I tried to use the arc length formula but there were too many letters in the problem.

X

What is the length of the parabola where dy/dx = (4Rx(L-2x)) / L^2? Here's the diagram:
<img src="/forum/uploads/parabola310.png" width="310" height"145" alt="parabola" />

Relevant page
<a href="/methods-integration/5-integration-other-trigonometric-forms.php">5. Integration: Other Trigonometric Forms</a>
What I've done so far
I tried to use the arc length formula but there were too many letters in the problem.

You are encouraged to use the math entry system so it's easier for you (and us) to read your math.

At least show us your first steps so we can guide you better.

Gentle warning: This has a really complicated solution. We'll get started and I'll stop you before it's too late.

X

Hi again
You are encouraged to use the math entry system so it's easier for you (and us) to read your math.
At least show us your first steps so we can guide you better.
<b>Gentle warning:</b> This has a really complicated solution. We'll get started and I'll stop you before it's too late.

I get it a bit more now. They are already giving me the `dy/dx` in the question. Here's what I've got:

`s=inta^b sqrt(1+(dy/dx)^2)`

`s=inta^b sqrt(1+((4R(L-2x))/L^2)^2)`

`s=inta^b sqrt(1+((16R^2(L^2-4Lx+4x^2))/L^4))`

I'm stuck there

X

I get it a bit more now. They are already giving me the `dy/dx` in the question. Here's what I've got:
`s=inta^b sqrt(1+(dy/dx)^2)`
`s=inta^b sqrt(1+((4R(L-2x))/L^2)^2)`
`s=inta^b sqrt(1+((16R^2(L^2-4Lx+4x^2))/L^4))`
I'm stuck there

Yes, that's good. But that's really the limit of reasonable calculation. The integral from there is really ugly.

For `a` and `b`, go back to your diagram.

Where is the origin? What are the `x`-coordinates of the far left and far right points of the parabola?

X

Yes, that's good. But that's really the limit of reasonable calculation. The integral from there is really ugly.
For `a` and `b`, go back to your diagram.
Where is the origin? What are the `x`-coordinates of the far left and far right points of the parabola?

If we have values for `R` and `L`, can it be done?

X

`a=0` and `b=L`, correct?
So the integral will be
`s=1/(L^2) int0^L sqrt( L^4+16R^2 L^2-64R^2Lx+ 64R^2 x^2)`
If we have values for `R` and `L`, can it be done?

Definitely not something we should be doing on paper!

X

I put it into Wolfram|Alpha with values `L=10` and `R=3`. The answer was `12.04`.
You can see the numeric solution here:
<a href="http://www.wolframalpha.com/input/?i=integral+from+0+to+10+0.01*%28sqrt%28+L^4%2B16*R^2*L^2-64*R^2*L*x%2B+64*R^2*x^2%29%29dx+where+L+%3D+10+and+R+%3D+3">Numeric solution with L=10 and R=3</a>
and the (horrible) algebraic answer here:
<a href="http://www.wolframalpha.com/input/?i=integral+1%2FL^2+%28sqrt%28+L^4%2B16*R^2*L^2-64*R^2*L*x%2B+64*R^2*x^2%29%29dx"> integral 1/L^2 (sqrt( L^4+16*R^2*L^2-64*R^2*L*x+ 64*R^2*x^2))dx - Wolfram|Alpha</a>
Definitely not something we should be doing on paper!