3. The attempt at a solution I tried solving this question a few different ways, and I got different answers every time. My first attempt I realized that the relationship between the changing x, changing y, and changing angle was tan([itex]\theta[/itex]) = y/x , so first of all I tried to figure out the final x, and I could do this because it gave me the angular velocity and the final angle, so I got the time to be ∏/2. Then I multiplied 2*(∏/2) to get the distance travelled. So the distance from the wall is 10 - ∏, now that I know the x I can figure out dy/dt, so I differentiate the first expression and get (sec([itex]\theta[/itex]))^2*d[itex]\theta[/itex]/dt = dy/dt * 1/(10-∏). Then the answer would be (sqrt(2))^2 * 0.5 * (10-∏) = dy/dt, so that was my first answer, now here to my second answer. I differentiate that first expression straight from the beginning, I won't post it but I used the quotient rule, then I figured out that I needed x and y, so I figured out the time the same way as before, it was ∏/2, then I figured out the distance from the wall which was also 10 - ∏, but I also figured out that tan(∏/4) = 1 = O/A so x = y, so y = 10 - ∏ also, but when I plug everything in then, I get 8.86 m/s instead of 6.86 m/s which was what I got in the first attempt. But both of these answers are wrong according to the prof.