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#1
hellooo

i don't have an fp2 text book, so i can;t look this up for myself, and if i google it i usually end up somewhere aiming at a much more advanced audience than myself so i just get more confused

the question is from a past paper

A curve has polar equation for where is a positive constant.

Part A) Sketch the curve

I've done this part, and it's a very pretty petally thingy with 3 loops, sort of in the shape of " > " ie two big ones on the left pointing up and away from each other, one in the middle facing right. i'm fairly confident about this part i got it right, it looks like the one in the mark scheme (which isn't labelled, annoyingly).

Part B) Find the area enclosed by one of the loops.

i found online you do integral of "1/2 times the function squared" (i have no idea why since i have no text book and hence no proofs of anything so i'm reluctantly ploughing on semi-blindly ) so i've done the integral and i agree with the mark scheme, woot, BUT my left hand loops are half way along (kinda being bisected down the spine) the backwardsly-extended lines pi/6 and -pi/6, while my right hand loop is enclosed by both of those lines.

either:

a) the integral only takes the area of positive r values, since technically the lines extending backwards are really the lines 5pi/6 and 7pi/6, so only my right hand loop is included in the area

b) it does take half the area from each left hand loop, so i should divide by 2 af the end when i get my area (in terms of a)

c) i did my graph wrong and the loops shouldn't be in those lines on the left hand side

WHICH ONE IS IT?!

sorry if this is unclear, it's hard without the diagram but yeah, thanks for any help!
0
11 years ago
#2
I'm fairly sure it's a. Loving the thread title btw...
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#3
(Original post by meatball893)
I'm fairly sure it's a. Loving the thread title btw...
ok great i checked the mark scheme (which didn't even occur to me, despite checking it for part A) and i did get the right answer, but i want to make sure the reason is because only the positive r values were used, and not because i drew my graph wrong and the correct answer is as a result of first getting the graph a bit wrong, then treating the graph wrongly in my integration, cancelling out my mistakes and accidentally getting it right...if you know what i mean :P

but thanks
0
11 years ago
#4
Here's the graph plotted with the values you gave -pi/2 to pi/2, and the lines +/-pi/6

I set "a" to 5 for the purposes of plotting it.
0
11 years ago
#5
(Original post by zdo0o)
ok great i checked the mark scheme (which didn't even occur to me, despite checking it for part A) and i did get the right answer, but i want to make sure the reason is because only the positive r values were used, and not because i drew my graph wrong and the correct answer is as a result of first getting the graph a bit wrong, then treating the graph wrongly in my integration, cancelling out my mistakes and accidentally getting it right...if you know what i mean :P

but thanks
I'd wait for a second opinion, but I'm fairly sure it is because the integral only deals with r>0.
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#6
(Original post by ghostwalker)
Here's the graph plotted with the values you gave -pi/2 to pi/2, and the lines +/-pi/6

I set "a" to 5 for the purposes of plotting it.
ah lol, i did that, but my lines were so messy i went over the loops and got confused when looking back on it... fml thanks though

however, if the left hand ones DID cross the lines, would integrating between the limits +/-pi/6 also take the left loops into account, or are they only found using 5pi/6 and 7pi/6 ?
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11 years ago
#7
since r>=0 for theta between +/-pi/6 you will only get the righthand loop between the two lines where "x" is >= 0 . Anything that is happening outside the +/- pi/6 values is irrelevant.

Not sure what you'd need for the other loops off hand.
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#8
(Original post by ghostwalker)
since r>=0 for theta between +/-pi/6 you will only get the righthand loop between the two lines where "x" is >= 0 . Anything that is happening outside the +/- pi/6 values is irrelevant.

Not sure what you'd need for the other loops off hand.
thanks for clearing that up
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