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Chapter 12: Problem 65

An object moves clockwise around a circle centered at the origin with radius \(5 \mathrm{m}\) beginning at the point (0,5) a. Find a position function \(\mathbf{r}\) that describes the motion if the object moves with a constant speed, completing 1 lap every 12 s. b. Find a position function \(\mathbf{r}\) that describes the motion if it occurs with speed \(e^{-t}\)

Short Answer

Expert verified
Answer: a. For the constant speed scenario, the position function is \( \mathbf{r}(t) = (5\cos(-\frac{\pi}{6}t),\; 5\sin(-\frac{\pi}{6}t))\). b. For the varying speed scenario, the position function is \( \mathbf{r}(t) = (5\cos(-(-\frac{1}{5}e^{-t} + \frac{1}{5})),\; 5\sin(-(-\frac{1}{5}e^{-t} + \frac{1}{5})))\).

Step by step solution

01

Constants and parameters

Take note of the constants and parameters given. The radius of the circle is 5 meters, the starting position is at point (0, 5), and for part a, the object completes 1 lap every 12 seconds.
02

Find the angular speed for constant speed scenario

For the constant speed scenario, we need to compute the object's angular speed (\(\omega\)), which is the rate at which the angle \(\theta\) changes with respect to time. As the object completes 1 lap in 12 seconds, and one lap corresponds to \(2\pi\) radians, we have: \(\omega = \frac{2\pi}{12} = \frac{\pi}{6}\;\text{rad/s}\).
03

Find the position function for the constant speed scenario

Since the object moves clockwise around the circle, its angular position will decrease instead of increase. This results in a negative angular speed: \(\omega = -\frac{\pi}{6}\;\text{rad/s}\). We can express the position function as \(\mathbf{r}(t) = (5\cos(-\frac{\pi}{6}t),\; 5\sin(-\frac{\pi}{6}t))\). This function gives the position of the object on the circle at any time t.
04

Find speed function for the varying speed scenario

For part b, we are given that the object's speed varies as \(e^{-t}\). Since speed is the magnitude of the velocity vector, we first need to find the tangential velocity. We know that the speed is equal to the radius times the angular speed (\(v = r\omega\)). Therefore, we have \(\omega(t) = \frac{e^{-t}}{5}\).
05

Integrate the angular speed to find the angular position

Integrate the angular speed function with respect to time to find the angular position function: \(\theta(t) = \int \frac{e^{-t}}{5} dt = -\frac{1}{5}e^{-t} + C\). We must find the constant of integration C to satisfy the initial position at (0,5) when t=0: \(\theta(0) = -\frac{1}{5}e^{0} + C = 0 \Rightarrow C = \frac{1}{5}\). So, \(\theta(t) = -\frac{1}{5}e^{-t} + \frac{1}{5}\).
06

Find the position function for the varying speed scenario

Now that we have the angular position function, we can find the position function for the varying speed scenario. Remember that the object moves clockwise around the circle, which means the angle will decrease: \(\mathbf{r}(t) = (5\cos(-\theta(t)),\; 5\sin(-\theta(t))) = (5\cos(-(-\frac{1}{5}e^{-t} + \frac{1}{5})),\; 5\sin(-(-\frac{1}{5}e^{-t} + \frac{1}{5})))\). The position functions for both scenarios are: a. Constant speed: \(\mathbf{r}(t) = (5\cos(-\frac{\pi}{6}t),\; 5\sin(-\frac{\pi}{6}t))\) b. Varying speed: \(\mathbf{r}(t) = (5\cos(-(-\frac{1}{5}e^{-t} + \frac{1}{5})),\; 5\sin(-(-\frac{1}{5}e^{-t} + \frac{1}{5})))\)

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Most popular questions from this chapter

a. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+2 \mathbf{u} \cdot \mathbf{v}+|\mathbf{v}|^{2}\). b. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}+|\mathbf{v}|^{2}\) if \(\mathbf{u}\) is perpendicular to \(\mathbf{v}\). c. Show that \((\mathbf{u}+\mathbf{v}) \cdot(\mathbf{u}-\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}\).

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