Problem 23 - Calculus Single Variable

Question

Calculate the length of the given parametric curve. $$ x=e^{t} \cos (t) \quad y=e^{t} \sin (t) \quad 0 \leq t \leq 1 $$

Step-by-Step Solution

Answer

The length of the curve is \( \sqrt{2}(e - 1) \).

Step 1: Understand the Formula for Arc Length

The formula to find the arc length of a parametric curve given by \( x(t) \) and \( y(t) \) over interval \( [a, b] \) is \( L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \). This formula will be crucial for solving the given problem.

Step 2: Differentiate the Parametric Equations

First, find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x(t) = e^t \cos(t) \), use the product rule to get \( \frac{dx}{dt} = e^t \cos(t) - e^t \sin(t) \). For \( y(t) = e^t \sin(t) \), again apply the product rule to obtain \( \frac{dy}{dt} = e^t \sin(t) + e^t \cos(t) \).

Step 3: Simplify the Derivative Expressions

Simplify the derivatives: \( \frac{dx}{dt} = e^t (\cos(t) - \sin(t)) \) and \( \frac{dy}{dt} = e^t (\sin(t) + \cos(t)) \).

Step 4: Square the Derivatives

Square each derivative: \( \left( \frac{dx}{dt} \right)^2 = e^{2t} (\cos(t) - \sin(t))^2 \) and \( \left( \frac{dy}{dt} \right)^2 = e^{2t} (\sin(t) + \cos(t))^2 \).

Step 5: Add the Squared Derivatives

Add the squared derivatives: \( \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 = e^{2t}((\cos(t) - \sin(t))^2 + (\sin(t) + \cos(t))^2) \). Simplify inside the parentheses: this results in \( e^{2t} (2) = 2e^{2t} \).

Step 6: Set Up the Integral for Arc Length

Substitute into the arc length formula: \( L = \int_0^1 \sqrt{2e^{2t}} \, dt = \sqrt{2} \int_0^1 e^t \, dt \).

Step 7: Evaluate the Integral

Calculate the integral: \( \int e^t \, dt = e^t \). So, \( L = \sqrt{2} [e^t]_0^1 = \sqrt{2} (e - 1) \).

Step 8: Write the Final Answer

The length of the parametric curve over the interval \( 0 \leq t \leq 1 \) is \( L = \sqrt{2}(e - 1) \).

Key Concepts

Parametric Equations

Parametric equations are a pair or set of equations that express the coordinates of the points that make up a geometric object as functions of a variable, usually denoted by \( t \) (called a parameter). In simpler terms, parametric equations allow us to describe a curve using one or more variables. For the exercise at hand, we have the parametric equations:

• \( x(t) = e^t \cos(t) \)
• \( y(t) = e^t \sin(t) \)

These parametric equations define a curve in the plane where the x and y coordinates are functions of \( t \), and \( t \) ranges from 0 to 1. Understanding how to describe a curve using parametric equations gives us a flexible way to analyze and compute various properties of the curve, such as its arc length.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a quantity changes. When dealing with parametric equations, we often need to find the derivative of each component with respect to the parameter \( t \). This gives us insight into how the rate of change in the x-direction and y-direction is affected as \( t \) varies. For instance:

• For \( x(t) = e^t \cos(t) \), we differentiate to get \( \frac{dx}{dt} = e^t \cos(t) - e^t \sin(t) \).
• For \( y(t) = e^t \sin(t) \), we differentiate to find \( \frac{dy}{dt} = e^t \sin(t) + e^t \cos(t) \).

The derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) are crucial for calculating the arc length of the parametric curve, as they tell us how steep the curve is at any given point.

Integrals

Integrals help us find quantities like areas under curves, accumulated change, and in this case, the arc length of a parametric curve. The integral used for calculating arc length takes into account the squared derivatives of the parametric equations. The formula for arc length \( L \) of a parametric curve \( x(t) \), \( y(t) \) over an interval \([a, b]\) is:\[L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt.\]Using this formula involves setting up the integral with the expression inside the square root, as obtained from the derivatives. In our problem, this simplifies and evaluates to \( L = \sqrt{2} [e^t]_0^1 = \sqrt{2} (e - 1) \), representing the arc length from \( t = 0 \) to \( t = 1 \). Integrals convert the slope information provided by derivatives into the actual "distance" along the curve, giving a concrete measure of arc length.

Product Rule

The product rule is a technique used to differentiate functions that are products of two more basic functions. It is vital here because our parametric equations consist of products of exponential and trigonometric functions. The product rule states:\[\frac{d}{dt} [ u(t) \cdot v(t) ] = u(t) \cdot \frac{dv}{dt} + v(t) \cdot \frac{du}{dt}.\]For the function \( x(t) = e^t \cos(t) \):

• Let \( u(t) = e^t \) and \( v(t) = \cos(t) \).
• Apply the product rule to get \( \frac{dx}{dt} = e^t \cdot (-\sin(t)) + \cos(t) \cdot e^t \).

Likewise, when working on \( y(t) = e^t \sin(t) \), the rule is applied similarly. Understanding the product rule is crucial when differentiating complex functions, ensuring each component's rate of change is accurately accounted for when calculating derivatives of parametric equations.