Let us consider a point is moving in a plane,

If (x,y) be the cartesian co-ordinates of the point and \( (r, \theta) \) be the plane polar co-ordinates in the two-dimensional system, then the relation between the speeds in cartesian and polar co-ordinates is given by,

the radial component of speed along \( \hat{r} \) is

\( \displaystyle{v_r=\frac{\dot{x}x+\dot{y}y}{\sqrt{x^2+y^2}}} \)

and the transverse component of the speed along \( \hat{\theta} \) is

\( \displaystyle{v_{\theta}=\frac{\dot{y}x-\dot{x}y}{\sqrt{x^2+y^2}}} \)

[ To know the derivation of the above equations (CLICK HERE) ]

Now given,

\( x=3 \), \( y=4 \), \( \frac{dx}{dt}=\dot{x}=5\ m/sec \) and \( \frac{dy}{dt}=\dot{y}=8\ m/sec \).

Putting these values in the above equations we get,

\( \displaystyle{v_r=\frac{\dot{x}x+\dot{y}y}{\sqrt{x^2+y^2}}} \)

\( \displaystyle{or,\ v_r=\frac{5\times{3}+8\times{4}}{\sqrt{3^2+4^2}}} \)

\( \displaystyle{or,\ v_r=9.4\ m/sec} \)

and

\( \displaystyle{v_{\theta}=\frac{\dot{y}x-\dot{x}y}{\sqrt{x^2+y^2}}} \)

\( \displaystyle{or,\ v_{\theta}=\frac{8\times{3}-5\times{4}}{\sqrt{3^2+4^2}}} \)

\( \displaystyle{or,\ v_{\theta}=0.8\ m/sec} \)