Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y)=e^{-x y} ; \mathbf{p}=(1,-1) ; \mathbf{a}=-\mathbf{i}+\sqrt{3} \mathbf{j}\)

Step-by-Step Solution

Answer

The directional derivative is \(-\frac{e}{2}(1+\sqrt{3})\).

Step 1: Find the Gradient of the Function

The gradient of a function \( f(x, y) \) is given by the vector of partial derivatives: \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For our function \( f(x, y) = e^{-xy} \), we find: \[ \frac{\partial f}{\partial x} = -y e^{-xy} \] \[ \frac{\partial f}{\partial y} = -x e^{-xy} \] Thus, the gradient is \( abla f(x, y) = (-y e^{-xy}, -x e^{-xy}) \).

Step 2: Evaluate the Gradient at the Point \(\mathbf{p}\)

Substitute the point \( \mathbf{p} = (1, -1) \) into the gradient equation: \[ abla f(1, -1) = (-(-1) e^{-(1)(-1)}, -(1) e^{-(1)(-1)}) = (e, -e) \] So, the gradient at point \( (1, -1) \) is \( (e, -e) \).

Step 3: Normalize the Direction Vector

The given direction vector is \( \mathbf{a} = -\mathbf{i} + \sqrt{3} \mathbf{j} \) or \( \mathbf{a} = (-1, \sqrt{3}) \). First, we find the magnitude of \( \mathbf{a} \): \[ \| \mathbf{a} \| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \] The unit vector in the direction of \( \mathbf{a} \) is: \[ \mathbf{a}_u = \left( \frac{-1}{2}, \frac{\sqrt{3}}{2} \right) \]

Step 4: Calculate the Directional Derivative

The directional derivative of \( f \) at \( \mathbf{p} \) in the direction of \( \mathbf{a} \) is given by the dot product of the gradient \( abla f(1, -1) \) and the unit vector \( \mathbf{a}_u \): \[ D_{\mathbf{a}}f = abla f(1, -1) \cdot \mathbf{a}_u = (e, -e) \cdot \left( \frac{-1}{2}, \frac{\sqrt{3}}{2} \right) \] \[ D_{\mathbf{a}}f = e \cdot \frac{-1}{2} + (-e) \cdot \frac{\sqrt{3}}{2} = -\frac{e}{2} - \frac{\sqrt{3}e}{2} = -\frac{e}{2}(1 + \sqrt{3}) \]

Step 5: State the Result

The directional derivative of \( f(x, y) = e^{-xy} \) at the point \( \mathbf{p} = (1, -1) \) in the direction of \( \mathbf{a} = -\mathbf{i} + \sqrt{3}\mathbf{j} \) is \( -\frac{e}{2}(1 + \sqrt{3}) \).

Key Concepts

Gradient

A gradient is a fundamental concept in multivariable calculus. It provides important information about the rate and direction of change in a function. For a function \(f(x, y)\), the gradient is represented by a vector. This vector consists of the function's partial derivatives with respect to each variable.

• Finding gradients will help you understand how a function changes at any given point.
• Gradients point in the direction of the steepest ascent of a function.

In our example, \(f(x, y) = e^{-xy}\), the gradient \(abla f(x, y)\) is calculated as \((-y e^{-xy}, -x e^{-xy})\). This tells us how \(f\) changes with \(x\) and \(y\). Evaluating at point \((1, -1)\), we find \(abla f(1, -1) = (e, -e)\), indicating the immediate directions of greatest increase or decrease of the function at that point.

Partial Derivatives

Partial derivatives reveal how a function changes as one of the input variables changes, while the others are held constant. In multivariable calculus, these derivatives extend the concept of a derivative to functions of several variables.

• Allow you to focus on how change in one direction impacts the function.
• Useful in obtaining the gradient vector.

For the function \(f(x, y) = e^{-xy}\), its partial derivative with respect to \(x\) is \(-y e^{-xy}\) and with respect to \(y\) is \(-x e^{-xy}\). These partial derivatives are components of the gradient and individually show how the function \(f\) changes as \(x\) and \(y\) independently vary.

Unit Vector

A unit vector is a vector with a magnitude of 1. It’s often used to specify a direction without contributing to changes in magnitude. To find a unit vector in the direction of any vector \(\mathbf{a}\), you need to divide \(\mathbf{a}\) by its magnitude.

• The unit vector keeps the direction of \(\mathbf{a}\) but with a standardized length.
• Used to analyze the direction of the function's rate of change.

In the exercise, our direction vector \(\mathbf{a} = (-1, \sqrt{3})\) has a magnitude of 2. The unit vector is then calculated as \(\mathbf{a}_u = \left( \frac{-1}{2}, \frac{\sqrt{3}}{2} \right)\), maintaining the original direction with normalized length.

Dot Product

The dot product is a mathematical operation between two vectors that returns a scalar. It is also known as the scalar product and is essential in finding a directional derivative.

• Calculates how much one vector goes in the direction of another.
• Combines magnitude and direction information of the vectors involved.

For vectors \(\mathbf{u} = (u_1, u_2)\) and \(\mathbf{v} = (v_1, v_2)\), the dot product is given as \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2\). In our directional derivative calculation \[D_{\mathbf{a}}f = abla f(1, -1) \cdot \mathbf{a}_u = (e, -e) \cdot \left( \frac{-1}{2}, \frac{\sqrt{3}}{2} \right)\]resulting in \(-\frac{e}{2}(1 + \sqrt{3})\), indicating the rate of change of the function in the desired direction.