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Chapter 12: Problem 1

Write a in the form \(\mathbf{a}=a_{\mathbf{T}} \mathbf{T}+a_{\mathrm{N}} \mathbf{N}\) without finding \(\mathbf{T}\) and \(\mathbf{N}\) $$\mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+b t \mathbf{k}$$

Short Answer

Expert verified
\( \mathbf{a} = a \mathbf{N} \) with \( a_{\mathbf{T}} = 0 \) and \( a_{\mathrm{N}} = a \).

Step by step solution

01

Find Velocity Vector

To write \( \mathbf{a} \) in the form \( \mathbf{a} = a_{\mathbf{T}} \mathbf{T} + a_{\mathrm{N}} \mathbf{N} \), we start by finding the velocity vector \( \mathbf{v}(t) \) by differentiating \( \mathbf{r}(t) \) with respect to \( t \). The velocity vector is \( \mathbf{v}(t) = \frac{d}{dt}((a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} + bt \mathbf{k}) = (-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + b \mathbf{k} \).
02

Find Acceleration Vector

Next, find the acceleration vector \( \mathbf{a}(t) \) by differentiating the velocity vector \( \mathbf{v}(t) \): \( \mathbf{a}(t) = \frac{d}{dt}((-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + b \mathbf{k}) = (-a \cos t) \mathbf{i} + (-a \sin t) \mathbf{j} + 0 \cdot \mathbf{k} = -a(\cos t\mathbf{i} + \sin t \mathbf{j}) \).
03

Calculate Magnitude of Velocity

Calculate the magnitude \( |\mathbf{v}(t)| \) necessary for determining \( a_{\mathbf{T}} \): \[ |\mathbf{v}(t)| = \sqrt{(-a \sin t)^2 + (a \cos t)^2 + b^2} = \sqrt{a^2 \sin^2 t + a^2 \cos^2 t + b^2} = \sqrt{a^2(\sin^2 t + \cos^2 t) + b^2} = \sqrt{a^2 + b^2} \].
04

Calculate Tangential Component

Determine the tangential component \( a_{\mathbf{T}} \) using the formula \( a_{\mathbf{T}} = \frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{v}|} \). Calculate the dot product: \( \mathbf{a} \cdot \mathbf{v} = [(-a \cos t) \mathbf{i} + (-a \sin t) \mathbf{j}] \cdot [(-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + b \mathbf{k}] = a^2 \sin t \cos t + a^2 \cos t \sin t = 0 \). Then, \( a_{\mathbf{T}} = \frac{0}{\sqrt{a^2 + b^2}} = 0 \).
05

Calculate Normal Component

Since \( a_{\mathbf{T}} = 0 \), the entire acceleration is in the normal direction, making \( a_{\mathrm{N}} = \sqrt{\mathbf{a} \cdot \mathbf{a}} = \sqrt{[-a(\cos t) \mathbf{i} + (-a \sin t) \mathbf{j}] \cdot [-a(\cos t) \mathbf{i} + (-a \sin t) \mathbf{j}]} = \sqrt{a^2 \cos^2 t + a^2 \sin^2 t} = \sqrt{a^2} = a \).
06

Write Acceleration in Given Form

Finally, express \( \mathbf{a} \) in the form \( \mathbf{a} = a_{\mathbf{T}} \mathbf{T} + a_{\mathrm{N}} \mathbf{N} \). Here, \( a_{\mathbf{T}} = 0 \) and \( a_{\mathrm{N}} = a \). Therefore, \( \mathbf{a} = 0 \cdot \mathbf{T} + a \cdot \mathbf{N} = a \mathbf{N} \).

