Question

Find the arc length of the given curve. \(x=t, y=t, z=2 t ; 0 \leq t \leq 2\)

Short answer

The arc length of the curve is \(2\sqrt{6}\).

Step-by-step solution

Parameterize the Curve

The given curve is parameterized by \(x = t\), \(y = t\), and \(z = 2t\) with \(0 \leq t \leq 2\). Since each coordinate is expressed in terms of \(t\), the parameterization is complete.

Determine the Derivative of the Parameterized Functions

Calculate the derivatives of each component with respect to \(t\). Compute \(\frac{dx}{dt} = 1\), \(\frac{dy}{dt} = 1\), and \(\frac{dz}{dt} = 2\).

Use the Arc Length Formula

The arc length \(L\) of a parametric curve is given by the formula: \[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\, dt\]

Substitute the Derivatives into the Formula

Substitute the derivatives into the arc length formula: \[L = \int_{0}^{2} \sqrt{(1)^2 + (1)^2 + (2)^2} \, dt = \int_{0}^{2} \sqrt{1 + 1 + 4} \, dt = \int_{0}^{2} \sqrt{6} \, dt\]

Calculate the Arc Length Integral

Calculate the integral: \[L = \int_{0}^{2} \sqrt{6} \, dt = \sqrt{6} \times [t]_{0}^{2} = \sqrt{6} \times (2 - 0) = 2\sqrt{6}\]

Conclusion of the Calculation

After performing the integral, we find that the arc length of the curve from \(t=0\) to \(t=2\) is \(2\sqrt{6}\).

Key concepts

Parametric Equations

Parametric equations are a powerful tool in calculus, allowing us to describe mathematical curves in a flexible and practical way. Unlike the typical Cartesian equation, which relates two variables such as x and y directly, parametric equations introduce a third variable, often t (representing time), to express both x and y in terms of this new parameter.
For example, in the problem, we have a space curve parameterized by \(x = t\), \(y = t\), and \(z = 2t\) where \(0 \leq t \leq 2\). Here, t acts as a timer, tracking the position of a point along the curve as t varies from 0 to 2. This way of representing curves is particularly useful in physics and engineering, where motions and trajectories need to be analyzed.
In general, parametric equations allow us to efficiently manage multi-variable functions and gain additional insights into the geometry and trajectory of objects.

Arc Length Formula

The arc length formula helps us compute the distance along a curve defined by parametric equations. The arc length L of a curve parameterized by t, with components \(x(t)\), \(y(t)\), and possibly \(z(t)\), is given by an integral:
\[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt\]
This integral formula adds the squares of the derivatives of each coordinate function with respect to t, providing a way to measure how far the curve extends in the 3D space between two bounds \(a\) and \(b\).
In simpler terms, the formula calculates how much the position changes in the x, y, and z directions, combines these changes, and then sums these changes over the interval.

Calculus Problem Solving

Calculus is central to tackling a wide range of mathematical problems, including finding arc lengths. It offers a set of tools and processes for measuring and analyzing continuous change, which includes the motion of objects described by parametric equations.
In solving this specific problem, calculus provides the method to break down the curve into infinitesimally small line segments which can be summed as an integral.
The process involves:

• Parameterizing the curve.
• Finding the derivatives of each parameterized component.
• Applying the arc length formula as an integral, incorporating these derivatives.

Using calculus, intricate shapes can be measured precisely, allowing one to answer questions about the total length of a path defined in a multi-dimensional space.

Calculus Integration

Integration is a fundamental concept in calculus used to calculate the accumulation of quantities, such as areas under curves or lengths of curves, like the arc length in our exercise. In the context of parametric equations, it helps in adding up an infinite number of infinitely small quantities, which in this problem are segments of the curve.
The actual process of integration takes the function inside the integral, such as \(\sqrt{6}\) in this exercise, and determines the total length along the curve between the lower bound \(t = 0\) and the upper bound \(t = 2\).
Here’s how it's applied:

• Substitute the derivatives into the arc length integral setup.
• Evaluate the integral over the specified interval.
• Obtain the result, in this case, \(2\sqrt{6}\) as the arc length.

This process demonstrates how powerful integration is in transforming a problem involving continuously changing values into a solvable, quantifiable result.