Problem 17 - The Calculus with Analytic Geometry

Question

In Exercises 17 through 20 , find \(D_{\mathbf{u}} f\) at the given point \(P\) for which \(\mathbf{U}\) is a unit vector in the direction of \(\overrightarrow{P Q}\). Also at \(P\) find \(D_{\mathrm{u}} f\), if \(\mathbf{U}\) is a unit vector for which \(D_{\mathbf{U}} f\) is a maximum. \(f(x, y)=e^{x} \tan ^{-1} y ; P(0,1), Q(3,5)\)

Step-by-Step Solution

Answer

Find the vector \(\

Step 1: Calculate the vector \(\overrightarrow{PQ}\)

Subtract the coordinates of point \(P\) from the coordinates of point \(Q\): \(\overrightarrow{PQ} = (3 - 0, 5 - 1) = (3, 4)\).

Step 2: Normalize \(\overrightarrow{PQ}\)

Find the magnitude of \(\overrightarrow{PQ}\): \| \overrightarrow{PQ} \| = \sqrt{3^2 + 4^2} = 5\). Then, the unit vector \(\mathbf{U}\) in the direction of \(\overrightarrow{PQ}\) is \(\

Key Concepts

Gradient Vector

The gradient vector of a function helps you understand how that function changes as you move through space. For a function of two variables, like they have here with \(f(x, y) = e^{x} \tan^{-1} y\), the gradient vector, denoted as \(abla f\), is a vector that consists of the partial derivatives of the function with respect to each variable.

In other words, if you want to find the gradient vector of \(f(x, y)\):

• Take the partial derivative of \(f\) with respect to \(x\): \(\frac{\partial f}{\partial x}\)
• Take the partial derivative of \(f\) with respect to \(y\): \(\frac{\partial f}{\partial y}\)

So, if we find these partial derivatives for our function:

\(\frac{\partial f}{\partial x} = e^x \tan^{-1} y\)
\(\frac{\partial f}{\partial y} = \frac{e^x}{1+y^2}\)

Thus, the gradient vector at any point \((x, y)\) is:

\(abla f = \left( e^{x} \tan^{-1} y, \frac{e^{x}}{1+y^2} \right)\)

This vector points in the direction of the steepest ascent of the function at any particular point.

Unit Vector

A unit vector is a vector with a magnitude of 1, which means it only indicates direction, not length. Unit vectors are very useful when we want to describe direction without worrying about the length.

In the given exercise, we need to find a unit vector in the direction of the vector \(\overrightarrow{PQ}\), which we found to be \((3, 4)\).

• First, we must determine the magnitude (or length) of \(\overrightarrow{PQ}\).
• The magnitude of \(\overrightarrow{PQ}\) is \(\|\overrightarrow{PQ}\| = \sqrt{3^2 + 4^2} = 5\).

To create a unit vector, we divide each component of \(\overrightarrow{PQ}\) by its magnitude:

• Unit vector \(\mathbf{U} = \left(\frac{3}{5}, \frac{4}{5}\right)\)

This unit vector \(\mathbf{U}\) in the direction of \(\overrightarrow{PQ}\) will be used to calculate the directional derivative in the given direction.

Exponential Functions

Exponential functions are functions in which the variable appears in the exponent. The general form is \(f(x) = a^x\), where \(a\) is a constant. In our exercise, the function \(f(x, y) = e^x \tan^{-1} y\) includes an exponential function part \(e^x\).

The key properties of exponential functions include:

• They grow rapidly as the variable increases.
• Their derivatives are proportional to the functions themselves.

When we take derivatives involving exponential functions, we need to remember that the derivative of \(e^x\) is simply \(e^x\) again. This property significantly simplifies calculations involving exponential functions and is crucial when finding the gradient vectors like we did in the previous section.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \tan^{-1} y \), are functions that 'reverse' the tangent operation. These functions take a ratio and return an angle.

For the function \(f(x, y) = e^x \tan^{-1} y\), you need to be familiar with the derivative of \( \tan^{-1} y\), which is \( \frac{1}{1+y^2} \).

This is essential when calculating the gradient vector, especially the partial derivative with respect to y. Remember:

• For any function \(u(y)\), if \(g(y) = \tan^{-1} (u(y))\), then \(g'(y) = \frac{u'(y)}{1+u(y)^2}\).

In our specific case, \( \frac{\partial f}{\partial y} \) uses this derivative property directly. Understanding these inverse trigonometric functions and their derivatives is important for solving many calculus problems involving angles and ratios.