Problem 17 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the function. $$f(x, t)=e^{-t} \cos \pi x$$

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial x} = -\pi e^{-t}\sin(\pi x) \) and \( \frac{\partial f}{\partial t} = -e^{-t}\cos(\pi x) \).

Step 1: Understand the Function

Given the function \( f(x, t) = e^{-t} \cos(\pi x) \), we need to find its first partial derivatives with respect to \( x \) and \( t \).

Step 2: Determine the Partial Derivative with Respect to x

To find the partial derivative \( \frac{\partial f}{\partial x} \), treat \( t \) as a constant. The derivative of \( \cos(\pi x) \) with respect to \( x \) is \( -\pi \sin(\pi x) \). Thus, \( \frac{\partial f}{\partial x} = e^{-t} \cdot -\pi \sin(\pi x) = -\pi e^{-t} \sin(\pi x) \).

Step 3: Calculate the Partial Derivative with Respect to t

For the partial derivative \( \frac{\partial f}{\partial t} \), treat \( x \) as a constant. The derivative of \( e^{-t} \) with respect to \( t \) is \( -e^{-t} \). So, \( \frac{\partial f}{\partial t} = -e^{-t} \cos(\pi x) \).

Step 4: Combine Results

The first partial derivatives are \( \frac{\partial f}{\partial x} = -\pi e^{-t} \sin(\pi x) \) and \( \frac{\partial f}{\partial t} = -e^{-t} \cos(\pi x) \).

Key Concepts

Multivariable Calculus

Multivariable calculus is the extension of calculus to functions of several variables. While single-variable calculus focuses on functions of a single independent variable, multivariable calculus introduces the concept of partial derivatives. These derivatives are a way to measure how a function changes as one of its variables changes while holding the others fixed.

For example, in the function given, \( f(x, t) = e^{-t} \cos(\pi x) \), there are two variables: \( x \) and \( t \). Partial derivatives help us understand the sensitivity of the function to individual variables.

In multivariable calculus, you will often look at surfaces in three-dimensional spaces defined by equations like \( z = f(x, y) \). By computing partial derivatives, you get the slope of the surface in the direction of one of the variables. This is crucial for analyzing and interpreting multivariable functions.

Derivative Rules

Working with partial derivatives involves applying common derivative rules, but with slight modifications to account for multiple variables. Here, we use the rule that when taking a partial derivative with respect to a particular variable, all other variables are treated as constants during the differentiation process.

In the original exercise, we calculated the partial derivative of \( f(x, t) = e^{-t} \cos(\pi x) \) with respect to \( x \) by treating \( t \) as constant. This means applying the chain rule to \( \cos(\pi x) \): the derivative is \( -\pi \sin(\pi x) \). With respect to \( t \), \( e^{-t} \) is differentiated to \( -e^{-t} \), treating \( x \) as constant.

Whenever you're computing derivatives, remember these rules:

• The constant rule: Constants differentiate to zero.


• The power rule: \( d/dx(x^n) = nx^{n-1} \).


• The product rule: For products, \( (uv)' = u'v + uv' \).


• The chain rule: For composites, \( (f(g(x)))' = f'(g(x))g'(x) \).

Exponential Functions

Exponential functions are an essential part of calculus and appear frequently in mathematical modeling. The general form is \( f(x) = a \cdot e^{bx} \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718. They are known for their unique properties, such as having the same rate of change (derivative) as the value of the function itself.

In the exercise problem, we encountered \( e^{-t} \). To differentiate \( e^{x} \), the derivative is simply \( e^{x} \), but for \( e^{-t} \), chain rule gives us \( -e^{-t} \) because of the negative exponent.

Some key properties of exponential functions to keep in mind include:

• The derivative of \( e^{x} \) is \( e^{x} \).


• Exponential growth: Growth rate is proportional to the current value.


• The presence of \( e \) ensures continuous compounding.

Understanding these properties aids in the efficient calculation and manipulation of functions involving exponentials, as seen in the problem.