# calculate the length of a curve

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hi there, I'm trying to calculate the length of a curve with a given equation between two points along the x axis (a and b). I know the standard integral for this is: sqrt(1 + (dy/dx)^2) dx, in range [a - b]. but my question is - if you have to differentiate the equation of the curve to get dy/dx, then when I find the anti-derivative of the above equation would it not just be equal to: 2/3(x + 1/3(A)^3)^3/2 where A equals the equation of the curve?

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Well, that's obviously wrong, as the value of a function at one particular point doesn't say anything about the values it had before, so it also doesn't say anything about the length of the parts before it.

Could you show how you got to that equation?

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the first equation he gave was correct:

$L=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$

his second is not :P

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I think that what you're doing wrong is that you assume that

when

a(x) = integral a'(x)

and

b(x) = integral b'(x)

that

integral a'(b'(x)) == a(b(x))

which is wrong.

If you would do it the other way around, ie. calc the derivative of a(b(x)), you could use the chain rule, but unfortunately we don't have such a simple formula for anti derivatives.

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so when I calculate the derivative (dy/dx) what do I do next?

from what I've read when you have the integral:

sqrt(1 + (dy/dx)^2)

you have to find the anti-derivate of this F(x)

for the two limits a and b and then subtract them:

area = F(b) - F(a)

I know this gives the area, but could someone explain how I use the above equation to get the length.

Quote:
 Could you show how you got to that equation?

I though the anti-derivative of dy/dx was y so I plugged this into equation one and then integrated.

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Quote:
 Original post by staticVoid2so when I calculate the derivative (dy/dx) what do I do next?from what I've read when you have the integral:sqrt(1 + (dy/dx)^2)you have to find the anti-derivate of this F(x)for the two limits a and b and then subtract them:area = F(b) - F(a)I know this gives the area, but could someone explain how I use the above equation to get the length.

this is correct.

Quote:
 I though the anti-derivative of dy/dx was y so I plugged this into equation one and then integrated.

The anti derivative of dy/dx is indeed y, but the way you are plugging things into your integral is wrong.

Maybe you could give the actual definition of your curve function, so we can solve it explicitely for that one? Then it should also become obvious why your current way doesn't work.

Thing is that there's not gonna be a general solution, that will work for any function, with the curve function and it's derivatives at fixed places. The way you solve it will really depend on the definition of your curve function.

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ok, the curve function is y = 25e^(-x^2/625)

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$y=25e^{-\frac{1}{625}x^2}$

$\frac{dy}{dx}=-0.08xe^{-x^2/625}$

$\\L=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx\\\\=\int_a^b\sqrt{1+0.0064x^2e^{-\frac{2}{625}x^2}}dx$

which according to wolfram's integrator, has no solution; not even a non analytic solution, so you would have to compute it with numerically.

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so do you mean I would just have to approximate the value using riemann sums or the trapezium rule?

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yes it has to be approximated

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This is why i'm content with my very novice understanding of calculus. I'll just burn some clock cycles rather than brain cells.

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Quote:
 Original post by staticVoid2so do you mean I would just have to approximate the value using riemann sums or the trapezium rule?

Don't use the trapezium rule, use simpson's rule, it reduces the number of subintervals to sum :)