Problem 22 - Calculus Early Transcendentals

Question

Compute the directional derivative of the following functions at the given point P in the direction of the given vector. Be sure to use a unit vector for the direction vector. $$f(x, y)=13 e^{x y} ; P(1,0) ;\langle 5,12\rangle$$

Step-by-Step Solution

Answer

Answer: The directional derivative of the given function at the point P(1, 0) in the direction of the vector $$\langle 5, 12 \rangle$$ is 12.

Step 1: Compute the gradient of the function

First, we need to find the gradient of the function, which is given by: $$\nabla f(x,y) = \begin{bmatrix} \frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y} \end{bmatrix}$$ Compute the partial derivatives with respect to x and y: $$\frac{\partial f}{\partial x} = \frac{d}{dx} \left( 13 e^{xy} \right) = 13 y e^{xy}$$ $$\frac{\partial f}{\partial y} = \frac{d}{dy} \left( 13 e^{xy} \right) = 13 x e^{xy}$$ So the gradient is: $$\nabla f(x,y) = \begin{bmatrix} 13 y e^{xy}\\ 13 x e^{xy} \end{bmatrix}$$

Step 2: Calculate the unit vector for the given direction vector

Given the direction vector $$\langle 5, 12 \rangle$$, we need to compute the unit vector in that direction. This is done by dividing each component of the vector by its magnitude: $$\text{Magnitude} = \| \langle 5, 12 \rangle \| = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$$ The unit vector is then given by: $$\frac{1}{13} \begin{bmatrix} 5 \\ 12 \end{bmatrix} = \begin{bmatrix} \frac{5}{13} \\ \frac{12}{13} \end{bmatrix}$$

Step 3: Take the dot product of the gradient and the unit vector

Now, we will find the directional derivative by taking the dot product of the gradient and the unit vector: $$D_{\text{direction}} f(x,y) = \nabla f(x,y) \cdot \text{unit vector} = \left<13 y e^{xy}, 13 x e^{xy}\right> \cdot \left<\frac{5}{13}, \frac{12}{13}\right>$$ It simplifies to: $$D_{\text{direction}} f(x,y) = 5 y e^{xy} + 12 x e^{xy}$$

Step 4: Evaluate the directional derivative at the given point P

Finally, substitute the coordinates of the point P(1, 0) into the directional derivative expression: $$D_{\text{direction}} f(1,0) = 5(0) e^{(1)(0)} + 12(1) e^{(1)(0)} = 0 + 12(1) = 12$$ The directional derivative of the given function at the point P(1, 0) in the direction of the vector $$\langle 5, 12 \rangle$$ is 12.

Key Concepts

Gradient

The gradient of a function is a vector that consists of all its partial derivatives. It is denoted by \( abla f \) and represents the direction of the steepest ascent for a multi-variable function. Think of it as a compass that guides us towards the highest elevation on a hill.
For a function \( f(x, y) \), the gradient is expressed as:

• \( abla f(x, y) = \begin{bmatrix} \frac{\partial f}{\partial x} \ \frac{\partial f}{\partial y} \end{bmatrix} \)

This method helps us understand how the function changes with respect to small changes in \( x \) and \( y \).
In the given exercise, we compute the gradient of \( f(x, y) = 13 e^{xy} \) by finding its partial derivatives with respect to \( x \) and \( y \). These derivatives form the components of the gradient vector. This results in:

• \( \frac{\partial f}{\partial x} = 13 y e^{xy} \)
• \( \frac{\partial f}{\partial y} = 13 x e^{xy} \)

Unit Vector

A unit vector is simply a vector with a magnitude (length) of 1. It maintains the same direction as the original vector but is scaled to this standard length, acting as a direction pointer with uniform strength.
To convert a vector into a unit vector, each component of the vector must be divided by its magnitude. The magnitude of a vector \( \langle a, b \rangle \) is calculated using:

• \( \|\langle a, b \rangle\| = \sqrt{a^2 + b^2} \)

This approach ensures the vector is scaled correctly. For instance, the direction vector \( \langle 5, 12 \rangle \) in our example, after calculation, reveals its unit vector:

• Magnitude = 13
• Unit Vector = \( \frac{1}{13} \begin{bmatrix} 5 \ 12 \end{bmatrix} = \begin{bmatrix} \frac{5}{13} \ \frac{12}{13} \end{bmatrix} \)

This step is crucial because the directional derivative requires a unit vector to measure how the function behaves precisely in that direction.

Partial Derivatives

Partial derivatives describe how a function changes as one of the variables is altered while keeping the others constant. They are foundational in multivariable calculus when determining the rate of change of quantities with multiple inputs.
The process involves differentiating a function with respect to one variable, while all other variables are treated as constants.
In the task at hand, we needed to compute the partial derivatives of \( f(x, y) = 13 e^{xy} \). Here's how it operates step by step:

• For \( x \): \( \frac{\partial f}{\partial x} \), treat \( y \) as constant. The result is \( 13 y e^{xy} \).
• For \( y \): \( \frac{\partial f}{\partial y} \), treat \( x \) as constant. The outcome is \( 13 x e^{xy} \).

These partial derivatives jointly form the gradient vector, which is vital in calculating the directional derivative.

Dot Product

The dot product (also known as the scalar product) is a way to multiply two vectors, resulting in a scalar value. It is used to find the angle relationship between two vectors and is particularly helpful in physics and engineering contexts.
The dot product of two vectors \( \langle a, b \rangle \) and \( \langle c, d \rangle \) is given by:

• \( a \cdot c + b \cdot d \)

This operation results in a single number, not another vector.
In the context of our problem, the dot product was essential in determining the directional derivative. By taking the dot product of the gradient vector \( \langle 13 y e^{xy}, 13 x e^{xy} \rangle \) and the unit vector \( \langle \frac{5}{13}, \frac{12}{13} \rangle \), we quantified the rate of function change in the specified direction. The entire operation boiled down to a simple scalar which was calculated as \( 5 y e^{xy} + 12 x e^{xy} \), giving insight into how steeply the function increases in that specific line.