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Solomon paper A - I need help

Can anyone help me with this question? IMG_20160329_010916.png
Original post by RUNSran
Can anyone help me with this question? IMG_20160329_010916.png


What have you done so far?
Reply 2
IMG_20160329_011643.png
Reply 3
Original post by undercxver
What have you done so far for part c?


i actually do not know how to start or where to start from
Reply 4
Original post by RUNSran
Can anyone help me with this question? IMG_20160329_010916.png


You know that the area is abydx=abydxdtdt\displaystyle \int_a^b y \, \mathrm{d}x = \int_a^b y \, \frac{\mathrm{d}x}{\mathrm{d}t} \, \mathrm{d}t

So you can replace yy and dxdt\frac{\mathrm{d}x}{\mathrm{d}t} by their respective expressions and the limits should follow from from x=0x=0 at which you need to find the value of tt that makes this so and then do the same for the other limit (which you've cropped out of the question, so...)
Original post by RUNSran
i actually do not know how to start or where to start from


You've been given parametric equations, you can make it dydx\frac{dy}{dx} by differentiating x=cos2tx=cos2t and y=cosecty=cosect.

Anyways, I'm assuming you should've done part a and b.

Find what tt equals when x=0x=0 .

This will be your limits for when you integrate to find the shaded area.

I hope this helps you start of?
Reply 6
Original post by Zacken
You know that the area is abydx=abydxdtdt\displaystyle \int_a^b y \, \mathrm{d}x = \int_a^b y \, \frac{\mathrm{d}x}{\mathrm{d}t} \, \mathrm{d}t

So you can replace yy and dxdt\frac{\mathrm{d}x}{\mathrm{d}t} by their respective expressions and the limits should follow from from x=0x=0 at which you need to find the value of tt that makes this so and then do the same for the other limit (which you've cropped out of the question, so...)


Oh, I get it - Thanks mate
what about 6 a? I tried solving it by parts but it seems I am missing something IMG_20160329_012520.png
Reply 7
Original post by undercxver
You've been given parametric equations, you can make it dydx\frac{dy}{dx} by differentiating x=cos2tx=cos2t and y=cosecty=cosect.

Anyways, I'm assuming you should've done part a and b.

Find what tt equals when x=0x=0 .

This will be your limits for when you integrate to find the shaded area.

I hope this helps you start of?


Yeah I figured out how to obtain the limits in terms of t but getting the integral terms was the challenging bit - someone has explain it to me.
i have posted posted another question and I will be very glad if you could help
Original post by RUNSran
Oh, I get it - Thanks mate
what about 6 a? I tried solving it by parts but it seems I am missing something IMG_20160329_012520.png


I tried integration by parts too and got .

Doesn't look right though. :s: is another answer I got. :lol:

I'm sorry but I'm not great at integration by parts yet. Maybe the way you did it is wrong? What exactly did you do step by step when integrating?
Original post by RUNSran
Oh, I get it - Thanks mate
what about 6 a? I tried solving it by parts but it seems I am missing something IMG_20160329_012520.png


expand cos(A+B) - cos(A-B) and see if that helps


dont open if youre doing step...

Spoiler

Reply 10
Original post by RUNSran
Oh, I get it - Thanks mate
what about 6 a? I tried solving it by parts but it seems I am missing something IMG_20160329_012520.png


'Factor formula' - cosAcosB=2sin(A+B2)sin(AB2) \displaystyle \cos A - \cos B = -2\sin \left (\frac{A+B}{2} \right )\sin \left (\frac{A-B}{2} \right ) .
Reply 11
Thank very much guys - I will try using the using the suggested Identities and see.
Reply 12
Original post by undercxver
I tried integration by parts too and got .

Doesn't look right though. :s: is another answer I got. :lol:

I'm sorry but I'm not great at integration by parts yet. Maybe the way you did it is wrong? What exactly did you do step by step when integrating?


I somehow can't see the pictures you uploaded .
This is what I did..IMG_20160329_081335.jpg

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