Understanding the Relationship between Velocity and Time in Calculus

  • Thread starter Thread starter sniffer
  • Start date Start date
  • Tags Tags
    Calculus
AI Thread Summary
The discussion focuses on understanding the relationship between velocity and time in calculus, specifically through the equation dv/dt = ∂v/∂r + ∂v/∂t. A participant suggests using the chain rule to clarify the relationship, leading to the equation dv/dt = ∂v/∂r * dr/dt + ∂v/∂t. This adjustment helps illustrate how changes in position (dr/dt) affect velocity. The original poster seeks clarification on whether dr/dt indicates a small velocity. Overall, the conversation emphasizes the importance of the chain rule in understanding the dynamics of velocity as a function of position and time.
sniffer
Messages
112
Reaction score
0
i just want simple explanation of this.
velocity v is a function of r and t.
then my professor, in order to derive relationship of something, he wrote:
\frac{dv}{dt}=\frac{\partial v}{\partial r}+\frac{\partial v}{\partial t}
then assuming small spatial variation,
\frac{dv}{dt} \approx \frac{\partial v}{\partial t}
my question: is the first equation above OK?
is it standard calculus? i am weak in this. please help.


thanks.
 
Physics news on Phys.org
look up the chain rule; when v = v(r(t),t),


\frac{dv}{dt}=\frac{\partial v}{\partial r}\frac{dr}{dt}+\frac{\partial v}{\partial t}

then you can see the following argument makes more sense?
 
thanks. i just want to clarify again, does it mean dr/dt is small, i.e. small velocity?
 
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top