Problem

The speed of a particle moving in a plane is equal to the magnitude of its instantaneous velocity,

v = |\mathbf{v}| = \sqrt{v_x^2 + v_y^2}.

(a) Show that the rate of change of the speed is

\frac{dv}{dt} = (v_x a_x + v_y a_y) / \sqrt{v_x^2 + v_y^2}.

(b) Show that the rate of change of speed can be expressed as

\frac{dv}{dt} = \mathbf{v} \cdot \mathbf{a} / v

, and use this result to explain why

\frac{dv}{dt}

is equal to

a_t

, the component of

\mathbf{a}

that is parallel to

\mathbf{v}