Question

Ok parametrics is screwing me over help.

Answer 1

Is the arc length of a parametric function equal to the distance a particle has traveled? Also, how come dy/dx of the parametric function no longer represent the velocity? The formula for speed is now this:\[v(t) = \sqrt{(x'(t))^2+y'((t))^2}\]
And all this got me to thinking, is there a difference between the derivative of position and the derivative of displacement?

Answer 2

To first question yes
The arc length of a parametric function equal to the distance of a particle has traveled

Second question yes
y=f(t) is the position function
y'=f'(t) is the velocity function
and
\[\sqrt{(y')^2}=\sqrt{(f'(t))^2} \text{ is the speed }\]
The speed is the distance of the derivative of f at t from the origin which is at t=0
This still translates over when you have two dependent variables instead of one
I'm talking about if you have (x(t),y(t))) where you have both x and y depending on t
you must first find the velocity point which is (x'(t),y'(t)) and then consider that point's distance from the origin which is at (x'(t),y'(t))=(0,0)

Answer 3

\[\text{ so the speed is } L>0 \\ L^2=(x'(t))^2+(y'(t))^2 \\ L=\sqrt{[x'(t)]^2+[y'(t)]^2}\]

Answer 4

Thank you for the explanation myininaya!