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Sorry, this is my first post ever and formatting is bad. I appreciate all assistance

Given the equation y= $ 3.8x^2-4.4x+1444 $ Find the definite arc length integral between 0 and 1200.

I am not sure on how to format $(\frac {dy}{dx})^2$ ,so I left it alone.

$$\int_0^{1200} \sqrt{1+ (\frac{dy}{dx}})^2 dx$$

Michael
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  • I think this ought to read $y=3.8x^2-4.4x+1444$ As it is, your equation has two complex solutions (no real solutions)...it does not define a curve. – lulu Nov 13 '15 at 00:47
  • OH my I am so sorry. I knew I would mess up the formatting. It is fine now, right? – Michael Nov 13 '15 at 00:50
  • Your formatting is fine! Can you solve the problem from here? $\frac {dy}{dx}$ isn't hard to compute. – lulu Nov 13 '15 at 00:52
  • $(7.6x-4.4)$. My issue is I dont know how to expand the integral with the square root. – Michael Nov 13 '15 at 00:54
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    Good. Can you compute $\int \sqrt {1+x^2} dx? – lulu Nov 13 '15 at 00:55
  • To be clear: that integral is doable, but it isn't exactly easy. Here's a discussion of the "arclength of a parabola" issue...not hard to adapt to your situation: http://math.stackexchange.com/questions/229642/length-of-a-parabolic-curve – lulu Nov 13 '15 at 00:59
  • Indeed it does, though as I say it gets you to an integral which, while standard, is not straightforward. It's unintuitive....I think one would expect the parabola to have a straightforward arclength. Mind you, things like ellipses are even worse. – lulu Nov 13 '15 at 01:03
  • Is there any website or place you recommend that I can learn such derivatives? I haven't really dealt with hyperbolic functions in high school yet. – Michael Nov 13 '15 at 01:08
  • You can avoid the hyperbolic functions. Use the substitution $z=tan(z)$. That gets you to $\int sec^3(z) dz$. Have you done that one yet? Here is a fairly clear discussion: http://hubpages.com/education/How-to-Integrate-Sqrt-1-x2-Sqrt-a2-x2 – lulu Nov 13 '15 at 01:10
  • Final question. How many pages would this take to solve and explain? – Michael Nov 13 '15 at 01:21
  • I think the link I just sent pretty well does it...so I'd say about one page. The integrals of $sec(x)$ and $sec^3(x)$ are worth studying, they do tend to come up. – lulu Nov 13 '15 at 01:24
  • So the equation I put up in the question would be one page right? :) – Michael Nov 13 '15 at 01:26
  • Absolutely. You can do it in less if you assume your audience knows $\int sec^3(x) dx$ (or if you are willing to simply write down the solution for that without discussion). – lulu Nov 13 '15 at 01:35

2 Answers2

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In the same spirit as David G. Stork'answer, consider the general case of $$y=ax^2+bx+c$$ $$\frac{dy}{dx}=2ax+b$$ So $$L=\int \sqrt{1+ (\frac{dy}{dx})^2}\, dx=\int \sqrt{1+ (2ax+b)^2}\, dx$$ Now, make a change of variable $$2ax+b=t \implies x=\frac{t-b}{2 a}\implies dx=\frac{dt}{2 a}$$ So $$L=\frac{1}{2 a}\int \sqrt{1+t^2}\,dt$$ Again, a standard change of variable $$t=\sinh(z)\implies dt=\cosh(z)\,dz$$ $$L=\frac{1}{2 a}\int \cosh^2(z)\,dz=\frac{1}{2 a}\int \frac 12 \left(1+\cosh(2z)\right)\,dz=\frac{1}{4 a}\left(z+\frac 12 \sinh(2z)\right)$$ Back to $x$, this gives $$L=\frac{(2 a x+b)\sqrt{1+(2 a x+b)^2} +\sinh ^{-1}(2 a x+b)}{4 a}$$

Claude Leibovici
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  • The final derivative is the indefinite, right? If so, do I need to calculate boundaries, or can I simply use $F(b)-f(a)$ to calculate the length? – Michael Nov 14 '15 at 03:09
  • Yes to the question. You have $L(x)$ and so the arc length is $L(b)-L(a)$ – Claude Leibovici Nov 14 '15 at 03:40
  • Oh sorry for asking this a bit late. Why change t into $sinh$? why not $tanh$? Or why not any other expression? Sorry, I am not well educated on hyperbolic functions and when to use them. In the future, when should input a hyperbole and when shouldn't I ? – Michael Nov 21 '15 at 19:50
  • @Michael. When I see $\int \sqrt{1+t^2},dt$, I always think about $t=\sinh(z)$ but, as you say, we could have used $t=\tan(z)$ for the same reason. In such a case, $$L=\frac{1}{2 a}\int \sec ^3(z),dz$$. If you want me to continue the calculations, let me know. – Claude Leibovici Nov 22 '15 at 03:56
  • The goal of substitution is to arrive to simpler expressions to be integrated. – Claude Leibovici Nov 22 '15 at 04:32
  • Integration by parts in also part of the story in many cases. Again, the goal of substitution is to arrive to something simpler for which we almost know the result. – Claude Leibovici Nov 22 '15 at 04:42
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Take the derivative and plug into your integral and use the fact that

$\int \sqrt{1 + (a + b x)^2} dx = \frac{\sqrt{(a+b x)^2+1} (a+b x)+\sinh ^{-1}(a+b x)}{2 b}$.

David G. Stork
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