Problem 129 - Calculus 3

Question

Evaluate the partial derivatives at point P(0, 1). Find \(\frac{\partial z}{\partial x}\) at \((0,1)\) for \(z=e^{-x} \cos (y)\)

Step-by-Step Solution

Answer

The partial derivative \(\frac{\partial z}{\partial x}\) at \((0,1)\) is \(-\cos(1)\).

Step 1: Understanding the Functions

We are given the function \(z = e^{-x} \cos(y)\) and need to find the partial derivative of \(z\) with respect to \(x\), \(\frac{\partial z}{\partial x}\), at the point \((0,1)\).

Step 2: Calculating the Partial Derivative

The partial derivative of \(z = e^{-x} \cos(y)\) with respect to \(x\) at any point is calculated by differentiating \(z\) only with respect to \(x\), treating \(y\) as a constant. The partial derivative is \[ \frac{\partial z}{\partial x} = \frac{d}{dx}(e^{-x} \cos(y)) = -e^{-x} \cos(y). \]

Step 3: Substituting the Given Point

Substitute \(x = 0\) and \(y = 1\) into the derivative formula obtained. Thus, \[ \frac{\partial z}{\partial x} \bigg|_{(0,1)} = -e^{0} \cos(1) = -1 \cdot \cos(1). \]

Step 4: Simplifying the Expression

Since \(e^0 = 1\), the expression simplifies to \(-\cos(1)\). Thus, the value of the partial derivative \(\frac{\partial z}{\partial x}\) at the point \((0,1)\) is \(-\cos(1)\).

Key Concepts

Functions

In mathematics, **functions** are fundamental building blocks that describe relationships between sets of input and output variables. Each function specifies how each input is mapped to an output. For instance, the function given in this exercise is \( z = e^{-x} \cos(y) \), where \( x\) and \( y\) are independent variables and \( z \) is a dependent variable.

• The exponential term \( e^{-x} \) involves \( x \), indicating that \( z \) changes when \( x \) changes, assuming \( y \) is constant.
• The \( \cos(y) \) term involves \( y \) only; this remains unchanged with variations in \( x \), illustrating how functions can contain elements dependent on various independent variables.

Understanding functions is crucial for grasping how changes in different variables affect the outcome, an essential skill in calculus and real-world problem-solving.

Differentiation

**Differentiation** is a core concept in calculus that determines the rate at which a function changes at any given point. In simpler terms, it involves calculating the function's derivative.There are two main types of derivatives:

• Ordinary derivatives: These are derivatives of functions of a single variable.
• Partial derivatives: These are used for functions with more than one variable.

In this exercise, we focused on calculating the partial derivative of the function \( z = e^{-x} \cos(y) \) with respect to \( x \). This requires differentiating \( z \) while keeping \( y \) constant, demonstrating the process of partial differentiation.

Evaluation of Derivatives

The **evaluation of derivatives** involves substituting specific values into the derivative to find the rate of change at a particular point.To evaluate the partial derivative \( \frac{\partial z}{\partial x} \) at the point \( (0,1) \):

• First, compute the partial derivative: \( \frac{\partial z}{\partial x} = -e^{-x} \cos(y) \).
• Then, plug in the values \( x = 0 \) and \( y = 1 \) into this derivative: \( \frac{\partial z}{\partial x} \bigg|_{(0,1)} = -e^0 \cos(1) \).
• Since \( e^0 = 1 \), simplify it to \( -\cos(1) \).

This substitution and simplification are critical steps in determining exact values, such as gradients or accelerations, at specific points in multivariable functions.