Problem 29 - Calculus for Biology and Medicine

Question

Suppose an organism moves down a sloped surface along the steepest line of descent, i.e., the direction in which the surface decreases most rapidly. If the surface is given by $$ f(x, y)=x^{2}-y^{2} $$ find the direction in which the organism will move at the point \((2,3)\)

Step-by-Step Solution

Answer

The organism moves in the direction \((-\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}})\) at point \((2, 3)\).

Step 1: Find the Gradient

The gradient \( abla f \) of the function \( f(x, y) = x^2 - y^2 \) is found by computing the partial derivatives: \( \frac{\partial f}{\partial x} = 2x \) and \( \frac{\partial f}{\partial y} = -2y \). Thus, \( abla f = (2x, -2y) \).

Step 2: Evaluate the Gradient at the Given Point

Substitute the coordinates \((2, 3)\) into the gradient function. This gives \( abla f(2, 3) = (2(2), -2(3)) = (4, -6) \).

Step 3: Determine the Direction of Steepest Descent

The direction of steepest descent is the negative of the gradient at that point. Thus, it is given by \(-abla f(2, 3) = (-4, 6)\).

Step 4: Normalize the Direction Vector

To find the unit vector in the direction of steepest descent, calculate the magnitude of \((-4, 6)\). The magnitude is \(\sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\). The unit vector is obtained by dividing each component by this magnitude: \(\left(-\frac{4}{2\sqrt{13}}, \frac{6}{2\sqrt{13}}\right) = \left(-\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}\right)\).

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a multivariable function changes with respect to one variable at a time, while keeping the other variables constant. When dealing with a function of two variables, such as

• f(x, y) = x^2 - y^2

we need to find its partial derivatives to explore its rate of change along the x-axis and y-axis independently.

The partial derivative of a function with respect to x, denoted as \( \frac{\partial f}{\partial x} \), gives us the slope of the function along the x-axis. For the function above, the partial derivative is:

• \( \frac{\partial f}{\partial x} = 2x \)

Likewise, the partial derivative with respect to y, \( \frac{\partial f}{\partial y} \), gives the slope along the y-axis and is found as follows:

• \( \frac{\partial f}{\partial y} = -2y \)

Both of these calculations result in the gradient vector \( abla f = (2x, -2y) \). This vector not only indicates the rate of change in each direction but also helps in determining how steep the slope is at any given point, which is essential for understanding the landscape of the function.

Direction of Steepest Descent

The direction of steepest descent is the path along which a function decreases most rapidly from a particular point. It is intuitively the direction that leads straight downhill. When you calculate the gradient vector of a function, it points in the direction of the greatest rate of increase.

To find the direction of steepest descent, you simply take the opposite direction of the gradient. Therefore:

• If \( abla f(2, 3) = (4, -6) \), then the direction of steepest descent is \( -abla f(2, 3) = (-4, 6) \).

This ensures that you are moving against the slope, essentially moving downhill as efficiently as possible.

This direction is very useful, especially in optimization problems, such as finding the minimum of a function, because it tells us the quickest way to decrease the function's value. In this context, understanding and utilizing the direction of steepest descent can help significantly in decision-making processes or improving algorithmic efficiency.

Unit Vector

A unit vector is a vector that has a magnitude of one and indicates direction only. This is important when you want to standardize a vector solely to indicate direction, without taking its magnitude into consideration.

To compute a unit vector in the direction of steepest descent, you need to normalize the vector \((-4, 6)\), which represents the direction of steepest descent. To do this, you first find the magnitude of the vector:

• Calculate the magnitude: \( \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13} \).

Next, you divide each component of the vector by its magnitude to obtain the unit vector:

• Unit vector: \( \left(-\frac{4}{2\sqrt{13}}, \frac{6}{2\sqrt{13}}\right) = \left(-\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}\right) \).

For practical purposes, such as when programming algorithms that deal with movement directions or optimizations, using a unit vector keeps calculations consistent and simple. It provides clear, normalized directions without having the magnitude component, which could skew interpretations or computations.