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Vaidehi09
 2 years ago
Best ResponseYou've already chosen the best response.0me neither...how can u find the average of just one quantity?

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.6You need \(<\sin^2\theta>\) right?

vikrantg4
 2 years ago
Best ResponseYou've already chosen the best response.1average must be 1/2 ? :P

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.6@Vaidehi09 basically that means average value of \(\sin^2\theta\) in its period  which is \(\pi\). (the graph for rekindling old memories:)dw:1343221614291:dw

shruti
 2 years ago
Best ResponseYou've already chosen the best response.0is this graph applicable for that one? @apoorvk

vikrantg4
 2 years ago
Best ResponseYou've already chosen the best response.1You can try integration

shruti
 2 years ago
Best ResponseYou've already chosen the best response.0please contineu how to do so.. ? need to integrate it from zero o pi. but in which period or how?

vikrantg4
 2 years ago
Best ResponseYou've already chosen the best response.1\[1\div(ba)\int\limits_{a}^{b}\sin^2\theta d \theta \]

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.6For this we need to integrate \(\sin^2\theta\) from 0 to \(\pi\),

vikrantg4
 2 years ago
Best ResponseYou've already chosen the best response.1take a = 0 and b = 2pi

vikrantg4
 2 years ago
Best ResponseYou've already chosen the best response.1no bro @apoorvk it should be 2pi

Vaidehi09
 2 years ago
Best ResponseYou've already chosen the best response.0@apoorvk i cant understand why the period is pi.

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.6\[\int\limits_0^{\pi}\sin^2\theta.d\theta\] and then I divide this by the difference of the limit, that is \(\pi 0=\pi\). Basically like the sum of the values by the 'domain covered'.

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.6@Vaidehi09, do you understand that the period sinx of 2pi?

apoorvk
 2 years ago
Best ResponseYou've already chosen the best response.6now, @Vaidehi09 when I square the function  two changes happen to curve  1. The curve's shape changes. 2. The negative part of the graph is flipped up in the positive axis. So, whereas the period was 2 pi for sinx, the \(\sin^2x\) graph repeats itself after every \(\pi\)distance. Hence the period is pi.

vikrantg4
 2 years ago
Best ResponseYou've already chosen the best response.1OmG ! Expert ! xD Keep it up brother.

Vaidehi09
 2 years ago
Best ResponseYou've already chosen the best response.0@apoorvk ah yes..of course! i was getting confused with the ve part and what kind of change will happen to the shape of the graph....but i get it now. thanks a lot! :))

vikrantg4
 2 years ago
Best ResponseYou've already chosen the best response.1I am new to calc.. so I am not sure about my replies :P

sami21
 2 years ago
Best ResponseYou've already chosen the best response.0@apoorvk good Explaination.
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