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    (Original post by Farhan.Hanif93)
    Even though this problem works by considering the equation of the tangent and the equation of the perpendicular to 2x+3y=0 through the origin, as you have shown. This is not the intended method and may lose you marks in the exam. You need to consider the NORMAL at P and the line 2x+3y=0.

    I think what you've posted may confuse a few as it's the roundabout way of doing it.
    Aah right; my bad, sorry.
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    (Original post by Femto)
    Aah right; my bad, sorry.
    No worries. No need to apologise. :p:
    It's good to see that you've taken my advice and decided to help out on here. Keep it up! :cool:
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      (Original post by Femto)
      Great, but why do you have to equate it when y=3/2 but not y=-2/3?

      Even so, surely the reciprocal of -2/3 is -3/2?
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      (Original post by Farhan.Hanif93)
      Even though this problem works by considering the equation of the tangent and the equation of the perpendicular to 2x+3y=0 through the origin, as you have shown. This is not the intended method and may lose you marks in the exam. You need to consider the NORMAL at P and the line 2x+3y=0.

      I think what you've posted may confuse a few as it's the roundabout way of doing it.
      Just out of interest, what do you mean by considering the normal? I don't want to lose marks on my exam

      I just thought that the method I used was right but if it isn't I can't carry on with it
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      (Original post by im so academic)
      Great, but why do you have to equate it when y=3/2 but not y=-2/3?

      Even so, surely the reciprocal of -2/3 is -3/2?
      Listen to Farhan he gives a more adequate explanation as to why - I think I'm confusing you. The reason I equated it to 3/2 was because 3/2 is the gradient of the tangent. Also, the negative reciprocal of -2/3 is 3/2.
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      (Original post by Femto)
      Just out of interest, what do you mean by considering the normal? I don't want to lose marks on my exam
      The gradient of the tangent to the curve y=k\sqrt x at x=4 is \frac{k}{4}.
      Therefore the gradient of the normal (which has a gradient perpendicular to k/4) is -\frac{4}{k}. Recall that if m is the gradient of a line then the gradient of any line perpendicular to this one has a gradient of -\frac{1}{m}.

      Then note that the question tells us that the line 2x+3y=0 is parallel to this normal. Which is another way of saying that they have the same gradient. As you noted, this line can be expressed as y=-\frac{2}{3}x. The gradient of this line is -\frac{2}{3}. You know that this gradient must be equal to the gradient of the normal above so setting them equal to each other:
      -\frac{2}{3}=-\frac{4}{k}.
      Solve for k and you're done.
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        (Original post by Femto)
        Listen to Farhan he gives a more adequate explanation as to why - I think I'm confusing you. The reason I equated it to 3/2 was because 3/2 is the gradient of the tangent. Also, the negative reciprocal of -2/3 is 3/2.
        Oh, the negative reciprocal. Thanks for that.

        The one thing I don't get now is why is the tangent gradient needed.
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        (Original post by Farhan.Hanif93)
        The gradient of the tangent to the curve y=k\sqrt x at x=4 is \frac{k}{4}.
        Therefore the gradient of the normal (which has a gradient perpendicular to k/4) is -\frac{4}{k}. Recall that if m is the gradient of a line then the gradient of any line perpendicular to this one has a gradient of -\frac{1}{m}.

        Then note that the question tells us that the line 2x+3y=0 is parallel to this normal. Which is another way of saying that they have the same gradient. As you noted, this line can be expressed as y=-\frac{2}{3}x. The gradient of this line is -\frac{2}{3}. You know that this gradient must be equal to the gradient of the normal above so setting them equal to each other:
        -\frac{2}{3}=-\frac{4}{k}.
        Solve for k and you're done.
        Aah yes, that makes sense; thanks for the information!
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        (Original post by im so academic)
        Oh, the negative reciprocal. Thanks for that.

        The one thing I don't get now is why is the tangent gradient needed.
        It's not, simply put - read Farhan's post (He just posted it). It explains it well.
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        (Original post by im so academic)
        Oh, the negative reciprocal. Thanks for that.

        The one thing I don't get now is why is the tangent gradient needed.
        Because, with C1 knowledge, you cannot find the gradient of the normal without first finding the gradient of the tangent.
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          OK this is my alternative version:

          dy/dx = k/4
          negative reciprocal of that (as it is the normal of that line) = -4/k

          equate this to y=-3/2

          i.e. -4/k = -3/2

          rearrange for k

          k=6



          Would this method be acceptable? As I find it easier to understand to take the negative reciprocal of k/4 as that is what the question states.
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            (Original post by Farhan.Hanif93)
            x
            Oh damn, you've already done it. Should've seen.

            Thank you, I totally get this explanation.

            Apologies for not seeing it before.
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              (Original post by Farhan.Hanif93)
              Because, with C1 knowledge, you cannot find the gradient of the normal without first finding the gradient of the tangent.
              The normal is -1/(dy/dx)

              I guess the gradient of tangent = dy/dx

              In that case, I understand it that way, thanks.
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              (Original post by im so academic)
              OK this is my alternative version:

              dy/dx = k/4
              negative reciprocal of that (as it is the normal of that line) = -4/k

              equate this to y=-3/2

              i.e. -4/k = -3/2

              rearrange for k

              k=6



              Would this method be acceptable? As I find it easier to understand to take the negative reciprocal of k/4 as that is what the question states.
              If you read above, that's what I posted. :p:
              That's the correct way to go about it.
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                (Original post by Farhan.Hanif93)
                If you read above, that's what I posted. :p:
                That's the correct way to go about it.
                Yes...

                And it is the way I understand it.

                Thank you very much for all your help. :hat:
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                Don't mean to burst all of your collective bubbles, but for C1 (if this is OCR MEI), no knowledge of calculus is assumed. I guess you wouldn't get the marks for doing it by differentiation? Although, come to think of it, surely you can't be penalised for knowing more than they assume.
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                  (Original post by pinky198)
                  Don't mean to burst all of your collective bubbles, but for C1 (if this is OCR MEI), no knowledge of calculus is assumed. I guess you wouldn't get the marks for doing it by differentiation? Although, come to think of it, surely you can't be penalised for knowing more than they assume.
                  There is in Edexcel and AQA, not sure about OCR specifications though.
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                  OCR B also includes differentiation for C1.
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                  (Original post by im so academic)
                  There is in Edexcel and AQA, not sure about OCR specifications though.
                  Bloody hell! I feel cheated - didn't start calculus until C2!
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                  Can I ask which paper this question is off, if anyone knows?
                 
                 
                 
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