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Magazine Chapter 14: Problem 25
Find the differential of the function. \(z=e^{-2 x} \cos 2 \pi t\)
Short Answer
Expert verified
The differential is \( dz = -2e^{-2x}\cos(2\pi t)\, dx - 2\pi e^{-2x}\sin(2\pi t)\, dt \).
Step by step solution
01
Identify the Function
The given function is a multi-variable function, specified as \( z = e^{-2x} \cos(2\pi t) \). It involves two independent variables \( x \) and \( t \).
02
Find Partial Derivative with respect to \( x \)
The partial derivative of \( z \) with respect to \( x \), denoted \( \frac{\partial z}{\partial x} \), involves differentiating \( e^{-2x} \) while treating \( \cos(2\pi t) \) as a constant. This yields:\[ \frac{\partial z}{\partial x} = \frac{d}{dx}(e^{-2x}) \cdot \cos(2\pi t) = -2e^{-2x} \cos(2\pi t) \].
03
Find Partial Derivative with respect to \( t \)
The partial derivative of \( z \) with respect to \( t \), denoted \( \frac{\partial z}{\partial t} \), involves differentiating \( \cos(2\pi t) \) while treating \( e^{-2x} \) as a constant. This yields:\[ \frac{\partial z}{\partial t} = e^{-2x} \cdot \frac{d}{dt}(\cos(2\pi t)) = -2\pi e^{-2x} \sin(2\pi t) \].
04
Write the Differential
The differential \( dz \) of a function of several variables is given by:\[ dz = \frac{\partial z}{\partial x} \, dx + \frac{\partial z}{\partial t} \, dt \].Substitute the derivatives found from Steps 2 and 3 into this expression:\[ dz = (-2e^{-2x} \cos(2\pi t)) \, dx + (-2\pi e^{-2x} \sin(2\pi t)) \, dt \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are used to find the rate of change of a function with respect to one variable, while keeping the other variables constant. This is especially important in multi-variable functions, where more than one independent factor is present. For example, in the function \(z = e^{-2x} \cos(2 \pi t)\), both \(x\) and \(t\) are variables.
- When finding the partial derivative with respect to \(x\), you treat \(\cos(2\pi t)\) as a constant and only differentiate \(e^{-2x}\). This results in \(\frac{\partial z}{\partial x} = -2e^{-2x} \cos(2\pi t)\).
- Similarly, when differentiating with respect to \(t\), \(e^{-2x}\) is treated as a constant, leading to an expression \(\frac{\partial z}{\partial t} = -2\pi e^{-2x} \sin(2\pi t)\).
Multi-variable Functions
A multi-variable function, as the name implies, involves more than one independent variable. This means the output depends on two or more inputs.
- In our problem, \(z = e^{-2x} \cos(2 \pi t)\) represents a multi-variable function with variables \(x\) and \(t\).
- Each independent variable can create a different dimension in the function's overall output, allowing the function to describe more complex systems and relationships.
- Understanding these functions is crucial in fields like physics, engineering, and economics, where interactions between different variables need to be precisely mapped.
Chain Rule
The chain rule is a principle in calculus used for differentiating compositions of functions. While the exercise didn't directly require the chain rule, understanding it is essential when dealing with more complex derivatives.
- It allows for differentiation when variables within a function are themselves functions of another variable.
- For instance, if you needed to differentiate \(e^{-2x}\) where \(x\) itself was some function of another variable, the chain rule would be necessary.
Differentials
Differentials express an incremental change in a function relative to changes in its variables. It's a way to articulate how tiny adjustments in input affect the overall outcome.
- The differential \(dz\) combines the partial derivatives with respect to each variable and their changes, represented as \(dx\) and \(dt\).
- In the exercise, the differential is \(dz = (-2e^{-2x} \cos(2\pi t)) \, dx + (-2\pi e^{-2x} \sin(2\pi t)) \, dt\).
- This shows the total effect on \(z\) for small changes in \(x\) and \(t\), employing the previously found partial derivatives.
