Is Velocity Constant if AxVx + AyVy Equals Zero?

[bob1182006]

Homework Statement

A particle is moving in the xy plane with velocity \vec{v}(t)=v_x(t)\vec{i}+v_y(t)\vec{j} and acceleration \vec{a}(t)=a_x(t)\vec{i}+a_y(t)\vec{j}. By taking the appropiate derivative show that the magnitude of v can be constant only if a_xv_x+a_yv_y = 0

Homework Equations

The Attempt at a Solution

So I know that in order for velocity to have a constant magnitude then acceleration must = 0.

Since acceleration is the derivative of velocity and dv/dt=0 iff v=some constant.
expanding the LHS:

\frac{d\vec{v}(t)}{dt}=0

\frac{dv_x(t)}{dt}\vec{i}+\frac{dv_y(t)}{dt}\vec{j} = 0

a_x(t)\vec{i}+a_y(t)\vec{j} = 0

since i,j are unit vectors and do not equal 0 the components of acceleration must = 0.

a_x(t)=a_y(t)=0

But it's also possible for a_x=-a_y in which case the acceleration would still = 0 so is this a proof really correct?

 

[learningphysics]

The velocity need not be constant. The magnitude of the velocity is constant. Get the magnitude of the velocity, and use the fact that it is constant... go from there..