91Ó°ÊÓ

The diagram shows a merry-go-round that is turning counterclockwise at a constant rate, making 2 revolutions in 1 minute. On the merry-go-round are horses \(A, B, C,\) and \(D\) at 4 meters from the center and horses \(E, F,\) and \(G\) at 8 meters from the center. There is a function \(a(t)\) that gives the distance the horse \(A\) is from the \(y\) -axis (this is the \(x\) -coordinate of the position \(A\) is in ) as a function of time \(t\) (measured in minutes). Similarly, \(b(t)\) gives the \(x\) -coordinate for \(B\) as a function of time, and so on. Assume that the diagram shows the situation at time \(t=0\). (Check your book to see figure) (a) Which of the following functions does \(a(t)\) equal? $$\begin{array}{ll}4 \cos t, & 4 \cos \pi t, \quad 4 \cos 2 t, \quad 4 \cos 2 \pi t \\\4 \cos \left(\frac{1}{2} t\right), & 4 \cos ((\pi / 2) t), \quad 4 \cos 4 \pi t\end{array}$$ Explain. (b) Describe the functions \(b(t), c(t), d(t),\) and so on using the cosine function: $$\begin{array}{l}b(t)=\longrightarrow(t)=\longrightarrow d(t)= \\\e(t)=\longrightarrow f(t)=\longrightarrow g(t)=\end{array}$$ (c) Suppose the \(x\) -coordinate of a horse \(S\) is given by the function \(4 \cos (4 \pi t-(5 \pi / 6))\) and the \(x\) -coordinate of another horse \(T\) is given by \(8 \cos (4 \pi t-(\pi / 3))\) Where are these horses located in relation to the rest of the horses? Mark the positions of \(T\) and \(S\) at \(t=0\) into the figure.

Step by step solution

01

(a) Find \(a(t)\) #

To find the correct expression for \(a(t)\), we can start by examining the diagram and noting that at \(t=0\), horse \(A\) is on the positive x-axis and 4 meters away from the center. We can see that the upper part will complete 2 revolutions in 1 minute which means that the angular velocity is 4Ï€ radians per minute or 2Ï€ radians per 30 seconds (one revolution). In general, the equation for the position of an object rotating in a circle with constant angular velocity is: $$x(t) = r \cos (\omega t - \phi)$$ where r is the radius (distance from the center), \(\omega\) is the angular velocity, and \(\phi\) is the phase shift. Horse \(A\) has a 0 radians phase shift at \(t=0\) because it is on the x-axis. Now, we can choose the correct expression for \(a(t)\). We know that: 1. The radius is 4 meters. 2. The angular velocity must be 4Ï€ radians per minute. 3. There should be no phase shift. From these conditions, we get a(t) = 4 cos(4Ï€t). So \(a(t)\) = \(4\cos 4\pi t\).

02

(b) Find the expressions for the rest of the horses #

Following the same logic as above, we derive the expressions for the remaining horses' x-coordinates. The angular velocities will be the same for all the horses on this merry-go-round, so all the cosine functions will have "\(4\pi t\)" in them. 1. Horse B, with a quarter-turn (Ï€/2) counterclockwise from Horse A: \(B(t) = 4\cos(4\pi t - \pi/2)\). 2. Horse C, with a half-turn (Ï€) counterclockwise from Horse A: \(C(t) = 4\cos(4\pi t - \pi)\). 3. Horse D, with a three-quarter turn (3Ï€/2) counterclockwise from Horse A: \(D(t) = 4\cos(4\pi t - 3\pi/2)\). 4. Horse E, with the same angle as Horse A and twice the radius (8 meters from center): \(E(t) = 8\cos(4\pi t)\). 5. Horse F, with a quarter-turn (Ï€/2) counterclockwise from Horse E: \(F(t) = 8\cos(4\pi t - \pi/2)\). 6. Horse G, with a half-turn (Ï€) counterclockwise from Horse E: \(G(t) = 8\cos(4\pi t - \pi)\).

