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FP3- polar coordinates help? Watch

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    Hey guys,

    Our teachers in school are so bad at teaching us stuff like this that I don't get anything except from what I revise at home or self teach.

    I was revising from past papers and I have no idea how to even start question 7bii)

    Since I know nothing of this chapter, it would be really really helpful if you posted your solution along with explanation. I've looked at mark schemes to figure out what's going on but I can't see how?!!?!?!

    Please help
    i've attached the past paper onto this thread thanks
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  1. File Type: pdf AQA-MFP3-P-QP-JUN15.PDF (189.2 KB, 41 views)
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    (Original post by liemluji)
    x
    alright so we can either go through the mark scheme together or i can guide you through the question step by step. Its up to you, which do you think youd learn better from?
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    (Original post by DylanJ42)
    alright so we can either go through the mark scheme together or i can guide you through the question step by step. Its up to you, which do you think youd learn better from?
    could we please go through the mark scheme together?
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    (Original post by liemluji)
    could we please go through the mark scheme together?
    sure, so here is the mark scheme numbered step by step

    Name:  Screenshot_1.png
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    1) in the cartesian plane if you wanted a straight horizontal line it would have the equatiion y = k, for some number k eg y = 2 is a straight horizontal line 2 units above the x axis and its parallel to the x axis.

    So you've learnt, or should have learnt (if you havent please say and ill show you this) that  \displaystyle y = r \sin \theta and  \displaystyle x = r \cos \theta , since we are looking for a straight horizontal line here we use  \displaystyle y = k and sub in  \displaystyle y = r \sin \theta which gives us  \displaystyle r \sin \theta = k .

    Now we must find k, you know A lies on this line with  \displaystyle (r=\frac{3}{2} , \theta = \frac{\pi}{6}) so sub this into  \displaystyle r \sin \theta = k and we get  \displaystyle \frac{3}{2} \sin (\frac{\pi}{6}) = k so  \displaystyle k = \frac{3}{4} , therfore we have the equation of the line as  \displaystyle r \sin \theta = \frac{3}{4}

    all good so far?
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    (Original post by DylanJ42)
    sure, so here is the mark scheme numbered step by step

    Name:  Screenshot_1.png
Views: 78
Size:  42.6 KB

    1) in the cartesian plane if you wanted a straight horizontal line it would have the equatiion y = k, for some number k eg y = 2 is a straight horizontal line 2 units above the x axis and its parallel to the x axis.

    So you've learnt, or should have learnt (if you havent please say and ill show you this) that  \displaystyle y = r \sin \theta and  \displaystyle x = r \cos \theta , since we are looking for a straight horizontal line here we use  \displaystyle y = k and sub in  \displaystyle y = r \sin \theta which gives us  \displaystyle r \sin \theta = k .

    Now we must find k, you know A lies on this line with  \displaystyle (r=\frac{3}{2} , \theta = \frac{\pi}{6}) so sub this into  \displaystyle r \sin \theta = k and we get  \displaystyle \frac{3}{2} \sin (\frac{\pi}{6}) = k so  \displaystyle k = \frac{3}{4} , therfore we have the equation of the line as  \displaystyle r \sin \theta = \frac{3}{4}

    all good so far?
    Yes! You explanation is so detailed and easy to follow! Thank you!
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    (Original post by liemluji)
    Yes! You explanation is so detailed and easy to follow! Thank you!
    Perfect okay youll be glad to hear that 2) and 3) of the mark scheme are short explanations so ill do 2) and 3) in this post

    2) Firstly here is a diagram for you to look at, its always a good idea to sketch all the important curves/lines for these types of questions

    http://imgur.com/a/NdHIv

    So you can see that A and more importantly B occur when curve1 meets the horizontal line. So we are going to find where  \displaystyle r = 1 + \cos(2\theta) and  \displaystyle r \sin \theta = \frac{3}{4} meet.

