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:
[tex](x-h)^2+(y-k)^2\text{ = }r^2[/tex]
At (h,k) = (0,0) and r = distance between (0,0) to (1,0).
We get,
[tex](x-0)^2+(y-k)^2=(\sqrt[]{(1-0)^2+(0-0)^2})^2[/tex][tex]x^2+y^2\text{ = }1[/tex]
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.
[tex]x^2+y^2\text{ = }1[/tex][tex]x^2+x^2\text{ = }1[/tex][tex]2x^2\text{ = 1 }\rightarrow\text{ }\frac{2x^2}{2}=\text{ }\frac{1}{2}[/tex][tex]x\text{ = }\sqrt[]{\frac{1}{2}}[/tex][tex]x\text{ = y = }\pm\frac{1}{\sqrt[]{2}}[/tex]
Therefore, the two points where the x and y will be equal is at:
[tex]\text{ x = y = +}\frac{1}{\sqrt{2}}\text{ and }-\frac{1}{\sqrt[]{2}}[/tex]
