Problem 36 - Calculus for Biology and Medicine

Question

In what direction does \(f(x, y)=e^{x} \cos y\) increase most rapidly at \((0, \pi / 2) ?\)

Step-by-Step Solution

Answer

The function increases most rapidly in the \(-y\) direction at the point \((0, \pi / 2)\).

Step 1: Identify the Gradient of the Function

The gradient of a function \(f(x, y)\) is given by the vector of its partial derivatives: \(abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). First, calculate the partial derivative with respect to \(x\): \(\frac{\partial f}{\partial x} = e^x \cos y\). Then, calculate the partial derivative with respect to \(y\): \(\frac{\partial f}{\partial y} = -e^x \sin y\). Thus, the gradient is \(abla f(x, y) = \left(e^x \cos y, -e^x \sin y\right)\).

Step 2: Evaluate the Gradient at the Given Point

Substitute the point \((0, \pi / 2)\) into the gradient: \(abla f(0, \pi / 2) = \left(e^0 \cos(\pi/2), -e^0 \sin(\pi/2)\right)\). This simplifies to \(abla f(0, \pi / 2) = (0, -1)\).

Step 3: Interpret the Result to Determine the Direction

The gradient vector \((0, -1)\) at the point \((0, \pi / 2)\) indicates the direction of the steepest ascent of the function. This vector points directly downwards on the \(y\)-axis, meaning the function increases most rapidly in the \(-y\) direction.

Key Concepts

Partial Derivatives

Partial derivatives are essential in understanding how a multivariable function changes with respect to each of its variables. When dealing with a function of two variables, like \( f(x, y) = e^x \cos y \), we compute the partial derivatives to pinpoint how the function value changes as either the variable \( x \) or \( y \) changes, while the other remains constant.
To find the partial derivatives, we treat all other variables as constants and differentiate as if it's a single-variable function.

• For \( x \): The partial derivative is \( \frac{\partial f}{\partial x} = e^x \cos y \).
• For \( y \): The partial derivative is \( \frac{\partial f}{\partial y} = -e^x \sin y \).

These derivatives form the components of the gradient vector, providing a crucial insight into the nature of the function. Each partial derivative tells us how the function moves in relation to its corresponding variable, which is foundational for problem-solving in calculus.

Direction of Steepest Ascent

The direction of steepest ascent of a function at any given point is determined by its gradient vector. The gradient vector consists of partial derivatives, pointing in the direction in which the function increases most rapidly.
In the case of \( f(x, y) = e^x \cos y \) at the point \((0, \pi/2)\), we computed the gradient vector as \((0, -1)\). This vector reveals the steepest ascent's direction.
Think of the gradient as an arrow indicating which way to "climb" the function surface for the height to increase fastest.

• Since the vector \((0, -1)\) points downward on the \( y \)-axis, it suggests the quickest increase occurs in the negative \( y \) direction.

Understanding this concept is invaluable not only in calculus but also in fields like machine learning, where gradient ascent/descent methods are used for optimization.

Function Evaluation

Function evaluation is integral in applying mathematical solutions to real-world problems. Evaluating a function at a specific point involves plugging the point's coordinates into the function and calculating the resulting value.
For example, with \( f(x, y) = e^x \cos y \), evaluating it at \((0, \pi/2)\) involves substituting these values into the function:

• Compute: \( e^0 \cos(\pi/2) \), which evaluates to \( 1 \times 0 = 0 \).

In our exercise, evaluating the gradient function at this point provided the vector \((0, -1)\). This evaluation step is key in numerous mathematical applications, such as determining the growth pattern or calculating potential changes in economics and engineering scenarios. Therefore, mastering function evaluation enhances problem-solving skills by blending theory with practical computation.