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Magazine Chapter 12: Problem 12
Find the arc length of the graph of \(\mathbf{r}(t)\) $$ \begin{array}{l}{\mathbf{r}(t)=t^{2} \mathbf{i}+(\cos t+t \sin t) \mathbf{j}+(\sin t-t \cos t) \mathbf{k}} \\ {0 \leq t \leq \pi}\end{array} $$
Short Answer
Expert verified
The arc length is \( \frac{\pi^2 \sqrt{5}}{2} \).
Step by step solution
01
Find the Derivative of the Vector
First, compute the derivative of the vector function \( \mathbf{r}(t) = t^{2} \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + (\sin t - t \cos t) \mathbf{k} \). The derivative is \( \mathbf{r}'(t) = \frac{d}{dt}(t^2)\mathbf{i} + \frac{d}{dt}(\cos t + t \sin t)\mathbf{j} + \frac{d}{dt}(\sin t - t \cos t)\mathbf{k} \). Compute each component's derivative separately.
02
Compute the Derivative Components
Compute each derivative: \( \frac{d}{dt}(t^2) = 2t \), \( \frac{d}{dt}(\cos t + t \sin t) = -\sin t + \sin t + t \cos t = t\cos t \), and \( \frac{d}{dt}(\sin t - t \cos t) = \cos t - \cos t + t \sin t = -t\sin t \). Hence, \( \mathbf{r}'(t) = 2t \mathbf{i} + t\cos t \mathbf{j} - t\sin t \mathbf{k} \).
03
Compute the Magnitude of the Derivative
Find the magnitude of the vector \( \mathbf{r}'(t) = 2t \mathbf{i} + t \cos t \mathbf{j} - t \sin t \mathbf{k} \). The magnitude is \( \|\mathbf{r}'(t)\| = \sqrt{(2t)^2 + (t \cos t)^2 + (-t \sin t)^2} \). Simplify to find \( \|\mathbf{r}'(t)\| = \sqrt{4t^2 + t^2 (\cos^2 t + \sin^2 t)} = \sqrt{5t^2} = \sqrt{5}|t| \).
04
Set Up the Integral for Arc Length
Since \( t \geq 0 \) over the interval \([0, \pi]\), we have \( |t| = t \). The arc length is given by \( L = \int_{0}^{\pi} \| \mathbf{r}'(t) \| \, dt = \int_{0}^{\pi} \sqrt{5} t \, dt \).
05
Evaluate the Integral
Integrate \( \sqrt{5} t \) over \([0, \pi]\). This is \( \sqrt{5} \int_{0}^{\pi} t \, dt \), which is \( \sqrt{5} \left.\frac{t^2}{2}\right|_{0}^{\pi} = \sqrt{5} \left(\frac{\pi^2}{2} - 0\right) = \frac{\pi^2 \sqrt{5}}{2} \).
06
Conclusion
The arc length of the curve is \( \frac{\pi^2 \sqrt{5}}{2} \). This represents the total length of the path traced by the vector function \( \mathbf{r}(t) \) from \( t = 0 \) to \( t = \pi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Functions
Vector functions are a fundamental concept in calculus and are used to describe paths or trajectories in space. A vector function like \( \mathbf{r}(t) = t^{2} \mathbf{i}+(\cos t+t \sin t) \mathbf{j}+(\sin t-t \cos t) \mathbf{k} \) represents a curve in a three-dimensional space.
- The variable \( t \) is called the parameter and defines the position on the curve at any given instant.
- The components \( t^2 \), \( \cos t + t \sin t \), and \( \sin t - t \cos t \) are the functions of \( t \) that describe motion along the \( x \), \( y \), and \( z \)-axes, respectively.
Magnitude of Derivative
The magnitude of the derivative of a vector function is a key step in finding the arc length of a curve. The derivative \( \mathbf{r}'(t) \) of the vector function represents the rate of change of the function and can be understood as the velocity of a moving object.For the given vector function, the derivative is calculated as:- \( \mathbf{r}'(t) = 2t \mathbf{i} + t\cos t \mathbf{j} - t\sin t \mathbf{k} \).To find the magnitude of this derivative, which describes the speed of the object along the curve, we compute:\[ \| \mathbf{r}'(t) \| = \sqrt{(2t)^2 + (t \cos t)^2 + (-t \sin t)^2} \].
- By simplifying this, we determine that the magnitude is \( \sqrt{5} |t| \).
- Since \( t \) is non-negative over the interval \([0, \pi]\), \( |t| = t \), making the magnitude \( \sqrt{5} t \).
Integral Calculation
The calculation of the arc length of a curve involves solving an integral that incorporates the magnitude of the derivative. In this case, the arc length \( L \) from \( t = 0 \) to \( t = \pi \) is determined with:\[ L = \int_{0}^{\pi} \| \mathbf{r}'(t) \| \, dt = \int_{0}^{\pi} \sqrt{5} t \, dt \].This integral describes the summation of infinitesimal sections of the curve, essentially adding up small straight segments of length that approximate the curve.
- By evaluating the integral, we compute: \( \sqrt{5} \int_{0}^{\pi} t \, dt \).
- This evaluates to \( \sqrt{5} \left.\frac{t^2}{2}\right|_{0}^{\pi} \), which after substitution yields the final result \( \frac{\pi^2 \sqrt{5}}{2} \).
