Problem 26 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to calculate the directional derivative of the function \( f(x, y) = x^2 \sin y \) at the point \( (-1, 0) \) in the direction of the vector \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \). Directional derivatives measure how a function changes as we move in a specific direction.

Step 2: Calculate the Gradient of the Function

The gradient of \( f \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Compute these partial derivatives:1. \( \frac{\partial f}{\partial x} = 2x \sin y \)2. \( \frac{\partial f}{\partial y} = x^2 \cos y \)Thus, \( abla f = \left( 2x \sin y, x^2 \cos y \right) \).

Step 3: Evaluate the Gradient at the Given Point

Substitute \( x = -1 \) and \( y = 0 \) into the gradient:\[abla f(-1,0) = \left( 2(-1) \sin 0, (-1)^2 \cos 0 \right) = (0, 1)\]

Step 4: Normalize the Direction Vector

The direction vector is \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \). Normalize it to get a unit direction vector. The magnitude of the vector is:\[\| \begin{bmatrix} 2 \ -1 \end{bmatrix} \| = \sqrt{2^2 + (-1)^2} = \sqrt{5}\]So the unit vector is:\[\begin{bmatrix} \frac{2}{\sqrt{5}} \ \frac{-1}{\sqrt{5}} \end{bmatrix}\]

Step 5: Compute the Directional Derivative

The directional derivative of \( f \) in the direction of a unit vector \( \mathbf{u} \) is given by:\[D_{\mathbf{u}}f = abla f \cdot \mathbf{u} \]Substitute \( abla f(-1,0) = (0, 1) \) and the unit vector:\[D_{\mathbf{u}}f = \begin{bmatrix} 0 \ 1 \end{bmatrix} \cdot \begin{bmatrix} \frac{2}{\sqrt{5}} \ \frac{-1}{\sqrt{5}} \end{bmatrix} = 0 \cdot \frac{2}{\sqrt{5}} + 1 \cdot \frac{-1}{\sqrt{5}} = \frac{-1}{\sqrt{5}}\]

Step 6: Final Result

The directional derivative of \( f(x, y) \) at \( (-1, 0) \) in the specified direction is \( \frac{-1}{\sqrt{5}} \).

Key Concepts

Gradient Calculation

The gradient of a function is a vital concept in calculus used to describe the direction and rate of the quickest increase of a function at a point. In our context, for the function \( f(x, y) = x^2 \sin y \), the gradient \( abla f \) is a vector that combines the partial derivatives with respect to \( x \) and \( y \). It's expressed as the vector:

\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]
Let’s break this down:

• \( \frac{\partial f}{\partial x} = 2x \sin y \) captures how the function changes as \( x \) varies, keeping \( y \) constant.
• \( \frac{\partial f}{\partial y} = x^2 \cos y \) describes the function's change with variations in \( y \) while \( x \) remains fixed.

Evaluating these partial derivatives at the specific point \((-1, 0)\) reveals the gradient to be \((0, 1)\). This tells us that at \( (-1,0) \), the function doesn't change along \( x \) and is increasing along \( y \). Understanding gradients helps in exploring how functions evolve in multi-dimensional spaces.

Partial Derivatives

Partial derivatives are a cornerstone technique in calculus, especially when dealing with functions of multiple variables. They show how a function changes as we vary one of the variables, keeping all others constant. For the function \( f(x, y) = x^2 \sin y \), two partial derivatives are crucial:

• The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} = 2x \sin y \), tells us how the function changes as we move along the \( x \)-axis, while \( y \) is held constant.
• The partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} = x^2 \cos y \), informs us about changes in the function along the \( y \)-axis, with \( x \) fixed.

Think of these as measuring the slope of a surface in \( x \) and \( y \) directions. Together, they formulate the gradient. Partial derivatives are powerful because they dissect the behavior of more complex functions, paving the way for further applications like optimization and studying multi-variable systems.

Unit Vector Normalization

Unit vector normalization is an essential method used in vector calculus to factor out the scale of a vector, focusing only on its direction. This process turns any vector into a unit vector, which means it has a length of one.

When dealing with directional derivatives, such as in the problem at hand, the direction vector provided is \( \begin{bmatrix} 2 \ -1 \end{bmatrix} \). To normalize it:

• First, calculate its magnitude using the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \). For our vector:
• \( \| \begin{bmatrix} 2 \ -1 \end{bmatrix} \| = \sqrt{2^2 + (-1)^2} = \sqrt{5} \)
• Now, divide each component by the magnitude: \( \begin{bmatrix} \frac{2}{\sqrt{5}} \ \frac{-1}{\sqrt{5}} \end{bmatrix} \)

This resulting vector maintains the same direction as the original but has a norm of one. Normalizing vectors is crucial for interpreting directions correctly when dealing with functions, as it ensures we focus solely on the slope of change in the function's surface along defined directions.