03

(c) Determine the locations of horses S and T #

To determine the locations of horses S and T, we can analyze their functions and compare them with the functions of the existing horses. 1. Horse S: \(4\cos(4\pi t - (5\pi/6))\) From this expression, we can observe that Horse S has a radius of 4 meters (same as A, B, C, and D) and a phase shift of \((5\pi/6)\). Since Horse S is on the same circle as A, B, C, and D, we can conclude that it is located between Horse B and Horse C, slightly closer to Horse B. 2. Horse T: \(8\cos(4\pi t - (\pi/3))\) Similar to Horse S, we can see that Horse T has a radius of 8 meters (same as E, F, and G) and a phase shift of \((\pi/3)\). Therefore, it is between Horse E and Horse F, slightly closer to Horse E. Mark the positions of T and S at \(t=0\) accordingly on the figure.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Velocity

Angular velocity is a measure of how quickly an object is revolving around a central point. In this context, it refers to how fast the merry-go-round is spinning. It can be expressed in radians per unit of time, such as seconds or minutes. For our exercise, the merry-go-round completes 2 revolutions per minute, which equates to an angular velocity of \(4 \pi\) radians per minute. This means the entire circle, equivalent to \(2\pi\) radians, is covered twice every minute.

• Defining Angular Velocity: The rate at which the angle is changing with respect to time in circular motion.
• Units: Commonly used units include radians per second or radians per minute.
• Expression in Equations: Used in calculating positions in circular motion, such as \(x(t) = r \cos(\omega t - \phi)\).

Phase Shift

Phase shift refers to the initial angle or offset at which a trigonometric function such as cosine begins. It can determine where a horse starts on the merry-go-round at time \(t=0\). In this exercise, Horse A starts on the positive x-axis, which indicates a phase shift of 0 radians. Each other horse's phase shift is based on its starting position relative to Horse A.

• No Phase Shift: Horse A starts from the x-axis without any offset, meaning \(\phi = 0\).
• Determining Phase Shift: Calculated based on the horses' starting positions relative to the x-axis and each other.
• Example Use: Equations like \(B(t) = 4 \cos(4\pi t - \pi/2)\) incorporate phase shifts for Horse B.

Circular Motion

Circular motion is the movement of an object along the circumference of a circle. In the context of the merry-go-round, horses are moving in a circular path at a constant rate. This introduces several dynamic aspects that can be analyzed using trigonometric functions like cosine.

• Uniform Circular Motion: Occurs when the speed of rotation is constant, as with the merry-go-round.
• Influence on Position Functions: Whichever point you're looking at, such as horse A's position in \(a(t)\), it will be influenced by the radius, angular velocity, and phase shift.
• More Than Movement: Circular motion analyses often extend to forces acting on the object, such as centripetal force which keeps the object in its path.

Cosine Function

The cosine function is a periodical function from trigonometry that is used to model wave-like or cyclical behaviors. In this exercise, cosine functions are used to define the x-coordinate positions of horses.

• Why Cosine: Useful for determining horizontal positions of points rotating in a circle, such as \(4\cos(4\pi t)\).
• Period Functionality: Relates to angular revolutions due to its periodic nature, repeating every \(2\pi\) radians.
• Real-life Application: Beyond theory, the cosine function helps solve problems involving oscillations and waves.

Revolutions

Revolutions measure how many complete turns an object makes around a center or axis within a specific period. For the merry-go-round exercise, it completes 2 revolutions every minute.

• Relation to Angular Velocity: The revolutions per minute directly translate to angular velocity, with 2 full turns being equivalent to \(4\pi\) radians.
• Periodic Nature: Revolutions describe the cyclic pattern of motion, essential for calculating positions using trigonometric functions.
• Practical Understanding: Knowing the number of revolutions helps predict positions at any time \(t\), by relating time and rotational speed.