Change in x and y for a change in position

  • Thread starter Thread starter NEWO
  • Start date Start date
  • Tags Tags
    Change Position
NEWO
Messages
93
Reaction score
0
Basically I have a pointer that starts at a particular reference point on a circle with cartisean coordinates,this pointer then moves around the circumference of the circle with an angle theta, the arc length is known but what I need to find is the change relative to x and y.

I know that the arc length is given by;

\delta c =r_{c}\theta_{a}

where \delta c, r_{c}, \theta_{a} are the arc length, radius of the circle and the angle of the arc respectively.

\delta c, r_{c}, and \theta_{a} are known

but what i don't know is how to find the new position of the pointer relative to x and y;

hence I need to find the change in x and the change in y.

I don't really know where to begin so if I could get some pointers rather than the answer then that would great.

Thanks for your time

N
 
Last edited:
Physics news on Phys.org
A point on a circle with radii r could be represent by
(rcos\theta ,rsin\theta)
in cartisean coodinate system...
Now you are asking to find the different of x, and y coordinate when \theta change from \theta_0 to \theta_1
 
chanvincent said:
A point on a circle with radii r could be represent by
(rcos\theta ,rsin\theta)
in cartisean coodinate system...
Now you are asking to find the different of x, and y coordinate when \theta change from \theta_0 to \theta_1

Yes that's what i need to do I think. I attach a crude representation of what I am trying to find

x and y are the unknowns

thanks
 

Attachments

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top