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Math Help - Differentiation of quotient

  1. #1
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    Differentiation of quotient

    i am a structural engineer analysing a frame. I am trying to find the ponit of maximum bending moment along a beam which is essentially the minimum point on a curve. i am following an example from a text which describe the distance to the max moment using the following expression:

    h=(90.9x-4.55x^2)/(10.16+0.4x)

    Putting dh/dx=0 he finds an expression from the min point on the curve

    The answer is quoted as x^2+50.7x-507.4=0 and then solves for x=8.65

    My calculus is rusty but I have tried to differentiate using the quotient rule quoted in maths text books. I cannot replicate this answer; the answer i get is not sensible. Can anyone help?
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  2. #2
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    Quote Originally Posted by parrot77 View Post
    i am a structural engineer analysing a frame. I am trying to find the ponit of maximum bending moment along a beam which is essentially the minimum point on a curve. i am following an example from a text which describe the distance to the max moment using the following expression:

    h=(90.9x-4.55x^2)/(10.16+0.4x)

    Putting dh/dx=0 he finds an expression from the min point on the curve

    The answer is quoted as x^2+50.7x-507.4=0 and then solves for x=8.65

    My calculus is rusty but I have tried to differentiate using the quotient rule quoted in maths text books. I cannot replicate this answer; the answer i get is not sensible. Can anyone help?
    Please show your attempt at using the quotient rule.
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  3. #3
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    I don't think the answer quoted is correct.

    When I tried differentiating the equation h=\dfrac{90.9x-4.55x^2}{10.16+0.4x}, i didn't get x^2+50.7x-507.4=0

    And when you put x^2+50.7x-507.4=0 into the quadratic formula, you don't get x=8.65, you get x = -13.72 and x = -36.98.

    If you recall the quotient rule, f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} you would see that the answer should come out as a fraction, not as a quadratic binomial. Not always, but in this case it should.
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  4. #4
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    If h=u/v dh/dx= vdu/dx-udv/dx/v^2

    =(10.16+0.4x)(90.9-9.1x)-(90.9x-4.55x^2)(0.4)/(10.16+0.4x)^2

    =(923.544-92.45x+36.36x-3.64x^2-36.36x+1.82X^2)/(0.16X^2+8.128+103.22)

    =(-1.82X^2-92.45X+923.544)/(0.16X^2+8.128X+103.22)

    =11.375X^2+11.37X-8.94=0
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  5. #5
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    Sorry guys it should have read x=8.56 not 8.65

    If i sub this answer into his formula it seems to work, ie:

    8.56^2 +50.7(8.56)- 507.4= 0
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  6. #6
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    Quote Originally Posted by parrot77 View Post
    If h=u/v dh/dx= vdu/dx-udv/dx/v^2

    =(10.16+0.4x)(90.9-9.1x)-(90.9x-4.55x^2)(0.4)/(10.16+0.4x)^2

    =(923.544-92.45x+36.36x-3.64x^2-36.36x+1.82X^2)/(0.16X^2+8.128+103.22)

    =(-1.82X^2-92.45X+923.544)/(0.16X^2+8.128X+103.22)

    =11.375X^2+11.37X-8.94=0
    taking from =(-1.82X^2-92.45X+923.544)/(0.16X^2+8.128X+103.22)=0

    (-1.82X^2-92.45X+923.544)=(0.16X^2+8.128X+103.22)*0

    (-1.82X^2-92.45X+923.544)=0

    can you finish now?
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  7. #7
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    Yes I can... divide accross by -1.82 gives

    x^2+50.7-507.4=0

    which is the answer given in the text. Where I went wrong was trying to divide out the fraction instead of bringing it across and mult by the zero

    Thanks very much BabyMilo, a lot of people talked about it but no one else could show me where i had gone wrong!!
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  8. #8
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    Quote Originally Posted by Educated View Post
    I don't think the answer quoted is correct.

    When I tried differentiating the equation h=\dfrac{90.9x-4.55x^2}{10.16+0.4x}, i didn't get x^2+50.7x-507.4=0

    And when you put x^2+50.7x-507.4=0 into the quadratic formula, you don't get x=8.65, you get x = -13.72 and x = -36.98.

    If you recall the quotient rule, f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} you would see that the answer should come out as a fraction, not as a quadratic binomial. Not always, but in this case it should.
    No, the "answer" is a number. Since a fraction is 0 if and only if the numerator is 0, it is quite possible that the binomial given is the numerator of the derivative fraction.
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  9. #9
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    thanks for the useful information guys. So we differentiate to determine an expression for dy/dx or dh/dx in this case? And for every case where we allow dy/dx=0 (ie at the minimum pont on the curve) the v^2 numerator in the quotient rule is obsolete, as it is in fact zero?
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  10. #10
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    Sorry to post multiple replys but i got the values for x using the formula x= -b+/-(b^2-4ac)^0.5/2a which worked out as x=8.56m and x= -59.22m. 8.56m is the only rational answer in this case (when the frame geometry is examined). HallsofIvy you get the values you have quoted above (x = -13.72 and x = -36.98) when you erroneously input c as 507.4 instead of -507.4.
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