Circumference, Limits, and Trig Functions... HELP PLEASE!

kingaaron08041991

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Suppose the circumfernce of a circle of radius r is divided into n equal pieces by n points, P[sub:1kwtep0i]1[/sub:1kwtep0i], P[sub:1kwtep0i]2[/sub:1kwtep0i], ... P[sub:1kwtep0i]n[/sub:1kwtep0i], where P[sub:1kwtep0i]1[/sub:1kwtep0i] is adjacent to P[sub:1kwtep0i]n[/sub:1kwtep0i] and P[sub:1kwtep0i]2[/sub:1kwtep0i], P[sub:1kwtep0i]2[/sub:1kwtep0i] is adjacent to P[sub:1kwtep0i]1[/sub:1kwtep0i] and P[sub:1kwtep0i]3[/sub:1kwtep0i], etc. Let l[sub:1kwtep0i]i[/sub:1kwtep0i] be the length of the line segment that connect P[sub:1kwtep0i]i[/sub:1kwtep0i] and P[sub:1kwtep0i]i+1[/sub:1kwtep0i] except l[sub:1kwtep0i]n[/sub:1kwtep0i] is the length of the line segment that connects P[sub:1kwtep0i]n[/sub:1kwtep0i] to P[sub:1kwtep0i]1[/sub:1kwtep0i]. Let L = l1+l2+...+l[sub:1kwtep0i]n[/sub:1kwtep0i]. (a.) express L as a function of n. (hint: use trig functions/radians). (b.) find the lim L [sub:1kwtep0i]n->infinity[/sub:1kwtep0i].
 
Re: NEED URGET HELP!!!

kingaaron08041991 said:
Suppose the circumfernce of a circle of radius r is divided into n equal pieces by n points, P[sub:24b4s0nm]1[/sub:24b4s0nm], P[sub:24b4s0nm]2[/sub:24b4s0nm], ... P[sub:24b4s0nm]n[/sub:24b4s0nm], where P[sub:24b4s0nm]1[/sub:24b4s0nm] is adjacent to P[sub:24b4s0nm]n[/sub:24b4s0nm] and P[sub:24b4s0nm]2[/sub:24b4s0nm], P[sub:24b4s0nm]2[/sub:24b4s0nm] is adjacent to P[sub:24b4s0nm]1[/sub:24b4s0nm] and P[sub:24b4s0nm]3[/sub:24b4s0nm], etc. Let l[sub:24b4s0nm]i[/sub:24b4s0nm] be the length of the line segment that connect P[sub:24b4s0nm]i[/sub:24b4s0nm] and P[sub:24b4s0nm]i+1[/sub:24b4s0nm] except l[sub:24b4s0nm]n[/sub:24b4s0nm] is the length of the line segment that connects P[sub:24b4s0nm]n[/sub:24b4s0nm] to P[sub:24b4s0nm]1[/sub:24b4s0nm]. Let L = l1+l2+...+l[sub:24b4s0nm]n[/sub:24b4s0nm]. (a.) express L as a function of n. (hint: use trig functions/radians). (b.) find the lim L [sub:24b4s0nm]n->infinity[/sub:24b4s0nm].
Let θk\displaystyle \theta_k be the central angle subtending the chord PkPk+1\displaystyle P_kP_{k+1} .
Then l=2rk=1Nsin(θk2)\displaystyle l = 2r\sum\limits_{k = 1}^N {\sin \left( {\frac{{\theta _k }}{2}} \right)}
 
Re: NEED URGET HELP!!!

Okay, but we haven't learned summations yet... so... how else could this be done? Could tangent be used somehow?
 
Re: NEED URGET HELP!!!

kingaaron08041991 said:
Okay, but we haven't learned summations yet... so... how else could this be done? Could tangent be used somehow?
You should have been able to figure this out for yourself. Did you work on it?

Because the points are equally spaced on the circle each θk=2πn\displaystyle \theta_k =\frac{2\pi}{n}.
Therefore, the summation is 2rnsin(πn)\displaystyle 2rn\sin\left(\frac{\pi}{n}\right).
 
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