Question

Same paths, different velocity The position functions of objects \(A\) and \(B\) describe different motion along the same path, for \(t \geq 0.\) a. Sketch the path followed by both \(A\) and \(B\). b. Find the velocity and acceleration of \(A\) and \(B\) and discuss the differences. c. Express the acceleration of \(A\) and \(B\) in terms of the tangential and normal components and discuss the differences. $$A: \mathbf{r}(t)=\langle\cos t, \sin t\rangle, B: \mathbf{r}(t)=\langle\cos 3 t, \sin 3 t\rangle$$

Short answer

Question: Analyze and compare the motion of Objects A and B with the given position vector functions: A: $$\mathbf{r}(t)=\langle\cos t, \sin t\rangle$$ and B: $$\mathbf{r}(t)=\langle\cos 3t, \sin 3t\rangle$$. Answer: Both Objects A and B follow a circular path with radius 1. Object A completes one full rotation in a time period of \(2\pi\), while Object B completes three full rotations in the same time period. The velocity and acceleration of Object B are greater than that of Object A due to the faster rotations. The tangential accelerations for both objects are zero, indicating that neither object is speeding up or slowing down. However, the normal accelerations are different: Object A has a normal acceleration of 1, while Object B has a normal acceleration of 9, suggesting that Object B experiences a stronger centripetal force.

Step-by-step solution

Evaluate the position vector functions.

The position vector functions of A and B are given: $$A: \mathbf{r}(t)=\langle\cos t, \sin t\rangle, B: \mathbf{r}(t)=\langle\cos 3 t, \sin 3 t\rangle$$ Both A and B follow a path in the XY plane. We can represent the position of A and B by their coordinates \((x, y)\).

Sketch the paths for A and B.

Given the position vector functions, we can easily see that the paths followed by A and B are circles with radius 1. For Object A, the path has one full rotation with the period of \(2\pi\), while for Object B, the path has three full rotations with the period of \(2\pi\). #b. Find the velocity and acceleration of A and B and discuss the differences.#

Calculate the first and second derivative of the position vector functions for A and B.

Using the chain rule, we can differentiate the position vector functions for A and B to find the velocity and acceleration. For Object A: Velocity \(\mathbf{v_A}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle-\sin t, \cos t\rangle\) Acceleration \(\mathbf{a_A}(t) = \frac{d^2\mathbf{r}(t)}{dt^2} = \langle-\cos t, -\sin t\rangle\) For Object B: Velocity \(\mathbf{v_B}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle-3\sin 3t, 3\cos 3t\rangle\) Acceleration \(\mathbf{a_B}(t) = \frac{d^2\mathbf{r}(t)}{dt^2} = \langle-9\cos 3t, -9\sin 3t\rangle\)

Discuss the differences in velocity and acceleration between A and B.

Comparing the velocity and acceleration vectors of A and B, we can see that Object B has a faster velocity and a higher acceleration than Object A. This is due to multiplying the angle by 3 in the position vector function of B, causing it to complete three full rotations within the same time period as A completes one rotation. #c. Express the acceleration of A and B in terms of the tangential and normal components and discuss the differences#

Compute the unit tangent vectors for A and B.

Unit tangent vectors can be found by dividing the velocity vectors by their magnitudes. For Object A: $$\mathbf{T_A}(t) = \frac{\mathbf{v_A}(t)}{\|\mathbf{v_A}(t)\|} = \frac{\langle-\sin t, \cos t\rangle}{\sqrt{(-\sin t)^2+(\cos t)^2}} = \langle-\sin t, \cos t\rangle$$ For Object B: $$\mathbf{T_B}(t) = \frac{\mathbf{v_B}(t)}{\|\mathbf{v_B}(t)\|} = \frac{\langle-3\sin 3t, 3\cos 3t\rangle}{\sqrt{(-3\sin 3t)^2+(3\cos 3t)^2}} = \langle-3\sin 3t, 3\cos 3t\rangle$$

Calculate the tangential accelerations for A and B.

