1. ## 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?

2. Originally Posted by parrot77
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?

3. I don't think the answer quoted is correct.

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

And when you put $\displaystyle x^2+50.7x-507.4=0$ 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, $\displaystyle f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$ 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.

4. 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

5. 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

6. Originally Posted by parrot77
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?

7. 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!!

8. Originally Posted by Educated
I don't think the answer quoted is correct.

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

And when you put $\displaystyle x^2+50.7x-507.4=0$ 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, $\displaystyle f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$ 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.

9. 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?

10. 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.