Hi, we just learned this new section today, and I am very confused....
A particle is moving around the unit circle (the circle of radius 1 centered at the origin). At the point (.6, .8) the particle has horizontal velocity dx/dt = 3. what is its vertical velocity dy /dt at that point?
AP Calculus - Damon, Tuesday, December 14, 2010 at 8:17pm
This is a 3,4,5 right triangle.
the sin of the angle between horizontal and the resultant tangent at (.6,.8) is dy/dt/3 = -3/5
so dy/dt = -9/5