Problem 112 - Calculus 3

Question

Find the length of the curve \(\mathbf{r}(t)=\langle 2 \sin t, 5 t, 2 \cos t\rangle\) for \(t \in[-10,10]\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the length of the curve defined by the vector function \(\mathbf{r}(t)=\langle 2 \sin t, 5 t, 2 \cos t \rangle\) over the interval \([-10, 10]\).

Step 2: Recall the Arc Length Formula

The arc length of a vector function \(\mathbf{r}(t)\) from \(t = a\) to \(t = b\) is given by the integral: \( L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \, dt \).

Step 3: Compute Derivatives

Calculate the derivatives of each component of \(\mathbf{r}(t)\): - \(\frac{dx}{dt} = \frac{d}{dt} [2 \sin t] = 2 \cos t\) - \(\frac{dy}{dt} = \frac{d}{dt} [5t] = 5\) - \(\frac{dz}{dt} = \frac{d}{dt} [2 \cos t] = -2 \sin t\)

Step 4: Substitute into the Arc Length Integral

Substitute the derivatives into the arc length formula:\[ L = \int_{-10}^{10} \sqrt{ (2 \cos t)^2 + (5)^2 + (-2 \sin t)^2 } \, dt \]

Step 5: Simplify the Expression Inside the Integral

Simplify the expression inside the square root:\[ (2 \cos t)^2 + 5^2 + (-2 \sin t)^2 = 4 \cos^2 t + 25 + 4 \sin^2 t \] which further simplifies to:\[ 4(\cos^2 t + \sin^2 t) + 25 = 4 \cdot 1 + 25 = 29 \]

Step 6: Evaluate the Integral

Substitute back into the integral:\[ L = \int_{-10}^{10} \sqrt{ 29 } \, dt \]Since \(\sqrt{29}\) is a constant, the integral simplifies to:\[ L = \sqrt{29} \times (10 - (-10)) = \sqrt{29} \times 20 \]

Step 7: Calculate the Final Answer

Calculate \( L = 20\sqrt{29} \). This is the length of the curve.

Key Concepts

Vector Calculus

Vector calculus is an essential branch of mathematics concerned with using vectors to describe and analyze physical quantities that have both magnitude and direction. These include forces, velocity, and acceleration. In vector calculus, we often deal with vector functions that give us the position of a point in space as a function of one or more parameters. These vector functions can be of any dimension, but in our exercise, we are dealing with a three-dimensional vector function:

• \(\mathbf{r}(t)=\langle 2 \sin t, 5t, 2 \cos t \rangle\)

With this vector function, each component describes a different axis, such as x, y, and z. By expressing physical situations in terms of vector functions, we can use calculus to solve complex problems involving motion and change. Calculations often involve taking derivatives or integrals, which help us to analyze the behavior and attributes of the function over time or space.

Parametric Equations

Parametric equations allow us to parameterize curves, surfaces, and more in mathematics, which means that we describe them in terms of parameters. These equations give us a way to express the coordinates of the points on a curve as functions of a third variable, often denoted as \(t\).

• For example, in the vector function above, \(t\) is the parameter.

The vector function \(\mathbf{r}(t)=\langle 2 \sin t, 5t, 2 \cos t \rangle\) is expressed in parametric form. Here, each coordinate is given as a function of \(t\):

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

By manipulating the parameter \(t\), we can trace out the complete path of the curve in space, allowing us to understand its geometric properties.

Arc Length Formula

The arc length formula helps us calculate the distance along a curve over a specified interval. In vector calculus, this is given by the integral:

• \( L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \, dt \)

For our curve \(\mathbf{r}(t)=\langle 2 \sin t, 5t, 2 \cos t \rangle\), the formula requires us to compute derivatives for each vector component:

• \(\frac{dx}{dt} = 2 \cos t\)
• \(\frac{dy}{dt} = 5\)
• \(\frac{dz}{dt} = -2 \sin t\)

These are substituted into the integral, which, upon simplifying, gives us an integrand of \(\sqrt{29}\). This simplifies the process, allowing us to compute the final length, \(L = 20\sqrt{29}\), which represents the complete path length of the specified curve over the interval \([-10, 10]\). Understanding this formula is crucial not only for this specific curve but also for determining the arc length of any parametric or vector function.