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 February 18th, 2009, 03:58 AM #1 Newbie   Joined: Feb 2009 Posts: 2 Thanks: 0 Calc length by integration So I have this problem: determine the length of the curve defined by $y= \frac{2}{3} \cdot x ^{ \frac{2}{3} }$ between the points $a= \left( 1, \frac{2}{3} \right)$ and $b= \left( 8, \frac{8}{3} \right)$. I've gotten this far: $\lim_{ n \to \infty } \sum_{ i= 1}^{n} \sqrt{ \left( i \left( \frac{x}{n} \right) - \left( i - 1 \right) \left( \frac{x}{n} \right) \right)^2 + \left( f \left( i \left( \frac{x}{n} \right) \right) - f \left( \left( i - 1 \right) \left( \frac{x}{n} \right) \right) \right)^2 }$. I still have to substitute $f$, but after that how do I turn this into an integral? I've been at this for an hour (-.-'). Help...
February 18th, 2009, 05:00 AM   #2
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Re: Calc length by integration

Quote:
 Originally Posted by bogged So I have this problem: determine the length of the curve defined by $y= \frac{2}{3} \cdot x ^{ \frac{2}{3} }$ between the points $a= \left( 1, \frac{2}{3} \right)$ and $b= \left( 8, \frac{8}{3} \right)$. I've gotten this far: $\lim_{ n \to \infty } \sum_{ i= 1}^{n} \sqrt{ \left( i \left( \frac{x}{n} \right) - \left( i - 1 \right) \left( \frac{x}{n} \right) \right)^2 + \left( f \left( i \left( \frac{x}{n} \right) \right) - f \left( \left( i - 1 \right) \left( \frac{x}{n} \right) \right) \right)^2 }$. I still have to substitute $f$, but after that how do I turn this into an integral? I've been at this for an hour (-.-'). Help...
This might help.

February 18th, 2009, 01:32 PM   #3
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Re: Calc length by integration

Quote:
 Originally Posted by bogged So I have this problem: determine the length of the curve defined by $y= \frac{2}{3} \cdot x ^{ \frac{2}{3} }$ between the points $a= \left( 1, \frac{2}{3} \right)$ and $b= \left( 8, \frac{8}{3} \right)$. I've gotten this far: $\lim_{ n \to \infty } \sum_{ i= 1}^{n} \sqrt{ \left( i \left( \frac{x}{n} \right) - \left( i - 1 \right) \left( \frac{x}{n} \right) \right)^2 + \left( f \left( i \left( \frac{x}{n} \right) \right) - f \left( \left( i - 1 \right) \left( \frac{x}{n} \right) \right) \right)^2 }$. I still have to substitute $f$, but after that how do I turn this into an integral? I've been at this for an hour (-.-'). Help...
The derivative of arc length is given by (1+(dy/dx)^2)^1/2 (integrate with respect to x) or (1+(dx/dy)^2)^1/2 (integrate with respect to y). They should both give the same answer, but one may be easier than the other.

 February 18th, 2009, 02:13 PM #4 Newbie   Joined: Feb 2009 Posts: 2 Thanks: 0 Re: Calc length by integration Should've done a quick search first, sorry. So we're at: $\int_1^8 \sqrt{ 1 + \left( \frac{2}{3} x^{ \frac{2}{3} } \right) ^{ \prime } } dx = \int_1^8 \sqrt{ 1 + \frac{4}{9} x^{ - \frac{1}{3} } } dx$ And I'm stumped again... There's no formula for the integral of g after f (as far as I know.) (I didn't really understand your explanation. Way too advanced for me...) (BTW, I have a test tomorrow. *shivers*)
February 18th, 2009, 03:09 PM   #5
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Re: Calc length by integration

Quote:
 Originally Posted by bogged Should've done a quick search first, sorry. So we're at: $\int_1^8 \sqrt{ 1 + \left( \frac{2}{3} x^{ \frac{2}{3} } \right) ^{ \prime } } dx = \int_1^8 \sqrt{ 1 + \frac{4}{9} x^{ - \frac{1}{3} } } dx$ And I'm stumped again... There's no formula for the integral of g after f (as far as I know.) (I didn't really understand your explanation. Way too advanced for me...) (BTW, I have a test tomorrow. *shivers*)
You forgot to square $\frac{4}{9} x^{ - \frac{1}{3}$.

I've tried working this one out using trig substitution and the like, but had no success. That's not to say there isn't an answer, though. Just seems to me to be quite complicated.

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