Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find the length of the parametric curve defined over the given interval. $$ x=2-t, y=2 t-3 ;-3 \leq t \leq 3 $$

Step-by-Step Solution

Answer

The length of the parametric curve is \( 6\sqrt{5} \).

Step 1: Understand the Parametric Equations

The parametric equations provided are:\( x = 2 - t \) and \( y = 2t - 3 \).These describe the path traced by a point as the parameter \( t \) varies over the interval \(-3 \leq t \leq 3\).

Step 2: Calculate Derivatives

To find the length of the curve, we need to calculate the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):- Derivative of \( x \): \( \frac{dx}{dt} = -1 \)- Derivative of \( y \): \( \frac{dy}{dt} = 2 \).

Step 3: Find the Differential Length Element

The formula for the differential arc length element \( ds \) is given by:\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]Substitute the derivatives:\[ ds = \sqrt{(-1)^2 + 2^2} \, dt = \sqrt{1 + 4} \, dt = \sqrt{5} \, dt \]

Step 4: Integrate over the Given Interval

We integrate \( ds \) from \( t = -3 \) to \( t = 3 \) to find the total length:\[L = \int_{-3}^{3} \sqrt{5} \, dt = \sqrt{5} \int_{-3}^{3} \, dt\]Calculate the integral:\[L = \sqrt{5} \times [t]_{-3}^{3} = \sqrt{5} \times (3 - (-3)) = \sqrt{5} \times 6 = 6\sqrt{5}\]

Step 5: State the Result

The length of the parametric curve over the interval from \( t = -3 \) to \( t = 3 \) is \( 6\sqrt{5} \).

Key Concepts

Parametric Equations

Parametric equations are a powerful tool in mathematics, often used to describe a curve in a plane by expressing the coordinates of the points on the curve as functions of a variable, called a parameter. For the problem at hand, the parametric equations are given as:

• \( x = 2 - t \)
• \( y = 2t - 3 \)

These equations articulate how both the \(x\) and \(y\) coordinates change as the parameter \(t\) changes within the given range from \(-3\) to \(3\). In contrast to Cartesian coordinates, which use a single equation, parametric equations offer clarity for problems where coordinates change independently.
They are particularly useful for tracing complex paths like circles or ellipses, which are difficult to describe using just \(x\) and \(y\) alone.

Arc Length Formula

The arc length formula is crucial when calculating the length of a curve described by parametric equations.
The basic idea is to sum up the small pieces of the curve between two points.
This summation gives us a good approximation of the actual curve length.For parametric curves, the differential arc length element \( ds \) is expressed as:

• \[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

This formula accounts for small changes in both the \(x\) and \(y\) directions, combining them to find the distance traveled along the curve.
In the given exercise, substituting the computed derivatives \(\frac{dx}{dt} = -1\) and \(\frac{dy}{dt} = 2\), we obtain:

• \[ ds = \sqrt{(-1)^2 + 2^2} \, dt = \sqrt{5} \, dt \]

This simplifies our task when integrating to find the total curve length over the defined interval.

Calculus

Calculus plays a significant role in studying parametric curves and finding arc lengths.
It provides us valuable techniques, both differentiation and integration, to solve these problems.To find the length of a parametric curve, we use integration to sum up an infinite number of infinitesimally small pieces of the curve.
This process requires prior steps of differentiation to establish the rate of change of the curve's coordinates (\(x\) and \(y\)) with respect to the parameter \(t\).In our exercise, after differentiating the equations to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), we integrated the result over the interval from \(-3\) to \(3\):

• \[ L = \int_{-3}^{3} \sqrt{5} \, dt \]

This represents the application of calculus, transforming our understanding of how the curve behaves over its entire length.

Differentiation

Differentiation is a cornerstone of calculus that allows us to determine the rate at which something changes.
In the context of parametric curves, we differentiate the parametric equations to discover how quickly the \(x\) and \(y\) coordinates shift as the parameter \(t\) changes.For the given equations:

• \( x = 2 - t \)
• \( y = 2t - 3 \)

The derivatives are:

• \( \frac{dx}{dt} = -1 \)
• \( \frac{dy}{dt} = 2 \)

These derivatives provide the necessary components for our arc length formula, \( ds = \sqrt{(-1)^2 + 2^2} \, dt = \sqrt{5} \, dt \).Differentiation not only helps calculate arc lengths but also describes the dynamics of motion along the curve such as velocity and acceleration, which are essential for comprehensive curve analysis.