Answer
Case 1: position in polar coordinates $\left( {r,\theta } \right) = \left( {t,t} \right)$.
The speed at $t=1$ is $\sqrt 2 $.
Case 2: position in rectangular coordinates $\left( {x,y} \right) = \left( {t,t} \right)$.
The speed at $t=1$ is $\sqrt 2 $.
The speed is increasing in the first case and constant in the second case.
Work Step by Step
Case 1: the coordinates of the particle in polar coordinates $\left( {r,\theta } \right) = \left( {t,t} \right)$.
Using the result in Exercise 41, the speed of a particle is
$speed = \sqrt {{{\left( {\frac{{dr}}{{dt}}} \right)}^2} + {r^2}{{\left( {\frac{{d{\rm{\theta }}}}{{dt}}} \right)}^2}} $
$speed = \sqrt {1 + {t^2}} $
This equation implies that the speed is increasing. The speed at $t=1$ is $\sqrt 2 $.
Case 2: the coordinates of the particle in rectangular coordinates $\left( {x,y} \right) = \left( {t,t} \right)$.
The velocity vector is $\left( {x',y'} \right) = \left( {1,1} \right)$. So, the speed is
$speed = \sqrt {{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}} = \sqrt 2 $
In this case, the speed is constant. The speed at $t=1$ is $\sqrt 2 $.
Note that in the first case, the position vector of the particle depends on trigonometric functions of time so that
$\left( {x\left( t \right),y\left( t \right)} \right) = \left( {r\left( t \right)\cos \theta \left( t \right),r\left( t \right)\sin \theta \left( t \right)} \right) = \left( {t\cos t,t\sin t} \right)$.
Whereas, in the second case, the position vector of the particle is linear in $t$. As a result, the speed is increasing in the first case and constant in the second case.
