Implicit Diff. / Stationary Points

Discussion in 'Math' started by Fraser_Integration, Jan 12, 2010.

  1. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    Hi there. Bit of a headscratcher here. I am perplexed by part (c) of the attached question.

    I understand that if the tangent makes an angle of pi/4 with the horizontal axis, this is equal to a gradient of 1. When I plug this into the value of dy/dx I worked out, I come out with the same co-ordinates as those in part b) ! This surely can't be.

    my answer to part a) : dy/dx = (2x - y) / (x + 1 - 2y)

    part b) stationary when 2x - y = 0, therefore sub y = 2x into first equation, and you come out with a quadratic 3x^2 - 2x -1 = 0 and this solves for x is -1/3 or + 1

    then c) as dy/dx is equal to 1, then 2x - y = x + 1 - 2y

    this re-arranges to y = 1 -x however when you sub that in you get the same quadratic as b) , please help!
     
    Last edited: Jan 12, 2010
  2. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
    142
    5
    oh, i figured it out. the original equation is not a function, but an oval shape map, so yes there can be two values for y. damn
     
  3. studiot

    AAC Fanatic!

    Nov 9, 2007
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    I didn't see this until after I finished my DVD tonight.

    I hope you figured correctly.

    Your working is correct as far as you have shown it, in that you have found the same quadratic for x at both y' = 0 and y' = 1 so x = +1 or -1/3 agreed.

    However

    when y' = 0, y = 2x so y = +2 or -2/3 ie the points (1,2) & (-1/3,-2/3)

    when y' = 1 y = 1-x so y = 0 or +4/3 ie the points (1,0) & (-1/3,4/3)

    So the points are different even if the quadratic is the same.
     
  4. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    awesome, I still couldn't figure out how "y = 2x" gave me 4 values for two values of y. Forgot about the other identity. Cheers guy.
     
  5. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Good to see someone taking a deeper interest.
    What does the integration stand for Maths or Social?

    For your interest, your example contained a quadratic that worked out OK for all values.

    However this is not always the case so you need always to be careful and check.

    The attachment contains an example of an implicit expression in x and y at 1.

    If you rearrange and use the quadratic formula you can gain an explicit expression. But beware Will Robinson you have to take the correct sign for it to work.

    You touched on this in your comment about functions. Strictly 1 is not a single function but the junction of two separate ones which are separated in 2.
     
  6. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    Thanks for that. It seems that there is a lot more to implicit functions and their derivatives than we were taught in our lectures (1st year electronic engineering incase you wondered), I guess they are wanting us to take an initiative!

    Yep the Integration is for Maths. I enjoy calculus, but starting to enjoy it less now it is getting tougher!

    If you're feeling clever, could you look at this question too please? I don't need a worked answer, just a clue about where to start. Cheers....
     
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Try this hint.

    If the expression is of the form (which yours is) \phi (x,y ) = a constant
    Where \phi is a function of x,y

    Then d\phi/dx = 0

    and this leads to ∂\phi/∂x + ∂\phi/∂y * dy/dx = 0

    which leads to dy/dx = - ∂\phi/∂x \div\phi/∂y (note the minus)

    I think this has all the ingredients you seek. Sorry it is difficult to format math here. Very happy for any TEX expert to improve it though.
     
  8. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
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    hmm I'll have a go!

    Where it gives you the values of u and v (1, 0) am I to put these values in at the start to eliminate the f^2 term, I take it?
     
  9. studiot

    AAC Fanatic!

    Nov 9, 2007
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    u, v are equivalent to my x, y
    f is a function
    f^2 is a function of a function so you need the chain rule for differentiation
    the whole left hand expression is equivalent to my \phi
     
  10. Fraser_Integration

    Thread Starter Member

    Nov 28, 2009
    142
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    nah, sorry, thanks for trying but I don't get it. I've never differentiated an implicit function in that form before, let alone the square of one!

    dont worry, I will bother someone who gets paid to help me!
     
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