Homework Help: Speed in radial coordinates and angular coordinates

1. Mar 11, 2014

mrchauncey

1. The problem statement, all variables and given/known data

First off, I am not a physics student. I am a math major taking a maple software course and there is a question that I can not figure out.

The question gives me a radial coordinates r

r:= $\frac{a*t^{2}*e^{-b*t}}{1+t^{2}}$

And angular coordinates:

θ:=b+c*t$^{2/3}$

Where a,b,c are real constants.

It then puts the radial and angular coordinates in the form

X:= r*cos(θ)
Y:=r*sin(θ) so it looks like this:

X:=$\frac{a*t^2*e^{-b*t}*cos(b+c*t^{2/3}}{1+t^{2}}$

Y:=$\frac{a*t^{2}*e^{-b*t}*sin(b+c*t^{2/3}}{1+t^{2}}$

Which I understand.

2. Relevant equations

Now the question is calculate speed V at an arbitrary time simplifying as much as possible. This is where I get confused. The question then says to find speed V at time t take

u=($\frac{dX}{dt}$)$^{2}$+($\frac{dY}{dt}$)$^{2}$

$\sqrt{u}$

It does not explain why they use that formula to calculate the speed. Just wondering if anyone can shed some light on this situation. This is an example in the book so I know this is how you do it or the way they want me to do it, so I am not looking for an answer, just an explanation.

3. The attempt at a solution

Thanks for looking.

Last edited: Mar 11, 2014
2. Mar 11, 2014

Staff: Mentor

Velocity is defined at the rate of change of position, which in one dimension gives
$$v = \frac{dx}{dt}$$
In 2D, as for you problem, velocity is a vector, $\vec{v} = (v_x, v_y)^T$, with
$$v_x = \frac{dx}{dt}, \quad v_y = \frac{dy}{dt}$$

Speed is simply the magnitude of the velocity (L2-norm), hence the formula you were given.

3. Mar 11, 2014

mrchauncey

Thanks. I actually just found that reading through my old calculus textbook on parametric equations, arc length, and speed. Its been a while since I took calculus and was never the best at speed, velocity and acceleration, etc. My math background has more to do with Actuarial Science and Stats. Once again thanks again. Have a nice day.