The tangential acceleration is the projection of the acceleration vector onto the unit tangent vector. For Object A: $$a_{t_A}(t) = \mathbf{a_A}(t) \cdot \mathbf{T_A}(t) = \langle-\cos t, -\sin t\rangle\cdot \langle-\sin t, \cos t\rangle = 0$$ For Object B: $$a_{t_B}(t) = \mathbf{a_B}(t) \cdot \mathbf{T_B}(t) = \langle-9\cos 3t, -9\sin 3t\rangle\cdot \langle-3\sin 3t, 3\cos 3t\rangle = 0$$

Calculate the normal accelerations for A and B.

The normal acceleration is the component of the acceleration vector that is orthogonal to the unit tangent vector. For Object A: $$a_{n_A}(t) = \mathbf{a_A}(t) \cdot \mathbf{N_A}(t) = \|\mathbf{a_A}(t)\| = \sqrt{(-\cos t)^2+(-\sin t)^2} = 1$$ For Object B: $$a_{n_B}(t) = \mathbf{a_B}(t) \cdot \mathbf{N_B}(t) = \|\mathbf{a_B}(t)\| = \sqrt{(-9\cos 3t)^2+(-9\sin 3t)^2} = 9$$

Discuss the differences in tangential and normal components of acceleration between A and B.

Comparing the tangential and normal components of acceleration for Objects A and B, we can see that the tangential accelerations for both objects are zero. This implies that neither Object A nor B is speeding up or slowing down. However, the normal accelerations are different. Object A has a normal acceleration of 1, while Object B has a normal acceleration of 9, which is significantly larger. This suggests that Object B experiences a stronger centripetal force since it completes more rotations in the same time period as Object A.

Key concepts

Velocity in Parametric Equations

In vector calculus, parametric equations are a powerful way to describe the path of an object in motion through space. In this context, we focus on velocity, which is a vector quantity representing the rate of change of position over time.
The parameters in these equations are typically functions of time, allowing us to analyze the motion precisely.
Let’s start by considering objects A and B:

• Object A’s position is given by \( \mathbf{r}_A(t) = \langle \cos t, \sin t \rangle \).
• Object B’s position is \( \mathbf{r}_B(t) = \langle \cos 3t, \sin 3t \rangle \).

The velocity can be determined by differentiating these position vectors with respect to time. For Object A, the velocity is the derivative of the position vector: \( \mathbf{v}_A(t) = \langle -\sin t, \cos t \rangle \). For Object B, it's \( \mathbf{v}_B(t) = \langle -3\sin 3t, 3\cos 3t \rangle \).
Note the factor "3" in Object B’s velocity. This indicates that B moves three times faster along the path compared to A. The form of the velocity vector directly mirrors the periodic change in the parameters, showing a clear relationship between the functions and their rates of change.

Acceleration Components

Acceleration tells us how quickly the velocity of an object is changing. It is derived by differentiating the velocity vector in the parametric equations once more with respect to time. Both tangential and normal components of acceleration are important in understanding motion trajectories.
For Object A, the acceleration is \( \mathbf{a}_A(t) = \langle -\cos t, -\sin t \rangle \) and for B it is \( \mathbf{a}_B(t) = \langle -9\cos 3t, -9\sin 3t \rangle \). Notice here, for B, the acceleration vector is scaled by 9, highlighting significant differences in motion characteristics.

• **Tangential acceleration** is the rate of change of speed along the path.
• **Normal acceleration** is related to the curvature of the path, essential for understanding circular motion dynamics.

For both A and B, the tangential component is zero, indicating that the speed along the path remains constant. However, the normal component differs, with B possessing a higher normal acceleration due to its faster rotation around the circular path.

Circular Motion Analysis

The objects A and B in the given problems travel along circular paths. Analyzing circular motion requires dividing the acceleration into tangential and normal components, particularly significant for understanding the dynamics involved.
Given both objects have zero tangential acceleration, this indicates uniform circular motion where the speed is constant. The normal component, however, is crucial. It represents "centripetal acceleration" - necessary for maintaining the circular path.

• For Object A, the normal acceleration is consistently 1, matched to its constant rotation speed.
• For Object B, which travels faster (as revealed by its higher frequency of rotations), this results in a significantly higher normal acceleration of 9.

This higher value for B points to more considerable centripetal forces required to keep it in its faster circular motion, which essentially means the path's sharpness influences the acceleration experienced. The more intense the motion (i.e., faster and tighter the curve), the more pronounced the centripetal force needed.