Problem 10 - Calculus for Biology and Medicine

Question

In Problems 9-14, compute the directional derivative of \(f(x, y)\) at the given point in the indicated direction. $$ f(x, y)=\sin (x+y) \text { at }(-1,0) \text { in the direction }\left[\begin{array}{r} 2 \\ -1 \end{array}\right] $$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{\cos(1)}{\sqrt{5}} \).

Step 1: Understand the directional derivative

The directional derivative of a function \( f(x, y) \) at a point \( (x_0, y_0) \) in the direction of a vector \( \mathbf{u} = \begin{bmatrix} a \ b \end{bmatrix} \) is defined as \( D_\mathbf{u} f = abla f(x_0, y_0) \cdot \mathbf{u} \), where \( abla f \) is the gradient of the function and \( \mathbf{u} \) is a unit vector.

Step 2: Compute the gradient of the function

To find the directional derivative, first compute the gradient \( abla f(x, y) \). For the function \( f(x, y) = \sin(x+y) \), the gradient is \( abla f = \begin{bmatrix} \frac{\partial f}{\partial x} \ \frac{\partial f}{\partial y} \end{bmatrix} \). Calculate the partial derivatives: \( \frac{\partial f}{\partial x} = \cos(x+y) \) and \( \frac{\partial f}{\partial y} = \cos(x+y) \). So, \( abla f = \begin{bmatrix} \cos(x+y) \ \cos(x+y) \end{bmatrix} \).

Step 3: Evaluate the gradient at the given point

Evaluate \( abla f(x, y) \) at \( (x, y) = (-1, 0) \). Since \( abla f = \begin{bmatrix} \cos(x+y) \ \cos(x+y) \end{bmatrix} \), substitute \( x = -1 \) and \( y = 0 \) to get \( abla f(-1, 0) = \begin{bmatrix} \cos(-1+0) \ \cos(-1+0) \end{bmatrix} = \begin{bmatrix} \cos(-1) \ \cos(-1) \end{bmatrix} \).

Step 4: Find the unit vector in the given direction

The direction is given by \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \). Convert this into a unit vector by dividing by its magnitude. The magnitude \( ||\mathbf{v}|| = \sqrt{2^2 + (-1)^2} = \sqrt{5} \). Thus, the unit vector \( \mathbf{u} = \frac{1}{\sqrt{5}} \begin{bmatrix} 2 \ -1 \end{bmatrix} \).

Step 5: Calculate the directional derivative

Using the formula \( D_\mathbf{u} f = abla f(x_0, y_0) \cdot \mathbf{u} \), plug in the values: \( abla f(-1, 0) = \begin{bmatrix} \cos(-1) \ \cos(-1) \end{bmatrix} \) and \( \mathbf{u} = \frac{1}{\sqrt{5}} \begin{bmatrix} 2 \ -1 \end{bmatrix} \). The dot product is: \[ D_\mathbf{u} f = \begin{bmatrix} \cos(-1) \ \cos(-1) \end{bmatrix} \cdot \frac{1}{\sqrt{5}} \begin{bmatrix} 2 \ -1 \end{bmatrix} = \frac{1}{\sqrt{5}}(2\cos(-1) - \cos(-1)) = \frac{1}{\sqrt{5}}\cos(-1). \]

Step 6: Final Calculation

Simplify the expression: \( \frac{\cos(-1)}{\sqrt{5}} \). Since \( \cos \) is an even function, \( \cos(-1) = \cos(1) \). Therefore, \[ D_\mathbf{u} f = \frac{\cos(1)}{\sqrt{5}} \].

Key Concepts

Gradient

To understand the concept of a gradient, imagine you're hiking up a hill. It's useful to know which direction is the steepest, so you can either climb quickly or descend safely. The steepest direction at any point is provided by the gradient. In mathematics, especially in multivariable calculus, the gradient is a vector that shows the direction and rate of steepest ascent of a function. It's denoted by \( abla f \) for a function \( f(x, y) \), which is really just a fancy way of saying the slope in different directions.

For our function \( f(x, y) = \sin(x+y) \), the gradient \( abla f \) at any point \( (x, y) \) is found by calculating the partial derivatives with respect to each variable. This gives us:

• \( \frac{\partial f}{\partial x} = \cos(x+y) \)
• \( \frac{\partial f}{\partial y} = \cos(x+y) \)

So, the gradient vector \( abla f \) becomes \( \begin{bmatrix} \cos(x+y) \ \cos(x+y) \end{bmatrix}. \) This tells us the direction and rate of change of \( f \) in the \( x \) and \( y \) directions.

Partial Derivatives

Partial derivatives are like finding changes in a specific path while considering multiple factors. When dealing with functions of several variables, such as \( f(x, y) \), it's crucial to examine how much the function changes as each variable individually changes, keeping the others constant.

To compute a partial derivative, you differentiate the function with respect to one variable while treating others as constants. In the given problem, the function is \( f(x, y) = \sin(x+y) \). We need to find:

• The partial derivative with respect to \( x: \frac{\partial f}{\partial x} = \cos(x+y) \)
• The partial derivative with respect to \( y: \frac{\partial f}{\partial y} = \cos(x+y) \)

These derivatives provide insights into how the function's value changes as you tweak the \( x \) or \( y \) values, separately. Hence, they're building blocks for creating the gradient vector, aiding in understanding the behavior of the function around any given point.

Unit Vector

The unit vector is like the navigator's compass in a sea of mathematics, providing a way to determine direction without concerning ourselves with magnitude. A unit vector is a vector with a magnitude of 1, pointing in a particular direction. This concept simplifies calculations in multivariable calculus, particularly for directional derivatives.

The directional derivative seeks the rate of change of \( f(x, y) \) in a specified direction. To ensure we only focus on direction and not distance, we convert any given direction vector, say \( \begin{bmatrix} 2 & -1 \end{bmatrix} \), into its corresponding unit vector. This is done by dividing it by its magnitude.

• Calculate the magnitude: \( ||\mathbf{v}|| = \sqrt{2^2 + (-1)^2} = \sqrt{5} \)
• Form the unit vector \( \mathbf{u} = \frac{1}{\sqrt{5}} \begin{bmatrix} 2 & -1 \end{bmatrix} \)

This process ensures that our calculations reflect just the direction of interest, allowing for a proper examination of how the function changes in the specified direction from the given point.