First year practise problem

1. Nov 14, 2013

Panphobia

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($\theta$) = 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($\theta$))^2*d$\theta$/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.

2. Nov 14, 2013

Dick

You've got some parts right. y=x*tan(θ). Differentiate both sides with respect to t and remember x AND θ are both functions of t. Use the product rule.

3. Nov 14, 2013

Panphobia

dy/dt = tan(θ) *dx/dt + x*(sec(θ))^2*dθ/dt

4. Nov 14, 2013

Dick

Ok, now use that.

5. Nov 14, 2013

Panphobia

dy/dt = 2 + (10 - ∏)*2*0.5
= 8.86 m/s
but my professor got 4.86 m/s...he used -2 instead of 2....sooo is he right?

6. Nov 14, 2013

Dick

Yes, he is. dx/dt isn't 2. x is getting smaller as time goes on.

7. Nov 14, 2013

Panphobia

ohh alright, thanks!