Length of a parabola? [Solved!]

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.

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.

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

X

Get everything under the square root, over `L^4`

X

OK. 

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

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

`=1/(L^2) inta^b sqrt( L^4+16R^2 L^2-64R^2Lx+ 64R^2 x^2)`

I'm stuck again. And what's `a` and `b`?

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?

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?

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&#x2f;L&#x5e;2 &#x28;sqrt&#x28; L&#x5e;4&#x2b;16&#x2a;R&#x5e;2&#x2a;L&#x5e;2-64&#x2a;R&#x5e;2&#x2a;L&#x2a;x&#x2b; 64&#x2a;R&#x5e;2&#x2a;x&#x5e;2&#x29;&#x29;dx - Wolfram|Alpha</a>

Definitely not something we should be doing on paper!

X

I agree!

Thanks a lot.