Problem 29 - Calculus for Biology and Medicine

Question

Compute the directional derivative of \(f(x, y)\) at the given point in the indicated direction. $$ f(x, y)=2 x y^{3}-3 x^{2} y \text { at }(1,-1) \text { in the direction }\left[\begin{array}{l} 3 \\ 1 \end{array}\right] $$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{15}{\sqrt{10}} \).

Step 1: Find the Gradient of f

First, we need to find the gradient \( abla f \) of the function \( f(x, y) = 2xy^3 - 3x^2y \). The gradient is a vector of the partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Compute these partial derivatives as follows: 1. \( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(2xy^3 - 3x^2y) = 2y^3 - 6xy \). 2. \( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(2xy^3 - 3x^2y) = 6xy^2 - 3x^2 \). Thus, the gradient is \( abla f = (2y^3 - 6xy, 6xy^2 - 3x^2) \).

Step 2: Evaluate the Gradient at the Point

Next, substitute the point \((1, -1)\) into the gradient we found: \( abla f(1, -1) = (2(-1)^3 - 6 \cdot 1 \cdot (-1), 6 \cdot 1 \cdot (-1)^2 - 3 \cdot 1^2) = (-2 + 6, 6 - 3) = (4, 3) \). This gives us the gradient at the point \((1, -1)\): \( abla f(1, -1) = (4, 3) \).

Step 3: Normalize the Given Direction Vector

The direction vector given is \( \mathbf{v} = \begin{bmatrix} 3 \ 1 \end{bmatrix} \). We need to normalize it to make it a unit vector. The magnitude of \( \mathbf{v} \) is \( \| \mathbf{v} \| = \sqrt{3^2 + 1^2} = \sqrt{10} \). Thus, the unit direction vector is \( \mathbf{u} = \frac{1}{\sqrt{10}} \begin{bmatrix} 3 \ 1 \end{bmatrix} \).

Step 4: Compute the Directional Derivative

The directional derivative of \( f \) at the point \((1, -1)\) in the direction of \( \mathbf{u} \) is calculated using the dot product \( D_{\mathbf{u}} f(x, y) = abla f \cdot \mathbf{u} \). Substituting the evaluated values gives: \( (4, 3) \cdot \frac{1}{\sqrt{10}} \begin{bmatrix} 3 \ 1 \end{bmatrix} = \frac{1}{\sqrt{10}} (4 \times 3 + 3 \times 1) = \frac{1}{\sqrt{10}} (12 + 3) = \frac{15}{\sqrt{10}} \). Therefore, the directional derivative is \( \frac{15}{\sqrt{10}} \).

Key Concepts

Gradient

In the context of multivariable calculus, the gradient plays a crucial role in understanding how a function behaves. It is a vector that essentially gives us the direction of the steepest ascent of a function at a particular point. The gradient of a function of two variables, such as \( f(x, y) = 2xy^3 - 3x^2y \), is found by computing the partial derivatives with respect to each variable.

• The symbol for the gradient is \( abla f \), and in two dimensions, it is represented as \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
• For our function, the partial derivative with respect to \( x \) is \( 2y^3 - 6xy \).
• The partial derivative with respect to \( y \) is \( 6xy^2 - 3x^2 \).

By combining these partial derivatives, we form the gradient vector \( abla f = (2y^3 - 6xy, 6xy^2 - 3x^2) \). This vector indicates how changes in \( x \) and \( y \) influence changes in the function \( f \). At a specific point, this gradient helps us predict the direction and rate of increase of the function.

Partial Derivatives

Partial derivatives are a fundamental concept when working with functions of multiple variables. They measure how the function changes as we change one variable while keeping the others constant. In the example function \( f(x, y) = 2xy^3 - 3x^2y \), we find two partial derivatives:

• The partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), is computed by treating \( y \) as a constant, which simplifies to \( 2y^3 - 6xy \).
• The partial derivative with respect to \( y \), denoted \( \frac{\partial f}{\partial y} \), is computed by treating \( x \) as a constant, resulting in \( 6xy^2 - 3x^2 \).

These partial derivatives tell us how the function value would change if we perturb either \( x \) or \( y \), respectively. Understanding partial derivatives is key to analyzing the behavior of complex functions over several variables, and it is essential for calculating the gradient.

Unit Vector

A unit vector is a vector that has a magnitude of 1. It is often used to indicate direction without regard to magnitude. To find a unit vector in the direction of any given vector, we "normalize" the original vector, which involves dividing it by its own magnitude.

• The magnitude of a vector \( \mathbf{v} \) with components \( (a, b) \) is calculated as \( \| \mathbf{v} \| = \sqrt{a^2 + b^2} \).
• For the vector \( \begin{bmatrix} 3 \ 1 \end{bmatrix} \), the magnitude becomes \( \sqrt{3^2 + 1^2} = \sqrt{10} \).
• Normalization gives us the unit vector \( \mathbf{u} = \frac{1}{\sqrt{10}} \begin{bmatrix} 3 \ 1 \end{bmatrix} \).

Unit vectors are particularly useful in physics and engineering for defining directions. In this context, they help us compute the directional derivative, indicating how fast a function changes in that particular normalized direction.