Problem 12 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 14, a function \(f\), a point \(P\), and a unit vector \(\mathbf{U}\) are given. Find (a) the gradient of \(f\) at \(P\), and (b) the rate of change of the function value in the direction of \(\mathrm{U}\) at \(P\). \(f(x, y)=e^{2 x y} ; P=(2,1) ; \mathbf{U}=\frac{4}{5} i-\frac{3}{5} \mathbf{j}\)

Step-by-Step Solution

Answer

The gradient at P is \((2 e^4, 4 e^4)\). The rate of change in the direction of \(\textbf{U}\) is \(-\frac{4 e^4}{5}\).

Step 1 - Find the partial derivatives

Calculate the partial derivatives of the function with respect to both variables. \(\frac{\text{∂}f}{\text{∂}x}\) and \(\frac{\text{∂}f}{\text{∂}y}\).oindent \(\frac{\text{∂}f}{\text{∂}x} = \frac{\text{∂}}{\text{∂}x}(e^{2xy}) = 2ye^{2xy}\)oindent \(\frac{\text{∂}f}{\text{∂}y} = \frac{\text{∂}}{\text{∂}y}(e^{2xy}) = 2xe^{2xy}\)

Step 2 - Compute the gradient at point P

Substitute the coordinates of point P (2,1) into the partial derivatives to find the gradient at point P.oindent \(abla f(P) = abla f(2,1) = abla f(x, y)|_{(x, y) = (2, 1)} = (2y e^{2xy}, 2x e^{2xy})|_{(2,1)} \ = (2(1) e^{2(2)(1)}, 2(2) e^{2(2)(1)}) = (2 e^4, 4 e^4)\)

Step 3 - Normalize the vector U

Given the unit vector: \(mathbf{U} = \frac{4}{5} i - \frac{3}{5} j \), confirm its length is 1.oindent The magnitude of \(\textbf{U}\) is \(\frac{4}{5}^2 + (- \frac{3}{5})^2 = \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1\).Since the magnitude is 1, \(\textbf{U}\) is already a unit vector.

Step 4 - Find the directional derivative

Calculate the rate of change of the function in the direction of \(\textbf{U}\) using the dot product of the gradient and \(\textbf{U}\).oindent \(abla f(P) \bullet \textbf{U} = (2 e^4, 4 e^4) \bullet \bigg(\frac{4}{5}, -\frac{3}{5}\bigg) = 2 e^4 \bigg(\frac{4}{5}\bigg) + 4 e^4 \bigg(-\frac{3}{5}\bigg) \ = \frac{8 e^4}{5} - \frac{12 e^4}{5} = -\frac{4 e^4}{5}\)

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They represent the rate of change of a function with respect to one variable, while keeping the other variables constant. In the given exercise, you need to find the partial derivatives of the function \( f(x, y) = e^{2xy} \). To do this, you differentiate \( f \) with respect to each variable separately. For example, the partial derivative \( \frac{\partial f}{\partial x} \) measures how \( f \) changes as \( x \) changes, while \( y \) is kept constant. Similarly, \( \frac{\partial f}{\partial y} \) measures the rate of change of \( f \) with respect to \( y \). Once you have the partial derivatives, you can evaluate them at the given point \( P \) to assist in finding the gradient vector.

Gradient Vector

The gradient vector is a vector that consists of all the partial derivatives of a function. For the function \( f(x, y) = e^{2xy} \), its gradient vector is denoted as \( abla f \) and is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). The gradient vector points in the direction of the greatest rate of increase of the function. In our exercise, the gradient at point \( P = (2, 1) \) needs to be calculated by substituting the coordinates of \( P \) into the partial derivatives. This gives the gradient vector at that specific point, which helps in determining the rate of change in any given direction.

Directional Derivative

The directional derivative of a function at a given point in a specified direction represents the rate of change of the function in that direction. To compute the directional derivative, you use the gradient vector and the direction vector \( \textbf{U} \). The formula for the directional derivative \( D_{\textbf{U}} f \) is \( abla f \cdot \textbf{U} \), where \( \cdot \) denotes the dot product. The exercise requires computing this derivative, which involves taking the dot product of the gradient vector at point \( P = (2, 1) \) and the unit vector \( \textbf{U} = \frac{4}{5} i - \frac{3}{5} j \). This calculation gives the rate of change of the function in the direction of \( \textbf{U} \).

Unit Vector

A unit vector is a vector of length 1 that indicates direction. In this exercise, \( \textbf{U} = \frac{4}{5} i - \frac{3}{5} j \) is provided as a unit vector. You must confirm that its length is 1. This is done by computing the magnitude: \[ |\textbf{U}| = \sqrt{\left( \frac{4}{5} \right)^2 + \left( - \frac{3}{5} \right)^2} = \sqrt{ \frac{16}{25} + \frac{9}{25} } = 1 \]. Since the magnitude is 1, \( \textbf{U} \) is indeed a unit vector. Unit vectors are essential because they simplify the computation of directional derivatives, ensuring the direction component is standardized to a length of 1.