Problem 1 - Calculus Early Transcendentals

Question

Find the length of the curve. \(\mathbf{r}(t)=\langle 2 \sin t, 5 t, 2 \cos t\rangle, \quad-10 \leqslant t \leqslant 10\)

Step-by-Step Solution

Answer

The length of the curve is \( 20 \sqrt{29} \).

Step 1: Understand the Formula for Arc Length

The arc length of a curve represented by a vector function \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) from \( t=a \) to \( t=b \) is given by the integral \[ L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \] where \( \mathbf{r}'(t) \) is the derivative of the vector function.

Step 2: Find the Derivative \( \mathbf{r}'(t) \)

First, calculate the derivative of the vector function \( \mathbf{r}(t) = \langle 2 \sin t, 5 t, 2 \cos t \rangle \). The derivative is \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \).

Step 3: Compute the Magnitude \( \| \mathbf{r}'(t) \| \)

Calculate the magnitude of the derivative vector function \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \). The magnitude is \( \| \mathbf{r}'(t) \| = \sqrt{(2 \cos t)^2 + 5^2 + (-2 \sin t)^2} = \sqrt{4 \cos^2 t + 25 + 4 \sin^2 t} \).

Step 4: Simplify the Magnitude Expression

Notice that \( 4 \cos^2 t + 4 \sin^2 t = 4(\cos^2 t + \sin^2 t) = 4 \), hence the magnitude simplifies to \( \| \mathbf{r}'(t) \| = \sqrt{4 + 25} = \sqrt{29} \).

Step 5: Set Up the Integral for Arc Length

Substitute the magnitude into the integral formula for arc length: \[ L = \int_{-10}^{10} \sqrt{29} \, dt \].

Step 6: Evaluate the Integral

Calculate the integral: \[ L = \sqrt{29} \int_{-10}^{10} dt = \sqrt{29} [t]_{-10}^{10} = \sqrt{29} \times (10 - (-10)) = \sqrt{29} \times 20 \].

Step 7: Compute the Final Answer

Compute the product: \[ L = 20 \sqrt{29} \]. This is the length of the curve over the interval \( t = -10 \) to \( t = 10 \).

Key Concepts

Understanding Parametric Equations

Parametric equations are a way to express geometric shapes or paths in multiple dimensions using parameters, often time denoted by the variable \( t \). In the context of three-dimensional space, these equations give us a set of functions that describe each coordinate.
For example, the parametric representation \( \mathbf{r}(t) = \langle 2 \sin t, 5t, 2 \cos t \rangle \) defines a path in space where:

• \( x(t) = 2 \sin t \)
• \( y(t) = 5t \)
• \( z(t) = 2 \cos t \)

These equations together represent a curve in 3D space, with each component changing as \( t \) changes.
The role of parametric equations is crucial in physics and engineering, as they can model any trajectory or path smoothly.

Exploring Vector Functions

Vector functions are pivotal when working with parametric equations in 3D space. They essentially assign a vector to every point along a curve.
In the example \( \mathbf{r}(t) = \langle 2 \sin t, 5t, 2 \cos t \rangle \), the vector function \( \mathbf{r}(t) \) provides comprehensive 3D context.

• Each function term (\( x(t), y(t), z(t) \)) can be seen as a part of a vector in space.
• The rate and direction in which these vectors change define the motion or growth along the curve.

Understanding vector functions involves recognizing their components and their derivatives, which tell us about the velocity and curvature of paths modeled in parametric form.

The Role of Integrals in Calculating Arc Length

Integrals are a cornerstone concept in calculus and are crucial for calculating the arc length of curves, especially those defined parametrically.
To determine the arc length, you'll often integrate the magnitude of a derivative vector function over a given interval. With \( \mathbf{r}'(t) = \langle 2 \cos t, 5, -2 \sin t \rangle \), finding the magnitude \( \| \mathbf{r}'(t) \| \) is the first step.
Here's how integrals play a role:

• Calculate the magnitude of the derivative: this represents instantaneous speed.
• Integrate this magnitude over the given interval to find total length.

Thus, \[ L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \] is the integral giving the arc length of a curve between parameters \( a \) and \( b \). Integrals allow us to piece together small elements of distance over the entire path, leading to the total curve length.

Connecting Calculus Concepts to Parametric Curves

Calculus provides a framework for analyzing motion and change, including how paths dictated by parametric equations behave.
When dealing with parametric curves, differentiation gives us tangent vectors, and integration helps find lengths. Both of these calculus operations come into play profoundly:

• Differentiation: By differentiating our vector function \( \mathbf{r}(t) \), we find \( \mathbf{r}'(t) \), which illustrates the velocity vector of the curve.
• Integration: Allows us to aggregate these differential changes over an interval to determine quantities like total arc length.

Through the lens of calculus, parametric curves become more than just abstract shapes—they link tightly with real-world contexts by quantifying properties like distance and direction over time. This deeper connection underscores calculus's ability to transform and translate geometrical concepts into functional understanding.