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arc length of a function

Petrus

Well-known member
Feb 21, 2013
739
Calculate the length of the curve

We got the formula \(\displaystyle \int_a^b\sqrt{1+[f'(x)]^2}\)
and \(\displaystyle f'(x)=\frac{x}{36}-\frac{9}{x} <=> \frac{x^2-324}{36x}\)
so now we got \(\displaystyle \int_9^{9e}\sqrt{1+(\frac{x^2-324}{36x})^2}\)
we can rewrite that as \(\displaystyle \int_9^{9e}\sqrt{1+\frac{(x^2-324)^2}{1296x^2}}\)
then do integration by part on it well I simply rewrite the function more too \(\displaystyle \int_9^{9e}\sqrt{\frac{1298x^2+(x^2-324)^2}{1296x^2}}\) and set \(\displaystyle u=1298x^2+(x^2-324)^2\) so we got \(\displaystyle dv=1296x^{-2}\) so our \(\displaystyle du=2592x+2x^2-648+x\) and \(\displaystyle v=\frac{1296x^{-3}}{-3}\)Is this correct?
 

earboth

Active member
Jan 30, 2012
74
Calculate the length of the curve

We got the formula \(\displaystyle \int_a^b\sqrt{1+[f'(x)]^2}\)
and \(\displaystyle f'(x)=\frac{x}{36}-\frac{9}{x} <=> \frac{x^2-324}{36x}\)
so now we got \(\displaystyle \int_9^{9e}\sqrt{1+(\frac{x^2-324}{36x})^2}\)
we can rewrite that as \(\displaystyle \int_9^{9e}\sqrt{1+\frac{(x^2-324)^2}{1296x^2}}\)
then do integration by part on it well I simply rewrite the function more too \(\displaystyle \int_9^{9e}\sqrt{\frac{1298x^2+(x^2-324)^2}{1296x^2}}\) and set \(\displaystyle u=1298x^2+(x^2-324)^2\) so we got \(\displaystyle dv=1296x^{-2}\) so our \(\displaystyle du=2592x+2x^2-648+x\) and \(\displaystyle v=\frac{1296x^{-3}}{-3}\)Is this correct?
I would do these calculations a little bit different:

\(\displaystyle \sqrt{1+\left(\frac x{36} - \frac9x \right)^2} = \)
\(\displaystyle \sqrt{1+\frac{x^2}{36^2} - \frac{1}{2} + \frac{81}{x^2} }= \)
\(\displaystyle \sqrt{\left(\frac {x}{36} + \frac{9}{x} \right)^2} = \left| \frac {x}{36} + \frac{9}{x} \right|\)
 

Petrus

Well-known member
Feb 21, 2013
739
I would do these calculations a little bit different:

\(\displaystyle \sqrt{1+\left(\frac x{36} - \frac9x \right)^2} = \)
\(\displaystyle \sqrt{1+\frac{x^2}{36^2} - \frac{1}{2} + \frac{81}{x^2} }= \)
\(\displaystyle \sqrt{\left(\frac {x}{36} + \frac{9}{x} \right)^2} = \left| \frac {x}{36} + \frac{9}{x} \right|\)
ehmm... Cant I do it my way?
edit: I have never done integrate with absolute value, how does that work?
 
Last edited:

earboth

Active member
Jan 30, 2012
74
ehmm... Cant I do it my way?
edit: I have never done integrate with absolute value, how does that work?
With the given domain your integral becomes:

\(\displaystyle \int_9^{9e}\left(\frac x{36} + \frac9x \right) dx\)
 

Petrus

Well-known member
Feb 21, 2013
739
I would do these calculations a little bit different:

\(\displaystyle \sqrt{1+\left(\frac x{36} - \frac9x \right)^2} = \)
\(\displaystyle \sqrt{1+\frac{x^2}{36^2} - \frac{1}{2} + \frac{81}{x^2} }= \)
\(\displaystyle \sqrt{\left(\frac {x}{36} + \frac{9}{x} \right)^2} = \left| \frac {x}{36} + \frac{9}{x} \right|\)
How do you cancel out 1? should it not be \(\displaystyle -\frac{1}{4}\) insted of \(\displaystyle -\frac{1}{2}\). I dont understand this part with integrate with absolute value can just cancel out?
 

earboth

Active member
Jan 30, 2012
74
How do you cancel out 1? should it not be \(\displaystyle -\frac{1}{4}\) insted of \(\displaystyle -\frac{1}{2}\). I dont understand this part with integrate with absolute value can just cancel out?
\(\displaystyle \left(\frac x{36} - \frac9x \right)^2 = \frac{x^2}{36^2} - \underbrace{2\cdot \frac x{36} \cdot \frac9x}_{\frac12} + \frac{9^2}{x^2}\)

... and \(\displaystyle 1-\frac12 = +\frac12\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Calculate the length of the curve

We got the formula \(\displaystyle \int_a^b\sqrt{1+[f'(x)]^2}\)
and \(\displaystyle f'(x)=\frac{x}{36}-\frac{9}{x} <=> \frac{x^2-324}{36x}\)
so now we got \(\displaystyle \int_9^{9e}\sqrt{1+(\frac{x^2-324}{36x})^2}\)
we can rewrite that as \(\displaystyle \int_9^{9e}\sqrt{1+\frac{(x^2-324)^2}{1296x^2}}\)
then do integration by part on it well I simply rewrite the function more too \(\displaystyle \int_9^{9e}\sqrt{\frac{1298x^2+(x^2-324)^2}{1296x^2}}\) and set \(\displaystyle u=1298x^2+(x^2-324)^2\) so we got \(\displaystyle dv=1296x^{-2}\) so our \(\displaystyle du=2592x+2x^2-648+x\) and \(\displaystyle v=\frac{1296x^{-3}}{-3}\)Is this correct?
Petrus, just a minor quibble...when you write integrals, you should include the differential at the end that indicates with respect to what variable you are integrating. (Happy)
 

Petrus

Well-known member
Feb 21, 2013
739
Hello,
I am still confused... Cant I just integrate with just subsitute?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would just integrate term by term:

\(\displaystyle s=\frac{1}{36}\int_9^{9e}x\,dx+9\int_9^{9e}\frac{1}{x}\,dx\)
 

Petrus

Well-known member
Feb 21, 2013
739
Thanks for the help MarkFL and earboth!:)