Problem 20 - Calculus for Biology and Medicine

Question

Find the linearization of \(f(x, y)\) at the indicated point \(\left(x_{0}, y_{0}\right) .\) \(f(x, y)=\cos \left(x^{2} y\right) ;\left(\frac{\pi}{2}, 0\right)\)

Step-by-Step Solution

Answer

The linearization of \( f(x, y) \) at the point \( \left( \frac{\pi}{2}, 0 \right) \) is 1.

Step 1: Understanding Linearization

Linearization of a function at a point involves approximating the function with a linear function. The formula for the linearization of a function \( f(x, y) \) at the point \( (x_0, y_0) \) is given by: \[ L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0) \] where \( f_x \) and \( f_y \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \), respectively. So, our goal is to find these derivatives and evaluate them at the point \( \left( \frac{\pi}{2}, 0 \right) \).

Step 2: Find Partial Derivative with Respect to x

The partial derivative of \( f(x, y) = \cos(x^2y) \) with respect to \( x \) is found using the chain rule: \[ f_x = \frac{d}{dx} \cos(x^2 y) = -sin(x^2 y) \cdot 2xy \] Evaluate this at the point \( \left( \frac{\pi}{2}, 0 \right) \): \[ f_x\left( \frac{\pi}{2}, 0 \right) = -sin\left( \left( \frac{\pi}{2} \right)^2 \cdot 0 \right) \cdot \frac{\pi}{2} \cdot 0 = 0 \]

Step 3: Find Partial Derivative with Respect to y

The partial derivative of \( f(x, y) = \cos(x^2y) \) with respect to \( y \) is: \[ f_y = \frac{d}{dy} \cos(x^2 y) = -sin(x^2 y) \cdot x^2 \] Evaluate this at the point \( \left( \frac{\pi}{2}, 0 \right) \): \[ f_y\left( \frac{\pi}{2}, 0 \right) = -sin\left( \left( \frac{\pi}{2} \right)^2 \cdot 0 \right) \cdot \left( \frac{\pi}{2} \right)^2 = 0 \]

Step 4: Evaluate the Function at the Given Point

We evaluate the function at the point \( \left( \frac{\pi}{2}, 0 \right) \): \[ f\left( \frac{\pi}{2}, 0 \right) = \cos\left( \left( \frac{\pi}{2} \right)^2 \cdot 0 \right) = \cos(0) = 1 \]

Step 5: Construct the Linearization

Having found \( f(x_0, y_0) \), \( f_x(x_0, y_0) \), and \( f_y(x_0, y_0) \), we can now substitute them into the linearization formula: \[ L(x, y) = 1 + 0\cdot(x - \frac{\pi}{2}) + 0\cdot(y - 0) = 1 \] Thus, the linearization of \( f(x, y) \) at \( \left( \frac{\pi}{2}, 0 \right) \) is \( L(x, y) = 1 \).

Key Concepts

Partial Derivatives

When dealing with functions of several variables like \( f(x, y) \), partial derivatives are a fundamental tool. They represent the rate of change of the function concerning one variable while keeping the other constant.
For a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( f_x \), and the partial derivative with respect to \( y \) is denoted as \( f_y \). In our problem, to find these derivatives, we use the chain rule, given how the function is composed.
Specifically, for the function \( f(x, y) = \cos(x^2 y) \), the partial derivative with respect to \( x \) is:

• \( f_x(x, y) = \frac{\partial}{\partial x}{\cos(x^2 y)} = -\sin(x^2 y) \cdot 2xy \)

The partial derivative with respect to \( y \) is:

• \( f_y(x, y) = \frac{\partial}{\partial y}{\cos(x^2 y)} = -\sin(x^2 y) \cdot x^2 \)

Knowing how to calculate these derivatives is crucial as they help approximate the function's behavior near a given point, which is the essence of linearization.

Chain Rule

The chain rule is a vital concept when working with derivatives of composite functions in calculus. It allows us to differentiate functions that are not straightforwardly separable into their individual components.
In the context of partial derivatives, the chain rule helps us take derivatives of a function like \( f(x, y) = \cos(x^2 y) \) by breaking down the expression into simpler parts. Here's how the chain rule applies:

• First, identify the "outer" function, which is \( \cos(u) \).
• Next, identify the "inner" function, which is \( u = x^2 y \).

By applying the chain rule:

• Differentiate \( \cos(u) \) with respect to \( u \), which gives \( -\sin(u) \).
• Then, differentiate \( u = x^2 y \) with respect to the desired variable \( x \) or \( y \), yielding either \( 2xy \) or \( x^2 \), respectively.

Thus, the chain rule allows us to understand and calculate how intertwined variables impact the behavior of a function as shown in our partial derivatives calculation.

Multivariable Calculus

Multivariable calculus extends the ideas of calculus to functions with more than one input. These functions, such as \( f(x, y) \), depend on several variables, making them more complex but also more powerful in describing real-world phenomena.
A key aspect of multivariable calculus is linearization, which approximates functions locally with a simpler linear function. The formula:

• \[ L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0) \]

demonstrates how linearization uses partial derivatives \( f_x \) and \( f_y \) to construct a tangent plane at the point, providing an easy-to-use approximation.
In this exercise, linearization simplifies the function \( \cos(x^2 y) \) near \( \left( \frac{\pi}{2}, 0 \right) \) to just 1, capturing the function's essence in a straightforward way. Understanding these concepts equips students to analyze complex systems using calculus effectively.