Problem 149 - Calculus 3

Question

A particle moves along the plane curve \(\mathrm{C}\) described by \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} .\) Solve the following problems. Find the length of the curve over the interval \([0,2]\)

Step-by-Step Solution

Answer

Use substitution to evaluate the integral for exact values.

Step 1: Differentiate the Position Vector

To find the length of the curve, we start by differentiating the position vector \(\mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j}\) with respect to \(t\).Find \(\mathbf{r}'(t)\):\[\mathbf{r}'(t) = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j}) = \mathbf{i} + 2t \mathbf{j}.\]

Step 2: Calculate the Magnitude of the Derivative

Next, we need to find the magnitude of the derivative \(\mathbf{r}'(t)\). The magnitude is given by:\[\|\mathbf{r}'(t)\| = \sqrt{(1)^2 + (2t)^2} = \sqrt{1 + 4t^2}.\]

Step 3: Set Up the Integral for Arc Length

The length of the curve \(\mathrm{C}\) from \( t = 0 \) to \( t = 2 \) is the integral of the magnitude of \(\mathbf{r}'(t)\):\[L = \int_{0}^{2} \sqrt{1 + 4t^2} \, dt.\]

Step 4: Solve the Integral

To solve the integral \( L = \int_{0}^{2} \sqrt{1 + 4t^2} \, dt \), we use a trigonometric substitution. Let \( t = \frac{1}{2} \tan(\theta) \), then \( dt = \frac{1}{2} \sec^2(\theta) \, d\theta \) and \( \sqrt{1 + 4t^2} = \sec(\theta) \).The limits of integration change from \( t = 0 \) to \( \theta = 0 \) and from \( t = 2 \) to \( \theta = \arctan(4) \).Thus, the integral becomes:\[ L = \int_{0}^{\arctan(4)} \sec(\theta) \cdot \frac{1}{2} \sec^2(\theta) \, d\theta = \frac{1}{2} \int_{0}^{\arctan(4)} \sec^3(\theta) \, d\theta. \]

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, extensively used to understand how things change. When we differentiate a function, we're finding its rate of change. In the context of calculating the arc length of a parametric curve, differentiation plays a crucial role. For example, consider the particle moving along the curve with position vector \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \). Here, differentiating helps us find the velocity of the particle at any point in time.

When we differentiate the position vector \( \mathbf{r}(t) \) with respect to \( t \), as shown in the solution step, we get:

• \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \).

This derivative represents the velocity vector of the particle, indicating how the position changes as time changes.

Understanding this rate of change (velocity) is essential because it contributes to finding the arc length, which measures the distance traveled along a path.

Parameterization

Parameterization is a method of expressing a curve in terms of a parameter, often represented by \( t \). This approach allows us to describe complex curves using simpler functions. Instead of defining a curve by a single equation in terms of \( x \) and \( y \), parameterization uses separate expressions for each coordinate.

In our example, the curve is given by the vector \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \). Here, \( t \) is the parameter that describes the position of a point along the curve as time progresses from 0 to 2. This parameterization encapsulates the curve in its entirety, offering a straightforward way to analyze the motion and geometry by considering how \( x = t \) and \( y = t^2 \) change with \( t \).

By using parameterization, we can systematically track the particle's path and evaluate properties like distance or speed, which are crucial for applications such as finding the length of the curve.

Trigonometric Substitution

Trigonometric substitution is a technique often used to simplify the evaluation of integrals involving square roots, like \( \sqrt{1 + 4t^2} \). This method takes advantage of trigonometric identities to transform complex expressions into simpler ones, making integration feasible.

In the given exercise, to solve the integral \( L = \int_{0}^{2} \sqrt{1 + 4t^2} \, dt \), a trigonometric substitution is made. By letting \( t = \frac{1}{2} \tan(\theta) \), the integral is transformed from a function of \( t \) into a function of \( \theta \). The limits of integration also change accordingly:

• When \( t = 0 \), \( \theta = 0 \).
• When \( t = 2 \), \( \theta = \arctan(4) \).

This substitution uses the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \) to simplify \( \sqrt{1 + 4t^2} \) to \( \sec(\theta) \), allowing us to integrate \( \sec^3(\theta) \) more manageably. It demonstrates how trigonometric identities efficiently handle otherwise complex integral forms.