#### Answer

The speed of the particle is
$speed = \sqrt {{{\left( {\frac{{dr}}{{dt}}} \right)}^2} + {r^2}{{\left( {\frac{{d{\rm{\theta }}}}{{dt}}} \right)}^2}} $

#### Work Step by Step

Using the relation $x = r\cos \theta $ and $y = r\sin \theta $, we have in rectangular coordinates the position vector of the particle:
$\left( {x\left( t \right),y\left( t \right)} \right) = \left( {r\left( t \right)\cos \theta \left( t \right),r\left( t \right)\sin \theta \left( t \right)} \right)$
By definition, the derivative of position vector with respect to time $t$ is the velocity vector. So, omitting $t$ we get
$\left( {x',y'} \right) = \left( {r'\cos \theta - r\theta '\sin \theta ,r'\sin \theta + r\theta '\cos \theta } \right)$,
where prime denotes differentiation with respect to $t$. So,
$r' = \frac{{dr}}{{dt}}$ and $\theta ' = \frac{{d{\rm{\theta }}}}{{dt}}$.
The speed of the particle is
$speed = \sqrt {\left( {x',y'} \right)\cdot\left( {x',y'} \right)} = \sqrt {{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}} $
So,
$speed = \sqrt {{{\left( {r'\cos \theta - r\theta '\sin \theta } \right)}^2} + {{\left( {r'\sin \theta + r\theta '\cos \theta } \right)}^2}} $
$speed = {\left( {{{\left( {r'} \right)}^2}{{\cos }^2} - 2r'r\theta '\cos \theta \sin \theta + {r^2}{{\left( {\theta '} \right)}^2}{{\sin }^2}\theta + {{\left( {r'} \right)}^2}{{\sin }^2}\theta + 2r'r\theta \sin \theta \cos \theta + {r^2}{{\left( {\theta '} \right)}^2}{{\cos }^2}\theta } \right)^{1/2}}$
$speed = \sqrt {{{\left( {r'} \right)}^2} + {r^2}{{\left( {\theta '} \right)}^2}} $
Hence,
$speed = \sqrt {{{\left( {\frac{{dr}}{{dt}}} \right)}^2} + {r^2}{{\left( {\frac{{d{\rm{\theta }}}}{{dt}}} \right)}^2}} $