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

[kingaaron08041991]

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].

 

[pka]

Re: NEED URGET HELP!!!

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 \(\displaystyle \theta_k\) be the central angle subtending the chord \(\displaystyle P_kP_{k+1}\) .
Then \(\displaystyle l = 2r\sum\limits_{k = 1}^N {\sin \left( {\frac{{\theta _k }}{2}} \right)}\)

 

[kingaaron08041991]

Re: NEED URGET HELP!!!

Okay, but we haven't learned summations yet... so... how else could this be done? Could tangent be used somehow?

 

[pka]

Re: NEED URGET HELP!!!

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 \(\displaystyle \theta_k =\frac{2\pi}{n}\).
Therefore, the summation is \(\displaystyle 2rn\sin\left(\frac{\pi}{n}\right)\).

 

[kingaaron08041991]

still dont get it