Problem 130 - 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: Identify the given function

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

Step 2: Find the partial derivative with respect to x

To find \(\frac{\partial z}{\partial x}\), we differentiate \(z = e^{-x} \cos(y)\) with respect to \(x\) while treating \(y\) as a constant. The derivative of \(e^{-x}\) with respect to \(x\) is \(-e^{-x}\), and since \(\cos(y)\) is constant with respect to \(x\), the result is: \[ \frac{\partial z}{\partial x} = -e^{-x} \cos(y)\].

Step 3: Evaluate the partial derivative at the point (0,1)

Substitute \(x = 0\) and \(y = 1\) into the partial derivative: \[-e^{-0} \cos(1) = -1 \times \cos(1) = -\cos(1)\].

Step 4: Simplify the expression

The expression \(-\cos(1)\) is the evaluated partial derivative at \((0, 1)\). It does not need further simplification as it is already in the simplest form.

Key Concepts

Understanding Multivariable Calculus

Multivariable calculus expands the concepts of single-variable calculus into more dimensions. It involves functions of several variables, like two or three, which can represent real-world scenarios more accurately. In our exercise, the function is based on two variables, allowing us to explore relationships between two dimensions.

Why is it important? Multivariable calculus is crucial for understanding systems where several factors interact. For instance:

• Physics and Engineering problems often involve multiple variables.
• Economics can model factors like supply and demand.
• Computer graphics use it for representing 3D objects.

This calculus branch teaches concepts like gradients and divergences, which provide insights into the behavior of functions in space.

The function given, \(z= e^{-x} \cos(y)\), demonstrates a relationship in a multivariable context, showing how \(z\) changes concerning \(x\) and \(y\). To handle such functions more efficiently, partial differentiation is one of the essential tools offered by multivariable calculus.

Partial Differentiation Explained

Partial differentiation allows us to understand how a multivariable function changes as one variable changes, keeping others constant. This process is vital when exploring a function's behavior in each dimension.

To find the partial derivative with respect to one variable:

• Treat all other variables as constants.
• Differentiate the function concerning the desired variable as you normally would in single-variable calculus.

In the exercise, you begin by tackling the function \(z = e^{-x} \cos(y)\) to find its partial derivative with respect to \(x\). Keep \(y\) constant while differentiating. The derivative of \(e^{-x}\) is \(-e^{-x}\), and since \(\cos(y)\) is constant relative to \(x\), it remains unchanged. Thus, your result is \(-e^{-x} \cos(y)\).

Partial differentiation uncovers unique insights:

• It helps identify the function's slope along a particular direction.
• Shows how the function evolves spatially at distinct points.

These insights prove invaluable in fields such as optimization and modeling.

Ease into Function Evaluation

Function evaluation is the process of calculating the value of a function at given points. It becomes essential after finding derivatives to determine the behavior of the function at specific points.

In our exercise, evaluating the partial derivative at the point \((0,1)\) is necessary for full understanding. The calculated partial derivative is \(-e^{-x} \cos(y)\). By substituting \(x = 0\) and \(y = 1\) into this derivative, you compute:

• \(-e^{0} \cos(1) = -1 \cos(1)\)
• This simplifies directly to \(-\cos(1)\)

Function evaluation provides:

• Specific values that can be used to analyze and predict behavior in a given system.
• Helps in comparing theoretical results with practical data or results.

In broader applications, assessing values at particular points helps determine equilibrium points and understand systems better, aiding in effective decision-making and problem-solving.