For the 20ft from doc. Set up triangle, with dr/dt on the hypotenuse = 2. 6 on the y, 20 on the x. Use x^2+ y^2= r^2. Differentiate and solve for dx/dt. 2x(dx/dt) + 2y(dy/dt)=2r(dr/dt); Since y value does not change, your y vector is equal to 0. Giving us 2x(dx/dt)=2r(dr/dt) ; dx/dt=(2r(dr/dt))/2x
Do the same thing for 10ft
No it is not surprising because as the angle between the dock and the boat decreases, more force is being used to pull up on the boat, not pull it inwards. Thus, the boats horizontal velocity will slow down as it beats the dock.
