text: # The trajectory of a particle moving in a plane is expressed in polar coordinates (r, θ) by the equations $$r=r_0 e^{β t} \text { and } \frac{d \theta}{d t}=ω$$, where the parameters r0, β  and ω are positive. Let vr and ar denote the velocity and acceleration, respectively, in the radial direction. For this trajectory
  
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  CSIR-UGC (NET) Physical Science: Held on (8 June 2023)
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  1. a< 0 at all times irrespective of the values of the parameters
  2. ar​ > 0 at all times irrespective of the values of the parameters
  3. $$\frac{d v_r}{d t}>0$$ and ar > 0 for all choices of parameters
  4. $$\frac{d v_r}{d t}>0$$ however, ar = 0 for some choices of parameters
  
  Option 4 : $$\frac{d v_r}{d t}>0$$ however, ar = 0 for some choices of parameters
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  CSIR-UGC (NET) Life Science: Held on (26 Dec 2023 Shift 1)
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  ## Detailed Solution
  
  Explanation:
  
  We will first write the velocity vector in polar co-ordinates and write the radial velocity and radial acceleration and by differentiating we get the desired solution.
  
  Given,-$$r=r_0 e^{β t} \text { and } \frac{d \theta}{d t}=ω$$     ----------------1
  
  Velocity in polar co-ordinates is given by the sum of radial and transverse velocity.
  
  $$v=v_r+v_t=\dot r \hat{r}+r\dot \theta \hat{\theta}$$
  
  Radial velocity $$v_r=\dot r$$
  
  Radial acceleration $$a_r=(\ddot {r}-r\dot \theta^2)$$
  
  Now, $$r=r_0 e^{β t}$$ and $$\dot r=r_0 \beta e^{β t}$$
  
  $$\frac{d v_r}{d t}=$$$$\beta ^2r_0 \beta e^{β t}$$ => $$\frac{d v_r}{d t}$$$$>0$$
  
  $$\ddot r=r_0 \beta^2 e^{β t}$$ and $$\ddot \theta^2=\omega^2$$
  
  $$a_r=$$$$\beta ^2r_0 \beta e^{β t}$$ $$-r\omega^2$$$$=$$$$\beta ^2r_0 \ e^{β t}$$$$-r_0 \omega^2e^{β t}$$$$=$$$$r_0 \ e^{β t} (\beta^2-\omega^2)$$
  
  If $$\beta=\omega$$$$a_r=0$$ for some choices of parameters.
  
  So, the correct answer is  -$$\frac{d v_r}{d t}>0$$ however, $$a_r=0$$ for some choices of parameters.