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Vaidehi09Best ResponseYou've already chosen the best response.0
me neither...how can u find the average of just one quantity?
 one year ago

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

vikrantg4Best ResponseYou've already chosen the best response.1
average must be 1/2 ? :P
 one year ago

apoorvkBest 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
 one year ago

shrutiBest ResponseYou've already chosen the best response.0
is this graph applicable for that one? @apoorvk
 one year ago

vikrantg4Best ResponseYou've already chosen the best response.1
You can try integration
 one year ago

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

vikrantg4Best ResponseYou've already chosen the best response.1
\[1\div(ba)\int\limits_{a}^{b}\sin^2\theta d \theta \]
 one year ago

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

vikrantg4Best ResponseYou've already chosen the best response.1
take a = 0 and b = 2pi
 one year ago

vikrantg4Best ResponseYou've already chosen the best response.1
no bro @apoorvk it should be 2pi
 one year ago

Vaidehi09Best ResponseYou've already chosen the best response.0
@apoorvk i cant understand why the period is pi.
 one year ago

apoorvkBest 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'.
 one year ago

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

apoorvkBest ResponseYou've already chosen the best response.6
now, @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.
 one year ago

vikrantg4Best ResponseYou've already chosen the best response.1
OmG ! Expert ! xD Keep it up brother.
 one year ago

Vaidehi09Best 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! :))
 one year ago

vikrantg4Best ResponseYou've already chosen the best response.1
I am new to calc.. so I am not sure about my replies :P
 one year ago

sami21Best ResponseYou've already chosen the best response.0
@apoorvk good Explaination.
 one year ago
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