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Chapter 13: Problem 11

In Exercises \(9-14, \mathrm{r}(t)\) 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. $$ \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(3 \sin t) \mathbf{j}+4 t \mathbf{k}, \quad t=\pi / 2 $$

Short Answer

Expert verified
Velocity: \(-2 \mathbf{i} + 4 \mathbf{k}\); Speed: \(2\sqrt{5}\); Direction: \(-\frac{1}{\sqrt{5}} \mathbf{i} + \frac{2}{\sqrt{5}} \mathbf{k}\).

Step by step solution

01

Find the Velocity Vector

To find the velocity vector of the particle, we need to differentiate the position vector \( \mathbf{r}(t) \) with respect to \( t \). The position vector is given by \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + 4t \mathbf{k} \). - The derivative of \( 2 \cos t \) with respect to \( t \) is \( -2 \sin t \). - The derivative of \( 3 \sin t \) with respect to \( t \) is \( 3 \cos t \). - The derivative of \( 4t \) with respect to \( t \) is \( 4 \).Thus, the velocity vector \( \mathbf{v}(t) \) is \( -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \).
02

Find the Acceleration Vector

To find the acceleration vector, differentiate the velocity vector \( \mathbf{v}(t) = -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \) with respect to \( t \). - The derivative of \( -2 \sin t \) is \( -2 \cos t \). - The derivative of \( 3 \cos t \) is \( -3 \sin t \). - The derivative of \( 4 \), which is constant, is \( 0 \).Thus, the acceleration vector \( \mathbf{a}(t) \) is \( -2 \cos t \mathbf{i} - 3 \sin t \mathbf{j} + 0 \mathbf{k} \).
03

Compute Velocity at \( t = \pi/2 \)

To compute the velocity vector at \( t = \pi/2 \), substitute \( \pi/2 \) into the velocity vector \( \mathbf{v}(t) = -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \).- \( -2 \sin(\pi/2) = -2(1) = -2 \).- \( 3 \cos(\pi/2) = 3(0) = 0 \).- The \( \mathbf{k} \) component stays \( 4 \).Thus, \( \mathbf{v}(\pi/2) = -2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} = -2 \mathbf{i} + 4 \mathbf{k} \).
04

Compute Speed at \( t = \pi/2 \)

The speed of the particle is the magnitude of the velocity vector \( \mathbf{v}(t) \). Use the formula \( |\mathbf{v}(t)| = \sqrt{v_x^2 + v_y^2 + v_z^2} \), where \( v_x, v_y, v_z \) are the components of \( \mathbf{v}(t) \).\[ |\mathbf{v}(\pi/2)| = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \].
05

Find Direction of Motion at \( t = \pi/2 \)

The direction of motion is given by the unit vector of the velocity vector \( \mathbf{v}(t) \). It is calculated by dividing \( \mathbf{v}(t) \) by its magnitude:\[ \text{Direction} = \frac{\mathbf{v}(\pi/2)}{|\mathbf{v}(\pi/2)|} = \frac{-2 \mathbf{i} + 4 \mathbf{k}}{2\sqrt{5}} = \left( -\frac{1}{\sqrt{5}} \right) \mathbf{i} + \left( \frac{2}{\sqrt{5}} \right) \mathbf{k} \].
06

Express Velocity as Product of Speed and Direction

Write the velocity at \( t = \pi/2 \) as the product of its speed and the direction:\[ \mathbf{v}(\pi/2) = 2\sqrt{5} \left( -\frac{1}{\sqrt{5}} \mathbf{i} + \frac{2}{\sqrt{5}} \mathbf{k} \right) = -2 \mathbf{i} + 4 \mathbf{k} \].