    Do you see how they are just simultaneous equations with unknowns of  \displaystyle r and  \displaystyle  \theta .

    We are not interested in theta so it makes sense to eliminate it;

    the line equation can be rearranged to give us;  \displaystyle \sin \theta = \frac{3}{4r}

    We can also get our curve equation in terms of r and sin by using  \displaystyle cos2\theta = 1 - 2sin^2\theta

    So the curve equation becomes  \displaystyle r = 1 + (1 - 2sin^2\theta)

    Now subbing  \displaystyle sin \theta = \frac{3}{4r} into that gives us;

     \displaystyle r = 1 + (1 - 2 \left( \frac{9}{16r^2} \right))

    tidying this up we get,  \displaystyle r = 2 - 2 \left( \frac{9}{16r^2} \right)

    part 2) done

    3) is literally just rearranging  \displaystyle r = 2 - 2 \left( \frac{9}{16r^2} \right) by multiplying everything by  \displaystyle 16r^2

    to give us  \displaystyle 16r^3 = 32r^2 - 18 , part 3) done
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    (Original post by DylanJ42)
    Perfect okay youll be glad to hear that 2) and 3) of the mark scheme are short explanations so ill do 2) and 3) in this post

    2) Firstly here is a diagram for you to look at, its always a good idea to sketch all the important curves/lines for these types of questions

    http://imgur.com/a/NdHIv

    So you can see that A and more importantly B occur when curve1 meets the horizontal line. So we are going to find where  \displaystyle r = 1 + \cos(2\theta) and  \displaystyle r \sin \theta = \frac{3}{4} meet.

    Do you see how they are just simultaneous equations with unknowns of  \displaystyle r and  \displaystyle  \theta .

    We are not interested in theta so it makes sense to eliminate it;

    the line equation can be rearranged to give us;  \displaystyle \sin \theta = \frac{3}{4r}

    We can also get our curve equation in terms of r and sin by using  \displaystyle cos2\theta = 1 - 2sin^2\theta

    So the curve equation becomes  \displaystyle r = 1 + (1 - 2sin^2\theta)

    Now subbing  \displaystyle sin \theta = \frac{3}{4r} into that gives us;

     \displaystyle r = 1 + (1 - 2 \left( \frac{9}{16r^2} \right))

    tidying this up we get,  \displaystyle r = 2 - 2 \left( \frac{9}{16r^2} \right)

    part 2) done

    3) is literally just rearranging  \displaystyle r = 2 - 2 \left( \frac{9}{16r^2} \right) by multiplying everything by  \displaystyle 16r^2

    to give us  \displaystyle 16r^3 = 32r^2 - 18 , part 3) done
    Yes I get it! for 4, you just factorise it right?
    and when you sketch the curves, do you use graphics calculator or do you plot values in?
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    (Original post by liemluji)
    Yes I get it! for 4, you just factorise it right?
    and when you sketch the curves, do you use graphics calculator or do you plot values in?
    yes! for 4) just factorise using long divison since you know  \displaystyle r = \frac{3}{2} is a root so  \displaystyle 2r - 3 is a factor.

    For 5) just use the quadratic formula to find r and take the positive value of r since the distance you require is positive. thats the whole part done

    I usually just put some values in, maybe every  \displaystyle \frac{\pi}{6} radians / 30 degrees and do a plot**, only spend like a few minutes on it though bc it helps but you dont want to waste too much time on it

    **also i learnt off a lot of things like this; http://schoolbag.info/mathematics/ba...s/image297.jpg just so i had an idea of what things such as  \displaystyle r = 1 +cos\theta or  \displaystyle r = 1 + sin\theta roughly look like

    so part ii) done, are you okay with iii) or do you need a hand with it too?
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    (Original post by DylanJ42)
    yes! for 4) just factorise using long divison since you know  \displaystyle r = \frac{3}{2} is a root so  \displaystyle 2r - 3 is a factor.

