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Chapter 12: Problem 5

An object moves on a trajectory given by \(\mathbf{r}(t)=\langle 10 \cos 2 t\) \(10 \sin 2 t\rangle,\) for \(0 \leq t \leq \pi .\) How far does it travel?

Short Answer

Expert verified
Answer: The object travels \(20\pi\) units along the trajectory.

Step by step solution

01

Find the derivative of the position vector

To find \(\mathbf{r}'(t)\), differentiate each component of the position vector \(\mathbf{r}(t)\) with respect to time: \(\mathbf{r}(t) = \langle 10\cos(2t), 10\sin(2t) \rangle\) Taking derivatives of each component, \(\frac{dx}{dt} = -20\sin(2t)\) and \(\frac{dy}{dt} = 20\cos(2t)\) So the derivative of the position vector is: \(\mathbf{r}'(t) = \langle -20\sin(2t), 20\cos(2t) \rangle\)
02

Find the magnitude of the derivative

To find the magnitude of the derivative, use the formula \(\lVert \mathbf{r}'(t) \rVert = \sqrt{(-20\sin(2t))^2 + (20\cos(2t))^2}\) Calculating the magnitude, we have: \(\lVert \mathbf{r}'(t) \rVert = \sqrt{(-20\sin(2t))^2 + (20\cos(2t))^2}\) \(= \sqrt{400\sin^2(2t) + 400\cos^2(2t)}\) Using the trigonometric identity \(\sin^2(x) + \cos^2(x) = 1\), we get: \(\lVert \mathbf{r}'(t) \rVert = \sqrt{400}\) Therefore, the magnitude of the derivative of the position vector is 20.
03

Evaluate the integral to find the arc length

Now that we have the magnitude of the derivative, we can find the arc length by evaluating the integral \(s = \int_0^{\pi} |\mathbf{r}'(t)| dt\): \(s = \int_0^{\pi} 20 \ dt\) Integrating with respect to \(t\) yields: \(s = 20t\big|_0^{\pi}\) Evaluating at the limits, we obtain: \(s = 20(\pi - 0) = 20\pi\) So the object travels a distance of \(20\pi\) units along the trajectory.

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Most popular questions from this chapter

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle-2,2\rangle, \mathbf{v}=\langle 2,1\rangle\)

Consider the lines $$\begin{aligned} \mathbf{r}(t) &=\langle 2+2 t, 8+t, 10+3 t\rangle \text { and } \\ \mathbf{R}(s) &=\langle 6+s, 10-2 s, 16-s\rangle. \end{aligned}$$ a. Determine whether the lines intersect (have a common point) and if so, find the coordinates of that point. b. If \(\mathbf{r}\) and \(\mathbf{R}\) describe the paths of two particles, do the particles collide? Assume \(t \geq 0\) and \(s \approx 0\) measure time in seconds, and that motion starts at \(s=t=0\).

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