Question

14. Imagine a particle starting at (1,0) and making one counterclockwise revolution on the unitcircle. Let t be the angle in standard position that corresponds to the particle's position. At howmany points along the path of the particle are the x and y coordinates equal?

Answer 1

Let's make a graph to better understand the question:

a.) A particle starting at (1,0) and making one counterclockwise revolution on the unit

circle.

In the given description, we can assume that the center of the circle when the particle makes a revolution is at the origin (0,0). Thus, the equation of the circle that the particle will make is:

`(x-h)^2+(y-k)^2\text{ = }r^2`

At (h,k) = (0,0) and r = distance between (0,0) to (1,0).

We get,

`(x-0)^2+(y-k)^2=(\sqrt[]{(1-0)^2+(0-0)^2})^2`
`x^2+y^2\text{ = }1`

Plotting the graph,

In conclusion, there will be two points along the path of the particle that the x and y coordinate equal.

At, x = y, let's substitute this to the formula of the graph of the circle to get the coordinates.

`x^2+y^2\text{ = }1`
`x^2+x^2\text{ = }1`
`2x^2\text{ = 1 }\rightarrow\text{ }(2x^2)/(2)=\text{ }(1)/(2)`
`x\text{ = }\sqrt[]{(1)/(2)}`
`x\text{ = y = }\pm\frac{1}{\sqrt[]{2}}`

Therefore, the two points where the x and y will be equal is at:

`\text{ x = y = +}(1)/(√(2))\text{ and }-\frac{1}{\sqrt[]{2}}`