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Key Concepts

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

Velocity Vector
The velocity vector is a fundamental concept in vector calculus, particularly when studying motion. It expresses how the position of a particle changes over time. To find the velocity vector from the given position vector \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} + bt \mathbf{k} \), you need to differentiate \( \mathbf{r}(t) \) with respect to time \( t \).
  • The calculation leads to \( \mathbf{v}(t) = (-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + b \mathbf{k} \).
  • This vector indicates that the particle's movement is circular in the \( xy \)-plane while moving linearly along the \( z \)-axis.
Understanding the velocity vector helps us determine the particle's speed and direction at any point in time. Remember, the velocity vector always tangents to the particle's path.
Acceleration Vector
The acceleration vector further elucidates how the velocity of an object changes over time. In our case, we obtain it by differentiating the velocity vector \( \mathbf{v}(t) \).
  • For the given velocity vector \( \mathbf{v}(t) = (-a \sin t) \mathbf{i} + (a \cos t) \mathbf{j} + b \mathbf{k} \), the derivative becomes \( \mathbf{a}(t) = (-a \cos t) \mathbf{i} + (-a \sin t) \mathbf{j} + 0 \cdot \mathbf{k} \).
  • The acceleration vector illustrates that the particle's acceleration is directed in the \( xy \)-plane and depends on \( t \).
Thus, this vector points to how the particle's speed changes in its path and plays a crucial role in describing motion, especially in circular movements.
Tangential Component
The tangential component of acceleration \( a_{\mathbf{T}} \) describes how fast the particle speeds up or slows down along its path. The formula to determine this component is \( a_{\mathbf{T}} = \frac{\mathbf{a} \cdot \mathbf{v}}{|\mathbf{v}|} \).
  • First, calculate the dot product \( \mathbf{a} \cdot \mathbf{v} \) which results in zero in this case.
  • Therefore, \( a_{\mathbf{T}} = \frac{0}{\sqrt{a^2 + b^2}} = 0 \).
This reveals that the acceleration does not have a tangential effect in this scenario. Instead, it implies a purely normal acceleration, which is common in uniform circular motion.
Normal Component
The normal component of acceleration \( a_{\mathrm{N}} \) is responsible for changing the direction of the velocity vector. When \( a_{\mathbf{T}} = 0 \), it indicates that all acceleration is normal to the path.
  • We compute \( a_{\mathrm{N}} \) as \( \sqrt{ \mathbf{a} \cdot \mathbf{a} } \).
  • Evaluating gives \( a_{\mathrm{N}} = \sqrt{a^2} = a \).
This means the whole acceleration of the particle is directed towards the center of the path of motion. In practical terms, it describes the centripetal force component that acts to change the direction of the particle's motion, keeping it in a curved path.

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Most popular questions from this chapter

\(\mathbf{r}(t)\) is the position of a particle in space at time \(t\) Find the particle's velocity and acceleration vectors. Then find the particle's speed and direction of motion at the given value of \(t .\) Write the particle's velocity at that time as the product of its speed and direction. $$\mathbf{r}(t)=\left(e^{-\eta} \mathbf{i}+(2 \cos 3 t) \mathbf{j}+(2 \sin 3 t) \mathbf{k}, \quad t=0\right.$$

Show that if \(\mathbf{u}\) is a unit vector, then the arc length parameter along the line \(\mathbf{r}(t)=P_{0}+t \mathbf{u}\) from the point \(P_{0}\left(x_{0}, y_{0}, z_{0}\right)\) where \(t=0,\) is \(t\) itself.

In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$\mathbf{r}(t)=(t \cos t) \mathbf{i}+(t \sin t) \mathbf{j}+(2 \sqrt{2} / 3) t^{3 / 2} \mathbf{k}, \quad 0 \leq t \leq \pi$$

You will explore graphically the behavior of the helix $$\mathbf{r}(t)=(\cos a t) \mathbf{i}+(\sin a t) \mathbf{j}+b t \mathbf{k}$$ as you change the values of the constants \(a\) and \(b .\) Use a CAS to perform the steps in each exercise. Set \(a=1 .\) Plot the helix \(\mathbf{r}(t)\) together with the tangent line to the curve at \(t=3 \pi / 2\) for \(b=1 / 4,1 / 2,2,\) and 4 over the interval \(0 \leq t \leq 4 \pi .\) Describe in your own words what happens to the graph of the helix and the position of the tangent line as \(b\) increases through these positive values.

Each of the following equations in parts (a)-(e) describes the motion of a particle having the same path, namely the unit circle \(x^{2}+y^{2}=1 .\) Although the path of each particle in parts (a)-(e) is the same, the behavior, or "dynamics," of each particle is different. For each particle, answer the following questions. i) Does the particle have constant speed? If so, what is its constant speed? ii) Is the particle's acceleration vector always orthogonal to its velocity vector? iii) Does the particle move clockwise or counterclockwise around the circle? iv) Does the particle begin at the point (1,0)\(?\) a. \(\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad t \geq 0\) b. \(\mathbf{r}(t)=\cos (2 t) \mathbf{i}+\sin (2 t) \mathbf{j}, \quad t \geq 0\) c. \(\mathbf{r}(t)=\cos (t-\pi / 2) \mathbf{i}+\sin (t-\pi / 2) \mathbf{j}, \quad t \geq 0\) d. \(\mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}, \quad t \geq 0\) e. \(\mathbf{r}(t)=\cos \left(t^{2}\right) \mathbf{i}+\sin \left(t^{2}\right) \mathbf{j}, \quad t \geq 0\)
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