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Key Concepts

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

Velocity Vector
In particle motion studies, the velocity vector is a crucial concept. The velocity vector provides the rate of change of the particle's position in space. It is derived by differentiating the position vector with respect to time. Given a position vector \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (3 \sin t) \mathbf{j} + 4t \mathbf{k} \), the velocity vector \( \mathbf{v}(t) \) can be determined by taking the derivative of each component:
  • The derivative of \( 2 \cos t \) is \( -2 \sin t \).
  • The derivative of \( 3 \sin t \) is \( 3 \cos t \).
  • The derivative of \( 4t \) is \( 4 \).
Thus, the velocity vector is \( \mathbf{v}(t) = -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \). This vector indicates both the speed and the direction in which the particle is moving at any given time.
Acceleration Vector
The acceleration vector reveals how the velocity of the particle changes over time. To find this, we differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \). Starting from the velocity vector \( \mathbf{v}(t) = -2 \sin t \mathbf{i} + 3 \cos t \mathbf{j} + 4 \mathbf{k} \), the acceleration vector \( \mathbf{a}(t) \) is calculated as follows:
  • The derivative of \( -2 \sin t \) is \( -2 \cos t \).
  • The derivative of \( 3 \cos t \) is \( -3 \sin t \).
  • The derivative of the constant \( 4 \) is zero.
Consequently, the acceleration vector is \( \mathbf{a}(t) = -2 \cos t \mathbf{i} - 3 \sin t \mathbf{j} \). This vector helps us understand how quickly and in which direction the velocity is being adjusted as the particle moves over time.
Speed and Direction of Motion
Speed and direction are key elements of a particle's motion. Speed is the magnitude of the velocity vector, which essentially measures how fast the particle is moving. For a particle at time \( t = \pi/2 \), the velocity vector is \( \mathbf{v}(t) = -2 \mathbf{i} + 0 \mathbf{j} + 4 \mathbf{k} \). You can calculate speed by finding the magnitude of this vector:\[ |\mathbf{v}(\pi/2)| = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{20} = 2\sqrt{5}.\]The direction of motion is represented by the unit vector in the direction of the velocity vector. This is found by dividing the velocity vector by its speed:\[\text{Direction} = \frac{-2 \mathbf{i} + 4 \mathbf{k}}{2\sqrt{5}} = \left( -\frac{1}{\sqrt{5}} \right) \mathbf{i} + \left( \frac{2}{\sqrt{5}} \right) \mathbf{k}.\]Ultimately, expressing the velocity vector as the product of its speed and the direction vector provides a clearer depiction of both how fast and in which direction the particle is moving at any moment.

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Most popular questions from this chapter

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? $$ \begin{array}{l}{\text { a. } \mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, \quad t \geq 0} \\ {\text { b. } \mathbf{r}(t)=\cos (2 t) \mathbf{i}+\sin (2 t) \mathbf{j}, \quad t \geq 0} \\ {\text { c. } \mathbf{r}(t)=\cos (t-\pi / 2) \mathbf{i}+\sin (t-\pi / 2) \mathbf{j}, \quad t \geq 0} \\ {\text { d. } \mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}, \quad t \geq 0} \\ {\text { e. } \mathbf{r}(t)=\cos \left(t^{2}\right) \mathbf{i}+\sin \left(t^{2}\right) \mathbf{j}, \quad t \geq 0}\end{array} $$

You will use a CAS to explore the osculating circle at a point \(P\) on a plane curve where \(\kappa \neq 0 .\) Use a CAS to perform the following steps: a. Plot the plane curve given in parametric or function form over the specified interval to see what it looks like. b. Calculate the curvature \(\kappa\) of the curve at the given value \(t_{0}\) using the appropriate formula from Exercise 5 or \(6 .\) Use the parametrization \(x=t\) and \(y=f(t)\) if the curve is given as a function \(y=f(x)\). c. Find the unit normal vector \(\mathbf{N}\) at \(t_{0} .\) Notice that the signs of the components of \(\mathbf{N}\) depend on whether the unit tangent vector \(\mathbf{T}\) is turning clockwise or counterclockwise at \(t=t_{0} .\) (See Exercise \(7 . )\) d. If \(\mathbf{C}=a \mathbf{i}+b \mathbf{j}\) is the vector from the origin to the center \((a, b)\) of the osculating circle, find the center \(\mathbf{C}\) from the vector equation $$\mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right)$$ The point \(P\left(x_{0}, y_{0}\right)\) on the curve is given by the position vector \(\mathbf{r}\left(t_{0}\right) .\) e. Plot implicitly the equation \((x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}\) of the osculating circle. Then plot the curve and osculating circle together. You may need to experiment with the size of the viewing window, but be sure it is square. \(\mathbf{r}(t)=t^{2} \mathbf{i}+\left(t^{3}-3 t\right) \mathbf{j}, \quad-4 \leq t \leq 4, \quad t_{0}=3 / 5\)

Constant Function Rule Prove that if \(\mathbf{u}\) is the vector function with the constant value \(\mathbf{C},\) then \(d \mathbf{u} / d t=\mathbf{0} .\)

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

In Exercises 9 and \(10,\) write a in the form \(a_{\mathrm{T}} \mathbf{T}+a_{\mathrm{N}} \mathbf{N}\) without finding \(\mathbf{T}\) and \(\mathbf{N} .\) $$ \mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}+b t \mathbf{k} $$
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