Question

Find the arc length of \(\vec{r}(t)\) on the indicated interval. $$ \vec{r}(t)=\langle 2 \cos t, 2 \sin t, 3 t\rangle \text { on }[0,2 \pi] $$

Short answer

The arc length of \( \vec{r}(t) \) from 0 to \( 2\pi \) is \( 2\pi\sqrt{13} \).

Step-by-step solution

Understand the Arc Length Formula

The arc length \( L \) of a space curve \( \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \) over the interval \([a, b]\) is given by the integral \[ L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt \].

Compute Derivatives

First, compute the derivatives of each component of \( \vec{r}(t) = \langle 2 \cos t, 2 \sin t, 3t \rangle \). - \( \frac{d}{dt}[2 \cos t] = -2 \sin t \)- \( \frac{d}{dt}[2 \sin t] = 2 \cos t \)- \( \frac{d}{dt}[3t] = 3 \)

Set up the Integral

Place the derivatives into the arc length formula. The integrand inside the square root becomes \[ \sqrt{(-2 \sin t)^2 + (2 \cos t)^2 + 3^2} = \sqrt{4 \sin^2 t + 4 \cos^2 t + 9} \].

Simplify the Integrand

Notice that \( \sin^2 t + \cos^2 t = 1 \), so the integrand simplifies to \[ \sqrt{4\left(\sin^2 t + \cos^2 t\right) + 9} = \sqrt{4 \times 1 + 9} = \sqrt{13} \].

Integrate

Now integrate over the interval \([0, 2\pi] \): \[ L = \int_0^{2\pi} \sqrt{13} \, dt = \sqrt{13} \times [t]_0^{2\pi} = \sqrt{13} \times (2\pi - 0) = 2\pi \sqrt{13} \].

Key concepts

Vector Calculus

In vector calculus, we explore the properties and applications of vector fields. Vectors are quantities with both magnitude and direction, making them ideal for representing physical phenomena like velocity or force.
For a vector function like \( \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \), each component function describes motion or change in a respective coordinate of space. Vector calculus allows us to analyze and understand these curves better.

We look at derivative and integral operations within vector contexts. When analyzing the arc length of a curve described by vectors, the focus is on computing how far a particle travels along this path.
By applying derivatives to each vector component, we measure how rapidly these positional values change with respect to time \( t \).
Ultimately, this leads to expressing how long a path is, by finding the rate at which this change occurs parallel to each dimension.
Calculating the arc length involves combining each directional derivative, bringing all changes together in the full description of the particle's movement path through three-dimensional space.

Integral Calculus

Integral calculus deals primarily with finding accumulated quantities, such as areas under curves or total values over intervals.
When applied to arc length, integral calculus allows us to calculate the distance along a curve, offering insights into the geometry of space curves.

In our specific case, the integral calculus component is crucial for computing the arc length of the vector curve \( \vec{r}(t) \). To find this length, we use the arc length formula:

• Compute the derivative of each vector component \( \vec{r}(t) = \langle 2 \cos t, 2 \sin t, 3t \rangle \).
• Substitute these into the arc length formula \( L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt \).
• Simplify the expression under the square root, noting simplifications such as \( \sin^2 t + \cos^2 t = 1 \).
• Calculate the integral over the specified interval to find the total arc length.

This step-by-step process reveals the accumulated spatial distance a particle travels along a vector-defined curve.

Parametric Equations

Parametric equations offer a way to express a set of quantities as functions of one or more independent parameters. In our exercise, we use \( t \) as the parameter to describe the position of a point in 3D space over time through the vector \( \vec{r}(t) = \langle 2 \cos t, 2 \sin t, 3t \rangle \).

These equations allow more flexibility compared to standard function representations, providing a natural way to describe complex curves and motions. Each part of a parametric equation explains how a specific spatial dimension changes over the parameter.
Through this representation, trigonometric functions like \( \cos t \) and \( \sin t \) describe circular or oscillating movements, while the linear function \( 3t \) offers constant linear translation.

Parametric equations are particularly handy when dealing with curves that loop or oscillate, as they often do not fulfill function requirements in Cartesian coordinates. They conveniently express movements or shapes that can't easily be described through simple function terms, making them a powerful tool in studying dynamic systems. Bullet points summarize steps, aiding understanding in lengthy processes, as employed in arc length calculation.