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    if they give us a polar equation and tell us to sketch it (2 or 3 marks usually)... will we lose marks if we just copy it from a graphical calculator and just work out and label the main points like where it crosses the initial line?

    Thanks
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    (Original post by S1M)
    if they give us a polar equation and tell us to sketch it (2 or 3 marks usually)... will we lose marks if we just copy it from a graphical calculator and just work out and label the main points like where it crosses the initial line?

    Thanks
    Course not; how would they know?
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    (Original post by Trinity Blume)
    Course not; how would they know?
    thanks...
    if you dont have a graphical calculator how are you meant to do it?
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    (Original post by S1M)
    thanks...
    if you dont have a graphical calculator how are you meant to do it?
    i plot points manually.

    it often goes wrong though
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    Just in a table format, same way you would do cartesian. Let theta (sp!) be pi/6 and pi/3 and pi/2 etc and then work out what r is in relation to the angle then just sketch. I always find it hard, so I use the graphics calculator to back me up.
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    Woah.. that will take a long time... for 2 marks only as well!!!
    Graphical calculator it is then

    thanks!
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    (Original post by kikzen)
    i plot points manually.

    it often goes wrong though
    Use the calculator, put the polar equation in parametric form.
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    (Original post by Nylex)
    Use the calculator, put the polar equation in parametric form.
    can't you just put the eqn in like r=...
    thats what i do and it works... just need to change type from y= to r=
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    (Original post by S1M)
    can't you just put the eqn in like r=...
    thats what i do and it works... just need to change type from y= to r=
    If you can do it that way, fine . On the calculator he has though, you can't.
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    (Original post by S1M)
    thanks...
    if you dont have a graphical calculator how are you meant to do it?
    Plotting points like other people have said isn't really the best way of sketching it as it's too time consuming and too easy to get wrong. I just think about or possibly even sketch y = f(theta), and then think about how the polar curve would look. It can also be useful to think about x = r*cos(theta), y = r*sin(theta) - if you think of r = sec(theta) then x = 1 so it's a vertical straight line through x = 1
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    (Original post by Bezza)
    Plotting points like other people have said isn't really the best way of sketching it as it's too time consuming and too easy to get wrong. I just think about or possibly even sketch y = f(theta), and then think about how the polar curve would look. It can also be useful to think about x = r*cos(theta), y = r*sin(theta) - if you think of r = sec(theta) then x = 1 so it's a vertical straight line through x = 1
    woah!!! too much thinking... im just gonna use the graphical calculator... i suppose its ok for some... ill be thinking too much about the other paper... thanks for letting us know how you do it
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    (Original post by S1M)
    woah!!! too much thinking... im just gonna use the graphical calculator... i suppose its ok for some... ill be thinking too much about the other paper... thanks for letting us know how you do it
    If I had one I would too! But I don't, so have to get by using my brain
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    (Original post by Bezza)
    If I had one I would too! But I don't, so have to get by using my brain
    Well at least you can manage without one... if my batteries die im fcuked...
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    (Original post by Bezza)
    If I had one I would too! But I don't, so have to get by using my brain
    explain your method!

    another way is to remember some standard forms i guess...
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    (Original post by kikzen)
    explain your method!

    another way is to remember some standard forms i guess...
    Explain how I use my brain?
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    (Original post by Bezza)
    Explain how I use my brain?
    lol... this is maths not biology...
    (Original post by kikzen)
    explain your method!

    another way is to remember some standard forms i guess...
    he sort of does it above (see below)... its more or less visualisation of the polar graph from the cartesian graphs i think... its not really suitable for all of us
    (Original post by Bezza)
    Plotting points like other people have said isn't really the best way of sketching it as it's too time consuming and too easy to get wrong. I just think about or possibly even sketch y = f(theta), and then think about how the polar curve would look. It can also be useful to think about x = r*cos(theta), y = r*sin(theta) - if you think of r = sec(theta) then x = 1 so it's a vertical straight line through x = 1
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    ok, I'll use an example: r = cos(2t) (t is theta)
    So I'd sketch (or think about) y = cos(2t) which is attached
    You can see that at t = 0, y and therefore r = 1 so it starts out at (1,0)
    As t increases, y (and r) decrease until at t = pi/4, y = r = 0, so the curve will be a tangent to the line theta = pi/4 at this point. As cos(2x) is an even function, you can do the same the other side of the initial half line.
    As t increases beyond pi/4, y and r become negative so are on the opposite side of the origin, at t = pi/2, r = -1 which is the same as point (1, -pi/2)
    Then just keep doing this and you end up with a four leaved shape. For edexcel you don't need to worry about r<0 so you'd only get the 2 "leaves" in the direction of the initial half line.

    The arrows on my polar plot (which you obviously wouldn't put in normally) show the route the general point takes as t increases from 0 so you can see what I was waffling about!!
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    (Original post by Bezza)
    ok, I'll use an example: r = cos(2t) (t is theta)
    So I'd sketch (or think about) y = cos(2t) which is attached
    You can see that at t = 0, y and therefore r = 1 so it starts out at (1,0)
    As t increases, y (and r) decrease until at t = pi/4, y = r = 0, so the curve will be a tangent to the line theta = pi/4 at this point. As cos(2x) is an even function, you can do the same the other side of the initial half line.
    As t increases beyond pi/4, y and r become negative so are on the opposite side of the origin, at t = pi/2, r = -1 which is the same as point (1, -pi/2)
    Then just keep doing this and you end up with a four leaved shape. For edexcel you don't need to worry about r<0 so you'd only get the 2 "leaves" in the direction of the initial half line.

    The arrows on my polar plot (which you obviously wouldn't put in normally) show the route the general point takes as t increases from 0 so you can see what I was waffling about!!
    hmm... thanks... i kind of get ya... but its still not gonna prevent me from using ma graphical calculator
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    (Original post by S1M)
    hmm... thanks... i kind of get ya... but its still not gonna prevent me from using ma graphical calculator
    Your way is definitely easier and quicker!!
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    (Original post by S1M)
    if they give us a polar equation and tell us to sketch it (2 or 3 marks usually)... will we lose marks if we just copy it from a graphical calculator and just work out and label the main points like where it crosses the initial line?

    Thanks
    Your graphics calculator can sketch polar? :eek: I want one...
 
 
 
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