Problem 27 - Calculus for Biology and Medicine

Question

Compute the directional derivative of \(f(x, y)\) at the given point in the indicated direction. $$ f(x, y)=e^{x+y} \text { at }(0,0) \text { in the direction }\left[\begin{array}{l} -1 \\ -1 \end{array}\right] $$

Step-by-Step Solution

Answer

The directional derivative is \(-\sqrt{2}\).

Step 1: Understand the Problem

We need to find the directional derivative of the function \( f(x, y) = e^{x+y} \) at the point \((0, 0)\) in the direction of the vector \( \begin{bmatrix} -1 \ -1 \end{bmatrix} \).

Step 2: Compute the Gradient

The gradient of the function \( f(x, y) = e^{x+y} \) is given by the vector of partial derivatives: \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Compute the partial derivatives: \[ \frac{\partial f}{\partial x} = e^{x+y}, \quad \frac{\partial f}{\partial y} = e^{x+y} \]. Thus, the gradient is \( abla f(x, y) = \begin{bmatrix} e^{x+y} \ e^{x+y} \end{bmatrix}. \) At the point \((0, 0)\), the gradient evaluates to \( abla f(0, 0) = \begin{bmatrix} e^{0+0} \ e^{0+0} \end{bmatrix} = \begin{bmatrix} 1 \ 1 \end{bmatrix}. \)

Step 3: Normalize the Direction Vector

The direction vector is \( \mathbf{v} = \begin{bmatrix} -1 \ -1 \end{bmatrix} \). We need to normalize it to obtain a unit vector in the direction we want: \[ \mathbf{u} = \frac{\mathbf{v}}{\| \mathbf{v} \|} = \frac{1}{\sqrt{(-1)^2 + (-1)^2}} \begin{bmatrix} -1 \ -1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} -1 \ -1 \end{bmatrix}. \]

Step 4: Calculate the Directional Derivative

The directional derivative \( D_{\mathbf{u}} f(0,0) \) is given by the dot product of the gradient vector and the unit direction vector:\[ D_{\mathbf{u}} f(0,0) = abla f(0,0) \cdot \mathbf{u} = \begin{bmatrix} 1 \ 1 \end{bmatrix} \cdot \frac{1}{\sqrt{2}} \begin{bmatrix} -1 \ -1 \end{bmatrix} = \frac{1}{\sqrt{2}}(1 \times -1 + 1 \times -1) = \frac{-2}{\sqrt{2}} = -\sqrt{2}. \]

Key Concepts

Gradient

The gradient is like a magical tool for understanding how a function changes at different points in space. For a function of two variables like \( f(x, y) \), the gradient \( abla f(x, y) \) is a vector that points in the direction of the greatest rate of increase of the function. This vector is constructed using the partial derivatives of the function with respect to each variable.

In simpler terms, the gradient tells us how the function slopes differently when moving in different directions. For the function \( f(x, y) = e^{x+y} \), the gradient is given by:

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

By evaluating the gradient at a specific point, such as \((0, 0)\), we obtain \( abla f(0, 0) = \begin{bmatrix} 1 & 1 \end{bmatrix} \). This result means that at the point \((0, 0)\), the function increases equally in the direction of both the x and y axes.

Partial Derivatives

Partial derivatives are the building blocks of the gradient. They measure how a function changes as we tweak just one variable, keeping others constant. For a function \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} \) shows how the function changes with small changes in \( x \) (while \( y \) stays fixed). Similarly, \( \frac{\partial f}{\partial y} \) shows changes with respect to \( y \).Considering the function \( f(x, y) = e^{x+y} \):

• The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = e^{x+y} \).
• The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = e^{x+y} \).

Both partial derivatives are equal, which indicates that changes in \( x \) or \( y \) contribute equally to changes in the function value. Hence, these derivatives provide the components of the gradient vector.

Unit Vector

A unit vector is a vector that has a length of one, which makes it super useful for indicating direction without worrying about magnitude. When finding a directional derivative, it’s important to use a unit vector to properly scale the direction.

In our problem, the direction vector is \( \mathbf{v} = \begin{bmatrix} -1 & -1 \end{bmatrix} \). This vector needs to be turned into a unit vector by dividing it by its magnitude (or length). The magnitude \( \| \mathbf{v} \| \) is calculated as:

• \( \| \mathbf{v} \| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \)

The unit vector \( \mathbf{u} \) is thus:

• \( \mathbf{u} = \frac{1}{\sqrt{2}} \begin{bmatrix} -1 & -1 \end{bmatrix} \)

This unit vector tells us that the desired direction stretches equally in both the negative x-axis and y-axis. When we compute the directional derivative, this unit vector ensures we consider only the change in direction.