HallsofIvy

dy/dx is the ratio of the y-component of the tangent vector to the x-component. Coupling that with the requirement that the length be 1 lets you determine both. For example, if dy/dt= v_y/v_x= 2 then v_y= 2v_x. If then [itex]\sqrt{v_x^2+ v_y^2}= 1[/itex], we have v_x²+ v_y²= v_x²+ 4v_x²= 5v[sup]x[/sub]²= 5 so [itex]v_x= 1/\sqrt{5}[/itex] and [itex]v_y= 2/\sqrt{5}[/itex].