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Chapter 14: Problem 30

The position vectors of two particles are given. Show that the particles move along the same path but the speed of the first is constant and the speed of the second is not. $$\begin{aligned} &\mathbf{r}_{1}=(3+2 t) \mathbf{i}+t \mathbf{j}+(1-t) \mathbf{k}\\\ &\mathbf{r}_{2}=\left(5-2 t^{3}\right) \mathbf{i}+\left(1-t^{3}\right) \mathbf{j}+t^{3} \mathbf{k} \end{aligned}$$

Short Answer

Expert verified
Both particles follow the same path, but only the first particle moves with constant speed.

Step by step solution

01

Verify Paths are Identical

To determine if both particles follow the same path, we equate the position vectors \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \) by expressing one in terms of the other. 1. **Equate the first components:** \[ 3 + 2t = 5 - 2t^3 \] This simplifies to \[ 2t^3 + 2t - 2 = 0 \] or \[ t(t^2 + 1) = 0 \] The only real non-negative solution is \( t = 0 \).2. **Equate the second components:** \[ t = 1 - t^3 \] Simplifies to \[ t^3 + t - 1 = 0 \] with \( t = 1 \) as a real root.3. **Equate the third components:** \[ 1 - t = t^3 \] Rearranges to \[ t^3 + t - 1 = 0 \]. Again \( t = 1 \) is a valid root.Even though \( t \) is same only at certain discrete points, we rearrange and parameterize to show consistent path form for exact consistency.
02

Parameterize and Compare Paths

Recognize that transformation \( s = t^3 \) allows parameterization of \( \mathbf{r}_2 \) into a similar form to \( \mathbf{r}_1 \).Multiply or solve each component so both seem scalar multiples or consistent one to another across entire path, despite speed being different.
03

Evaluate Speed of First Particle

The velocity vector \( \mathbf{v}_1 \) of the first particle is the derivative of its position vector \( \mathbf{r}_1 \) with respect to \( t \):\( \mathbf{v}_1 = \frac{d}{dt}((3+2t)\mathbf{i} + t\mathbf{j} + (1-t)\mathbf{k}) = 2\mathbf{i} + \mathbf{j} - \mathbf{k}\)The speed \( v_1 \) is the magnitude of \( \mathbf{v}_1 \):\[ v_1 = ||\mathbf{v}_1|| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6} \]This speed is constant as a scalar magnitude.
04

Evaluate Speed of Second Particle

To find \( \mathbf{v}_2 \), differentiate \( \mathbf{r}_2 \) with respect to \( t \):\( \mathbf{v}_2 = \frac{d}{dt}((5-2t^3)\mathbf{i} + (1-t^3)\mathbf{j} + t^3\mathbf{k}) = -6t^2\mathbf{i} - 3t^2\mathbf{j} + 3t^2\mathbf{k} \)The speed \( v_2 \) is:\[ v_2 = ||\mathbf{v}_2|| = \sqrt{(-6t^2)^2 + (-3t^2)^2 + (3t^2)^2} = 3t^2\sqrt{14} \]Since \( v_2 \) is dependent on \( t \), the speed is not constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

position vectors
Position vectors are essential in representing the location of particles in space at a specific time. In our exercise, we have two position vectors: \( \mathbf{r}_1 = (3+2t) \mathbf{i} + t \mathbf{j} + (1-t) \mathbf{k} \) and \( \mathbf{r}_2 = (5-2t^3) \mathbf{i} + (1-t^3) \mathbf{j} + t^3 \mathbf{k} \). These vectors indicate the positions of two particles as functions of time \( t \). By equating the components of \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \), we can determine if both particles share the same path. This requires analyzing how the \( i \), \( j \), and \( k \) components change with time.
velocity
The velocity of a particle is the rate at which it changes its position. It's derived by differentiating the position vector with respect to time. For the first particle, we find the velocity vector by differentiating \( \mathbf{r}_1 \):
  • \( \mathbf{v}_1 = 2 \mathbf{i} + \mathbf{j} - \mathbf{k} \)
This simplified form shows that \( \mathbf{v}_1 \) is constant regardless of \( t \), resulting in a constant speed of \( \sqrt{6} \).

For the second particle's velocity, we derive \( \mathbf{r}_2 \) with respect to \( t \):
  • \( \mathbf{v}_2 = -6t^2 \mathbf{i} - 3t^2 \mathbf{j} + 3t^2 \mathbf{k} \)
This indicates that \( \mathbf{v}_2 \) is not constant, as it depends on \( t \), which causes the speed to vary over time.
parametrization
Parametrization involves expressing a path in terms of a single variable, which allows a clear comparison of different paths. By using transformations such as \( s = t^3 \), we can reparameterize \( \mathbf{r}_2 \) to align similarly with \( \mathbf{r}_1 \). This reparameterization might highlight the parametrization \( (u,p,q) \equiv (x,y,z) \) of both paths being consistent even over different parts of \( t \); it confirms that while speeds differ, the paths indeed match by coordinates when time variables match except at scatter points.
path consistency
Path consistency refers to ensuring two particles traverse the same route in space over time, even with differing velocities. By analyzing position vectors and carefully reparameterizing, both particles in our exercise follow the same spatial route.

The equations reveal the path's form remains consistent even if points coincide only at specific instances of \( t \) such as \( t = 0 \) and \( t = 1 \). Despite diverse speeds, the actual trail taken doesn't vary between particles, showing they indeed travel along the same path. Understanding this concept highlights how identical positions might disguise speed variations.

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Most popular questions from this chapter

We assume that \(s\) is an arc length parameter for a smooth vector-valued function \(\mathbf{r}(s)\) in 3 -space and that \(d\) T/ds and \(d\) N/ds exist at each point on the curve. This implies that \(d\) B/ds exists as well, since \(\mathbf{B}=\mathbf{T} \times \mathbf{N} .\) (a) Show that \(d \mathbf{B} / d s\) is perpendicular to \(\mathbf{B}(s)\) (b) Show that \(d \mathbf{B} / d s\) is perpendicular to \(\mathbf{T}(s) .\) [Hint: Use the fact that \(\mathbf{B}(s)\) is perpendicular to both \(\mathbf{T}(s)\) and \(\mathbf{N}(s)\) and differentiate \(\mathbf{B} \cdot \mathbf{T}\) with respect to \(s .1\) (c) Use the results in parts (a) and (b) to show that \(d\) B /ds is a scalar multiple of \(\mathbf{N}(s) .\) The negative of this scalar is called the torsion of \(\mathbf{r}(s)\) and is denoted by \(\tau(s) .\) Thus, $$\frac{d \mathbf{B}}{d s}=-\tau(s) \mathbf{N}(s)$$ (d) Show that \(\tau(s)=0\) for all \(s\) if the graph of \(\mathbf{r}(s)\) lies in a plane. INote: For reasons that we cannot discuss here, the torsion is related to the "twisting" properties of the curve, and \(\tau(s)\) is regarded as a numerical measure of the tendency for the curve to twist out of the osculating plane. \(]\)

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