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a particle is located on the coordinate plane at $(5,0)$. define a ''move'' for the particle as a counterclockwise rotation of $\frac{\pi}{4}$ radians about the origin followed by a translation of $10$ units in the positive $x$-direction. find the particle's position after $150$ moves.

User Cybaek
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1 Answer

4 votes

Answer:after 150 moves, the particle will be located at the coordinates (10, 10).

Explanation:

To find the particle's position after 150 moves, we can break down each move into two steps: a counterclockwise rotation of π/4 radians followed by a translation of 10 units in the positive x-direction. We can repeat these two steps 150 times.

Step 1: Counterclockwise Rotation of π/4 Radians

A counterclockwise rotation of π/4 radians can be represented by the transformation matrix:

\cos(\theta) & -\sin(\theta) \\

\sin(\theta) & \cos(\theta)

\end{bmatrix}\]

In this case, θ = π/4:

\[R(\frac{\pi}{4}) = \begin{bmatrix}

\cos(\frac{\pi}{4}) & -\sin(\frac{\pi}{4}) \\

\sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4})

\end{bmatrix} = \begin{bmatrix}

\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\

\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}

\end{bmatrix}\]

Step 2: Translation of 10 Units in the Positive x-Direction

A translation in the positive x-direction can be represented by adding 10 to the x-coordinate:

\[T(10) = \begin{bmatrix}

1 & 0 \\

0 & 1

\end{bmatrix}\begin{bmatrix}

10 \\

0

\end{bmatrix} = \begin{bmatrix}

10 \\

0

\end{bmatrix}\]

Now, let's apply these two steps (rotation and translation) 150 times to the initial position (5, 0).

Starting position: (5, 0)

After one move: \[R(\frac{\pi}{4}) \cdot T(10) \cdot (5, 0) = \begin{bmatrix}

\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\

\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}

\end{bmatrix} \cdot \begin{bmatrix}

10 \\

0

\end{bmatrix} = \begin{bmatrix}

5\sqrt{2} \\

5\sqrt{2}

\end{bmatrix}\]

After two moves: \[R(\frac{\pi}{4}) \cdot T(10) \cdot \begin{bmatrix}

5\sqrt{2} \\

5\sqrt{2}

\end{bmatrix} = \begin{bmatrix}

\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\

\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}

\end{bmatrix} \cdot \begin{bmatrix}

10 \\

0

\end{bmatrix} = \begin{bmatrix}

10 \\

10

\end{bmatrix}\]

Now, you can continue this process for a total of 150 moves. After 150 moves, the particle's position will be:

\[\begin{bmatrix}

10 \\

10

\end{bmatrix} = (10, 10)\]

So, after 150 moves, the particle will be located at the coordinates (10, 10).

User Tschwab
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