Question

, find the length of the parametric curve defined over the given interval.$$ x=2 t-1, y=3 t-4 ; 0 \leq t \leq 3 $$

Short answer

The length of the curve is \( 3\sqrt{13} \).

Step-by-step solution

Write Down the Parametric Equations

The parametric equations for the curve are given by:\( x(t) = 2t - 1 \) and \( y(t) = 3t - 4 \). The curve is defined over the interval \(0 \leq t \leq 3\).

Derive the Functions

Find the derivatives of \( x(t) \) and \( y(t) \) with respect to \( t \). The derivative of \( x(t) = 2t - 1 \) is \( x'(t) = 2 \). The derivative of \( y(t) = 3t - 4 \) is \( y'(t) = 3 \).

Write the Arc Length Formula for Parametric Curves

The formula to find the length of a parametric curve \((x(t), y(t))\) from \( t = a \) to \( t = b \) is \( L = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2} \, dt \).

Substitute the Derivatives into the Arc Length Formula

Substitute \( x'(t) = 2 \) and \( y'(t) = 3 \) into the formula: \( L = \int_0^3 \sqrt{(2)^2 + (3)^2} \, dt \). Simplify the expression inside the square root, \( L = \int_0^3 \sqrt{4 + 9} \, dt = \int_0^3 \sqrt{13} \, dt \).

Evaluate the Definite Integral

The expression becomes a constant multiplying the length of the interval. Thus, \( L = \sqrt{13} \int_0^3 \, dt = \sqrt{13} \times (3 - 0) = 3\sqrt{13} \).

Write the Final Answer

The length of the parametric curve is \( 3\sqrt{13} \).

Key concepts

Arc Length Calculation

Calculating the arc length of a curve is an essential concept in calculus. It involves finding the distance along a curve from one point to another. To determine the arc length of a parametric curve, there's a special formula that comes in handy: \[L = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2} \, dt\]Here, the integral sums up the infinitesimal distances along the curve between two points, where each tiny distance is calculated using the Pythagorean theorem.In practice, once you determine the derivatives of the parametric functions, you just substitute them into this formula, evaluate the integral over the defined interval, and you'll have your arc length. In our original exercise, with the parametric equations \(x=2t-1\) and \(y=3t-4\) over the interval \(0 \leq t \leq 3\), we computed the derivatives first and then assessed their squares within the integral to obtain the result as \(3\sqrt{13}\).

Parametric Equations

Parametric equations express a set of related quantities as explicit functions of an independent parameter, usually denoted as \(t\). For curves, parametric equations separate the traditional \(x\) and \(y\) expressions into two distinct equations:

• \(x(t)\) describes the curve in terms of \(t\).
• \(y(t)\) describes another curve component in terms of the same \(t\).

These equations define a curve by specifying the coordinates of points on the curve as \(t\) varies over a given interval.In the original exercise, the parametric equations \(x(t) = 2t - 1\) and \(y(t) = 3t - 4\) offer a simple linear representation of how \(x\) and \(y\) change with \(t\). Understanding these forms is crucial because it enables us to translate a simple parameter into a position in the Cartesian coordinate system.

Integration in Calculus

Integration is a fundamental tool in calculus, used for adding up infinitely small quantities to get a total amount. In the context of arc length, integration helps determine the total distance along a curve by summing up all the tiny segments of the curve.When we integrate a function over a certain interval, it essentially totals up the area under a curve described by the function within the specified bounds. In the original exercise, we used integration to calculate the arc length of the curve derived from the parametric equations over the interval \(0 \leq t \leq 3\).The task involved integrating the constant \(\sqrt{13}\) over the interval, which became simply multiplying \(\sqrt{13}\) by the length of the interval \(3\). This use of integration shows its power in handling continuous sums over a range, making it crucial for tasks like finding arc length.

Derivatives of Parametric Functions

Derivatives of parametric functions are key to finding the arc length and understanding the behavior of parametric curves. The derivative of a parametrically defined function tells us how the function changes as the parameter \(t\) changes:

• \(x'(t)\) gives the rate of change of \(x\) with respect to \(t\).
• \(y'(t)\) gives the rate of change of \(y\) with respect to \(t\).

In our original problem, we found that the derivatives of the parametric equations \(x(t) = 2t - 1\) and \(y(t) = 3t - 4\) were constants, \(x'(t) = 2\) and \(y'(t) = 3\), indicating constant rates of change.These derivatives played a crucial role in the arc length calculation. They were used in the formula to evaluate the integral which summed up all instantaneous distances along the curve. Understanding how to derive these components is fundamental in parametric calculus, offering insights not just into the geometry of curves but also into their dynamic characteristics.