The coordinates of a particle moving in $$x-y$$  plane at any instant of time $$t$$ are $$x = 4{t^2};y = 3{t^2}.$$    The speed of the particle at that instant is

Solution :
According to the question, at any instant $$t,$$
$$\eqalign{ & x = 4{t^2},y = 3{t^2} \cr & \therefore {v_x} = \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {4{t^2}} \right) = 8t \cr & {\text{and}}\,\,{v_y} = \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {3{t^2}} \right) = 6t \cr} $$
The speed of the particle at instant $$t.$$
$$v = \sqrt {v_x^2 + v_y^2} = \sqrt {{{\left( {8t} \right)}^2} + {{\left( {6t} \right)}^2}} = 10\,t$$