1, 2) \cdot \widehat{\mathbf{v}} \]Substitute the values:\[D_{\mathbf{v}}g(1, 2) = \left( \frac{2}{5}, \frac{1}{5} \right) \cdot \left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right)\]Calculate the dot product:\[D_{\mathbf{v}}g(1, 2) = \frac{2}{5} \cdot \frac{1}{\sqrt{5}} + \frac{1}{5} \cdot \frac{2}{\sqrt{5}} = \frac{2}{5\sqrt{5}} + \frac{2}{5\sqrt{5}} = \frac{4}{5\sqrt{5}}\]Thus, the directional derivative is \( \frac{4}{5\sqrt{5}} \).

Key Concepts

Gradient

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

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

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.

Dot Product

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 \( \mathbf{v} \).