Question

For the following exercises, find the arc length of the curve over the given interval. $$ x=3 t+4, y=9 t-2,0 \leq t \leq 3 $$

Short answer

The arc length of the curve is \(9\sqrt{10}\).

Step-by-step solution

Understanding Parametric Equations

The problem is given in terms of parametric equations where \( x \) and \( y \) are functions of a parameter \( t \). We have \( x = 3t + 4 \) and \( y = 9t - 2 \). Our goal is to find the arc length over the interval \( 0 \leq t \leq 3 \).

Formula for Arc Length of Parametric Equations

The arc length \( L \) of a curve given by parametric equations \( x(t) \) and \( y(t) \) over an interval \( [a, b] \) is calculated using \[ L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt. \]

Finding Derivatives

Calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). \( \frac{dx}{dt} = \frac{d}{dt}(3t + 4) = 3 \) \( \frac{dy}{dt} = \frac{d}{dt}(9t - 2) = 9 \).

Substitute and Simplify

Substitute \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 9 \) into the arc length formula: \[ L = \int_0^3 \sqrt{3^2 + 9^2} \, dt = \int_0^3 \sqrt{9 + 81} \, dt = \int_0^3 \sqrt{90} \, dt. \] Simplify the expression: \( \sqrt{90} = 3\sqrt{10} \), so \[ L = \int_0^3 3 \sqrt{10} \, dt. \]

Evaluate the Integral

Integrate the expression: \[ L = \int_0^3 3 \sqrt{10} \, dt = 3 \sqrt{10} \times [t]_0^3 = 3 \sqrt{10} \times (3 - 0). \] Therefore, \[ L = 9 \sqrt{10}. \]

Key concepts

Parametric Equations

Parametric equations are a way to describe a curve in terms of a parameter, usually denoted as \( t \). In a parametric form, both \( x \) (the horizontal component) and \( y \) (the vertical component) are expressed as separate functions of \( t \). For example, with the equations \( x = 3t + 4 \) and \( y = 9t - 2 \), we can find any point on the curve by simply plugging a value of \( t \) into both equations. This method is particularly useful for representing more complex curves that cannot be easily described by a single function \( y = f(x) \).

Using parametric equations allows for:

• Easy representation of curves that loop or have vertical tangents, which are difficult in normal function form.
• Efficient computation when the curve is defined by movement or pathways, such as in physics.

Understanding parametric equations is fundamental in calculating arc length because these equations help in breaking down the curve into manageable components.

Integral Calculus

Integral calculus is a branch of calculus that focuses on the concept of integration, which is essentially the process of finding the whole from knowing the rate of change. In the context of arc length, integration is utilized to compute the total length of a curve described by parametric equations.

The integral calculus enables us to sum an infinite number of infinitesimally small pieces of the curve to obtain the total arc length. The formula used for arc length calculation involves:

• Squaring the derivatives of the parametric equations, \( \left( \frac{dx}{dt} \right)^2 \) and \( \left( \frac{dy}{dt} \right)^2 \).
• Adding them together within a radical sign, forming part of the integrand of the arc length formula.
• Integrating this expression over the specified interval \([a, b]\).

This process is pivotal in many real-world applications, such as determining the actual path length traveled by an object or designing precise mechanical components where exact dimensions are crucial.

Derivatives

Derivatives are a key concept in calculus representing an instantaneous rate of change or the slope of a function at any given point. When working with parametric equations like \( x = 3t + 4 \) and \( y = 9t - 2 \), finding the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) is a crucial step because they describe how \( x \) and \( y \) change with respect to the parameter \( t \).

The derivative serves different purposes:

• Part of the arc length formula, where you need to take the square of the derivatives of each component, \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
• Giving information on the changing speed and direction of the moving point described by the parametric equations.

In our specific problem, since \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 9 \), these derivatives are constants, which simplifies the arc length calculation considerably. Understanding derivatives in this context allows one to transition smoothly from differential equations to their integral counterparts to solve real-world problems.