    For 5) just use the quadratic formula to find r and take the positive value of r since the distance you require is positive. thats the whole part done

    I usually just put some values in, maybe every  \displaystyle \frac{\pi}{6} radians / 30 degrees and do a plot**, only spend like a few minutes on it though bc it helps but you dont want to waste too much time on it

    **also i learnt off a lot of things like this; http://schoolbag.info/mathematics/ba...s/image297.jpg just so i had an idea of what things such as  \displaystyle r = 1 +cos\theta or  \displaystyle r = 1 + sin\theta roughly look like

    so part ii) done, are you okay with iii) or do you need a hand with it too?
    Thank you so much for all the time and effort! I really appreciate it!
    I think I can manage to do the last one myself
    Btw, do you think getting a graphics calculator for fp3 and fp4 will greatly help and save time?
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    (Original post by liemluji)
    Thank you so much for all the time and effort! I really appreciate it!
    I think I can manage to do the last one myself
    Btw, do you think getting a graphics calculator for fp3 and fp4 will greatly help and save time?
    no problem at all

    i never used one and im not even sure if youre allowed them in the exam, but check if they are allowed and if you think itll help you out then go for it.
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    (Original post by DylanJ42)
    no problem at all

    i never used one and im not even sure if youre allowed them in the exam, but check if they are allowed and if you think itll help you out then go for it.
    great! Thanks for your help!
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    (Original post by DylanJ42)
    no problem at all

    i never used one and im not even sure if youre allowed them in the exam, but check if they are allowed and if you think itll help you out then go for it.
    Hi! Sorry to bother you again but I think you are the best person to ask

    I was doing another question on polar coordinates and I'm not sure how this works.

    It's very short! you don't have to go through each step, could you please tell me why the limits work in different ways and when it's integrated by substitution the 1/2 disappears?

    Thank you!

    It's number 8a)
    Attached Images
  2. File Type: pdf AQA-MFP3-QP-JUN14.PDF (192.0 KB, 35 views)
  3. File Type: pdf AQA-MFP3-W-MS-JUN14.PDF (200.8 KB, 29 views)
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    (Original post by liemluji)
    Hi! Sorry to bother you again but I think you are the best person to ask

    I was doing another question on polar coordinates and I'm not sure how this works.

    It's very short! you don't have to go through each step, could you please tell me why the limits work in different ways and when it's integrated by substitution the 1/2 disappears?

    Thank you!

    It's number 8a)
    ah the second line is confusing you?

    basically on the first line they are integrating between  \displaystyle -\frac{\pi}{4} and  \displaystyle \frac{\pi}{4} but what they have done on the second line is realise you can also just integrate between  \displaystyle 0 and  \displaystyle \frac{\pi}{4} and then multiply that area by 2

    so Area =  \displaystyle \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} r^2 \: d\theta =  \displaystyle  2 \times \frac{1}{2} \int_0^{\frac{\pi}{4}} r^2 \: d\theta  =  \displaystyle \int_0^{\frac{\pi}{4}} r^2  d\theta <-- pay close attention to how the limits change on this line
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    (Original post by DylanJ42)
    ah the second line is confusing you?

    basically on the first line they are integrating between  \displaystyle -\frac{\pi}{4} and  \displaystyle \frac{\pi}{4} but what they have done on the second line is realise you can also just integrate between  \displaystyle 0 and  \displaystyle \frac{\pi}{4} and then multiply that area by 2

    so Area =  \displaystyle \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} r^2 \: d\theta =  \displaystyle  2 \times \frac{1}{2} \int_0^{\frac{\pi}{4}} r^2 \: d\theta  =  \displaystyle \int_0^{\frac{\pi}{4}} r^2  d\theta <-- pay close attention to how the limits change on this line
    OH! Thank you so much! You are such a great help Thank you!!!
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    (Original post by liemluji)
    OH! Thank you so much! You are such a great help Thank you!!!
    my pleasure
 
 
 
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