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    (Original post by economicss)
    Thanks so much! It says in the answers though that the answer for part c is 1, but not sure whether it's wrong or not?!
    Sorry, yes. I rushed this so didn't double check! The answer is 1 because the lowest point wasn't the correct one, it was the point at which the locus is at a normal to the radii. I can draw a diagram up if you want?
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    can someone explain in words, how to solve question 8b on June 2013 (R) Paper thanks

    Also is there a 2015R paper??- if there is can someone set me a link
    link to paper June 2013 R :
    https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

    edit: I don't understand what or why we differentiate
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    (Original post by Cpj16)
    can someone explain in words, how to solve question 8b on June 2013 (R) Paper thanks

    Also is there a 2015R paper??- if there is can someone set me a link
    link to paper June 2013 R :
    https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

    edit: I don't understand what or why we differentiate
    Find the point on the curve such that the tangent is parallel to the initial line, then calculate it's y coordinate.
    Can you see what to do next?
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    (Original post by Alby1234)
    Can anyone telling me what I'm doing wrong in part b? I'm separating the variables but it gives a nasty reciprocal on on of the integrals which proves difficult to solve. Am I going about it the right way?Attachment 544565
    do you have the answers?
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    (Original post by ImJared)
    Sorry, yes. I rushed this so didn't double check! The answer is 1 because the lowest point wasn't the correct one, it was the point at which the locus is at a normal to the radii. I can draw a diagram up if you want?
    If you could please that would be really helpful, thanks
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    (Original post by EricPiphany)
    Find the point on the curve such that the tangent is parallel to the initial line, then calculate it's y coordinate.
    Can you see what to do next?
    Thanks for your help.
    I don't understand what the y co-ordinate there is
    because the polar co-ordinate (r, theta)
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    (Original post by Cpj16)
    Also is there a 2015R paper??- if there is can someone set me a link
    There isn't.
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    (Original post by Cpj16)
    Thanks for your help.
    I don't understand what the y co-ordinate there is
    because the polar co-ordinate (r, theta)
    We have y = r sin(θ).
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    (Original post by economicss)
    If you could please that would be really helpful, thanks
    Just drew this up quickly - a bit scruffy but it demonstrates what I'm talking about. The lowest point is where the line from the origin to the circle is at a tangent (as the gradient no longer gets steeper). Therefore, you can use simple Pythagoras to determine the length from the origin to the point.

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    (Original post by EricPiphany)
    We have y = r sin(θ).
    Ahh thanks

    Also when we have an modulus inequality question with which have the greater or equal to sign.
    Then what do we do?!
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    (Original post by Cpj16)
    Ahh thanks

    Also when we have an modulus inequality question with which have the greater or equal to sign.
    Then what do we do?!
    With complex numbers or reals?
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    (Original post by EricPiphany)
    With complex numbers or reals?
    real

    x+2≥3/x
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    I have added an example of what I mean
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    (Original post by Cpj16)
    real
    Solve exactly as you do the ones with a strictly greater than sign. Then check both sides of any interval behaves as expected, and if not, remove it from the solution.
    Sorry this is a really bad explanation, either someone else will help, or send an example that I can do.
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    (Original post by EricPiphany)
    Solve exactly as you do the ones with a strictly greater than sign. Then check both sides of any interval behaves as expected, and if not, remove it from the solution.
    Sorry this is a really bad explanation, either someone else will help, or send an example that I can do.
    I posted one above
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    (Original post by Cpj16)
    I posted one above
    Name:  Capture.PNG
Views: 126
Size:  43.1 KB
    ( I'm assuming you've been taught how to do this in the past.)T here are other ways too, you could do case work.
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    (Original post by EricPiphany)
    Name:  Capture.PNG
Views: 126
Size:  43.1 KB
    ( I'm assuming you've been taught how to do this in the past.)T here are other ways too, you could do case work.
    thanks this was a really good explanation
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    Will it always follow the same pattern
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    (Original post by ImJared)
    Just drew this up quickly - a bit scruffy but it demonstrates what I'm talking about. The lowest point is where the line from the origin to the circle is at a tangent (as the gradient no longer gets steeper). Therefore, you can use simple Pythagoras to determine the length from the origin to the point.

    Thanks very much! I get that the 2 is the radius but how did you find root 5 please, bit confused about how you know where to draw the lines to find the angle, thanks
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    (Original post by economicss)
    Thanks very much! I get that the 2 is the radius but how did you find root 5 please, bit confused about how you know where to draw the lines to find the angle, thanks
    Distance from centre to origin. Use the distance formula.
 
 
 
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