Problem 54 - Calculus III

Question

A particle moves along the plane curve \(C\) 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

The length of the curve over the interval \([0,2]\) is approximately \(4.65\).

Step 1: Find the Derivative of the Position Vector

The position vector of the particle is given by \( \mathbf{r}(t) = t\mathbf{i} + t^2 \mathbf{j} \). To find the velocity, we need the derivative of \( \mathbf{r}(t) \). So, compute \( \mathbf{r}'(t) \):\[ \mathbf{r}'(t) = \frac{d}{dt}(t\mathbf{i} + t^2 \mathbf{j}) = \mathbf{i} + 2t \mathbf{j}. \]

Step 2: Calculate the Magnitude of the Derivative

The magnitude of the velocity vector \( \mathbf{r}'(t) \) 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 from \( t = 0 \) to \( t = 2 \) is found by integrating the magnitude of the derivative:\[ L = \int_{0}^{2} \|\mathbf{r}'(t)\| \, dt = \int_{0}^{2} \sqrt{1 + 4t^2} \, dt. \]

Step 4: Integrate to Find the Arc Length

This integral requires a trigonometric substitution. Set \( 2t = \tan \theta \). Then \( dt = \frac{1}{2}\sec^2 \theta \, d\theta \). The integral transforms as:\[ L = \int \sqrt{1 + \tan^2 \theta} \cdot \frac{1}{2} \sec^2 \theta \, d\theta = \int \frac{1}{2} \sec^3 \theta \, d\theta, \]which is simplified using substitution or a table of integrals, and evaluated from \( t = 0 \) to \( t = 2 \). After substitution back to \( t \), it results in:\[ L = \frac{1}{4} [\sinh^{-1}(2t) + 2t\sqrt{1 + 4t^2}] \Big|_0^2. \]

Step 5: Evaluate the Integral

Substitute the limits into the result:\[ L = \frac{1}{4} \left[\sinh^{-1}(4) + 4\sqrt{17} - \sinh^{-1}(0) - 0\right]. \]This yields the final length. Approximate using a calculator if necessary.

Key Concepts

Derivative of a Vector Function

A vector function represents a curve in space, and for many applications, like finding the path a particle takes, we often need to compute the derivative of this vector function. The derivative of a vector function is a crucial concept, as it represents the velocity of the particle at any given point in time. It's like finding how fast and in what direction the particle is moving. When we take the derivative of each component of the position vector separately, we obtain the vector function's derivative. In this exercise, for the position vector \( \mathbf{r}(t) = t\mathbf{i} + t^2 \mathbf{j} \), the derivative is \[ \mathbf{r}'(t) = \frac{d}{dt}(t\mathbf{i} + t^2 \mathbf{j}) = \mathbf{i} + 2t \mathbf{j}. \] This derivative is the velocity vector.

• A constant \( \mathbf{i} \) signifies a consistent movement along the \( x \)-axis.
• The \( 2t \mathbf{j} \) indicates that motion in the \( y \)-direction changes as \( t \) increases.

Understanding this derivative allows us to further explore the particle's motion by determining its speed and path length.

Magnitude of a Vector

The magnitude of a vector is a measure of its length or size. For a particle moving in space, it's equivalent to the speed of the particle. To compute the magnitude for a vector \( \mathbf{v} = a\mathbf{i} + b\mathbf{j} \), we use the formula:\[ \|\mathbf{v}\| = \sqrt{a^2 + b^2}. \]Breaking it down, you're essentially applying the Pythagorean theorem to find the resultant vector's length. For our example:- The velocity vector is \( \mathbf{r}'(t) = \mathbf{i} + 2t \mathbf{j} \).- Plugging values into the magnitude formula gives us\[ \|\mathbf{r}'(t)\| = \sqrt{(1)^2 + (2t)^2} = \sqrt{1 + 4t^2}. \]This magnitude, \( \sqrt{1 + 4t^2} \), tells us how fast the particle is moving along its path at any time \( t \). This is key to determining the arc length of the curve, as the speed varies depending on \( t \).

Definite Integral

The definite integral is a mathematical concept used to calculate the total accumulation of a quantity, such as area under a curve or, in our case, the length of a path over a specified interval. It involves integrating a function over a specific interval to get a precise result. For the arc length of a curve, we integrate the magnitude of its velocity vector over the interval of \( t \).In this problem, the arc length \( L \) is found using the integral:\[ L = \int_{0}^{2} \sqrt{1 + 4t^2} \, dt. \]This computes the total path length from \( t = 0 \) to \( t = 2 \).The function inside the integral, \( \sqrt{1 + 4t^2} \), means the speed varies with \( t \), and finding the integral helps us accumulate all these individual speeds over the period. Remember, the definite integral gives a numerical result that describes the entire journey between the two points.

Trigonometric Substitution

Trigonometric substitution is a technique commonly used to evaluate integrals containing quadratic expressions, like \( \sqrt{1 + 4t^2} \). The idea is to use a trigonometric identity to simplify the integral, making it easier to handle. In this exercise, the substitution \( 2t = \tan \theta \) is used, which transforms our original integral:- The expression \( \sqrt{1 + \tan^2 \theta} \) becomes \( \sec \theta \) because of the identity \( 1 + \tan^2 \theta = \sec^2 \theta \).Therefore, the integral \[ L = \int \sqrt{1 + 4t^2} \, dt \]transforms to \[ L = \int \frac{1}{2} \sec^3 \theta \, d\theta. \]After making the substitution, solve the integral using known results from a table of integrals or advanced substitution methods.Finally, once the integral in terms of \( \theta \) is evaluated, convert back to \( t \) to find the actual arc length between our bounds.