Problem 16 - Calculus for Biology and Medicine

Question

In Problems 15-18, compute the directional derivative of \(f(x, y)\) at the point \(P\) in the direction of the point \(Q .\) $$ f(x, y)=4 x y+y^{2}, P=(-1,1), Q=(3,2) $$

Step-by-Step Solution

Answer

The directional derivative is \(\frac{14}{\sqrt{17}}\).

Step 1: Understand the Problem

We need to find the directional derivative of the function \(f(x, y) = 4xy + y^2\) at the point \(P=(-1, 1)\) in the direction towards the point \(Q=(3, 2)\).

Step 2: Compute the Gradient of \(f\)

Calculate the gradient \(abla f(x, y)\). The partial derivatives are: \( \partial f/\partial x = 4y \) and \( \partial f/\partial y = 4x + 2y \). Therefore, \(abla f(x, y) = (4y, 4x + 2y)\).

Step 3: Evaluate the Gradient at Point \(P\)

Substitute \(x = -1\) and \(y = 1\) into the gradient: \(abla f(-1, 1) = (4 \times 1, 4(-1) + 2 \times 1) = (4, -2)\).

Step 4: Calculate the Direction Vector \(\mathbf{u}\)

Find the vector from \(P\) to \(Q\): \(\mathbf{v} = (3 + 1, 2 - 1) = (4, 1)\). Normalize \(\mathbf{v}\) to get \(\mathbf{u}\): \(||\mathbf{v}|| = \sqrt{4^2 + 1^2} = \sqrt{17}\), so \(\mathbf{u} = \left(\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}}\right)\).

Step 5: Compute the Directional Derivative

Using \(D_{\mathbf{u}} f = abla f \cdot \mathbf{u}\), compute: \[(4, -2) \cdot \left(\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}}\right) = \frac{4 \times 4 + (-2) \times 1}{\sqrt{17}} = \frac{16 - 2}{\sqrt{17}} = \frac{14}{\sqrt{17}}\].Thus, the directional derivative is \(\frac{14}{\sqrt{17}}\).

Key Concepts

Gradient

The gradient is a vector that shows the direction and rate of the steepest ascent of a function. It's like a multi-directional slope of a hill. For a function of two variables, like the one we have, it consists of the partial derivatives of the function with respect to each variable. Here’s how it breaks down:

• For a function \( f(x, y) \), the gradient is written as \( abla f(x, y) \).
• The gradient gives you a vector pointing in the direction of the steepest increase of the function.
• To find the gradient of \( f(x, y) = 4xy + y^2 \), we take the partial derivatives: \( \partial f/\partial x = 4y \) and \( \partial f/\partial y = 4x + 2y \).

Putting these together, the gradient vector is \( abla f(x, y) = (4y, 4x + 2y) \), which tells us how sensitive the function is to changes in \( x \) and \( y \). It’s a powerful tool for understanding how functions behave in multidimensional spaces.

Partial Derivative

Partial derivatives are a way to understand how a function changes as each of its variables changes, while keeping all other variables constant. Think of it as examining the slope of a road by focusing only on how steep the road is going west or north, separately instead of both directions at once. Here’s what you need to know:

• A partial derivative is noted with the symbol \( \partial \) rather than \( d \).
• For a function \( f(x, y) \), \( \partial f/\partial x \) measures how \( f \) changes with \( x \) when \( y \) is constant, while \( \partial f/\partial y \) measures how \( f \) changes with \( y \) when \( x \) is constant.

In our function \( f(x, y) = 4xy + y^2 \), the partial derivative with respect to \( x \) is \( 4y \). This tells us the slope in the direction of \( x \). Similarly, \( \partial f/\partial y = 4x + 2y \) gives the slope in the direction of \( y \). These derivatives help build the gradient vector, offering insights into the behavior of the function.

Function of Two Variables

A function of two variables, such as \( f(x, y) = 4xy + y^2 \), depends on two inputs rather than a single one. Imagine a surface in three-dimensional space formed by the combination of inputs \( x \) and \( y \). Understanding this concept is crucial for exploring how changes in inputs affect an output:

• Each pair \( (x, y) \) gives a value \( f(x, y) \), akin to the height of a landscape at the point \( (x, y) \).
• Analyzing such functions involves considering the interplay between both variables.

This analysis allows for the exploration of contours — akin to lines on a map reflecting altitude levels. Grasping the two-variable nature provides a more complex and rich understanding of real-world scenarios, like finding the maximum profit based on prices and marketing spend, or other simultaneous changes in different factors.

Vector Normalization

Vector normalization is the process of converting a vector into a unit vector — that is, a vector with a length, or magnitude, of 1, but pointing in the same direction. It's like scaling down a drawing without changing its proportions.

• The length of a vector \( \mathbf{v} = (a, b) \) is calculated as \( \sqrt{a^2 + b^2} \).
• To normalize \( \mathbf{v} \), divide each component by the vector’s length.

In our exercise, the direction vector from \( P \) to \( Q \) was \( \mathbf{v} = (4, 1) \). Calculating its length \( ||\mathbf{v}|| \) gives \( \sqrt{17} \). By dividing its components by \( \sqrt{17} \), we get the unit vector \( \mathbf{u} = \left(\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}}\right) \), which points in the same direction as \( \mathbf{v} \) but has a length of 1. This normalized vector is essential in finding the directional derivative, which measures how a function changes in a specific direction.