Question:

Find the length of the curve

http://gauss.vaniercollege.qc.ca/web...6819fae171.png = ?

Any help would be greatly appreciated!

Thanks in advance!

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- Mar 25th 2010, 10:50 AMs3aTough (for me) arc length question
__Question:__

Find the length of the curve

http://gauss.vaniercollege.qc.ca/web...6819fae171.png = ?

Any help would be greatly appreciated!

Thanks in advance! - Mar 25th 2010, 11:33 AMGeneral
You have $\displaystyle x=f(y)$ and $\displaystyle a\leq y \leq b$

$\displaystyle L=\int_a^b \sqrt{ 1 + \left[ f'(y) \right]^2} \, dy$ - Mar 25th 2010, 11:44 AMDeadstar
You want to solve...

$\displaystyle \int_{-125}^{512} \sqrt{1 + (f'(y))^2} dx$

= $\displaystyle \int_{-125}^{512} \sqrt{1 + \bigg{(}4y^{1/3} - \frac{1}{16} y^{-1/3}\bigg{)}^2} dx$

= $\displaystyle \int_{-125}^{512} \sqrt{1 + 16y^{2/3} - \frac{1}{2} + \frac{1}{256 y^{2/3}} } dx$

= $\displaystyle \int_{-125}^{512} \sqrt{\frac{1}{256} \frac{(64y^{2/3} + 1)^2}{y^{2/3}}} dx$

= $\displaystyle \int_{-125}^{512} \frac{1}{16} \frac{64y^{2/3} + 1}{y^{1/3}} dx$ - Mar 25th 2010, 11:49 AMGeneral
- Mar 25th 2010, 12:44 PMDeadstar
- Mar 26th 2010, 08:54 AMs3a
But when I do that I get final answer = 10416.65625 and it's wrong.

More specifically;

3(512)^(4/3) + 3/32 * (512)^(2/3) - 3(-125)^(4/3) - 3/32 *(-125)^(2/3) = 10416.65625. - Mar 26th 2010, 10:29 AMDeadstar
What should the answer be? I've done this a few times now and I always end up with roughly the same (10415.43750). Check your original equation, limits etc and go over the method yourself to see if you get the same final integral as me.

- Mar 26th 2010, 10:30 AMdrumist
A correction is needed. This is correct:

$\displaystyle \int_{-125}^{512} \sqrt{\frac{1}{256} \frac{(64y^{2/3} + 1)^2}{y^{2/3}}} \, dy$

However, this simplifies like so:

$\displaystyle \frac{1}{16} \int_{-125}^{512} \frac{64y^{2/3} + 1}{|y^{1/3}|} \, dy$

So we should split it into two intervals to evaluate it:

$\displaystyle \frac{1}{16} \int_{-125}^0 \frac{64y^{2/3} + 1}{-y^{1/3}} \, dy + \frac{1}{16} \int_0^{512} \frac{64y^{2/3} + 1}{y^{1/3}} \, dy $ - Mar 27th 2010, 05:55 AMs3a
Why is there only an absolute value only on the denominator?

Edit: I see the top will automatically get rid of the negative. I'm guessing I'll need to involve limits for this. I'll do it and say what happened. - Mar 27th 2010, 10:34 AMs3a
Ok so I did it and got it wrong again.

(My work is attached)

Edit: I saw a mistake on the L = (before L = 12) where it should be a negative instead of positive for the thing after the first thing that I slashed off but I still get L = 52 which is wrong as well! - Mar 27th 2010, 11:05 AMdrumist
Reducing it first helps avoid the need for a limit. That is:

$\displaystyle \frac{1}{16} \int_{-125}^0 \frac{64y^{2/3} + 1}{-y^{1/3}} \, dy + \frac{1}{16} \int_0^{512} \frac{64y^{2/3} + 1}{y^{1/3}} \, dy$

$\displaystyle = -\frac{1}{16} \int_{-125}^0 \frac{64y^{2/3} + 1}{y^{1/3}} \, dy + \frac{1}{16} \int_0^{512} \frac{64y^{2/3} + 1}{y^{1/3}} \, dy$

$\displaystyle = -\frac{1}{16} \int_{-125}^0 \left(64y^{1/3} + y^{-1/3}\right) dy + \frac{1}{16} \int_0^{512} \left(64y^{1/3} + y^{-1/3}\right) dy$

$\displaystyle = -\frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_{-125}^0 + \frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_0^{512}$

$\displaystyle = -\frac{1}{16} \left[-48(-125)^{4/3}-\tfrac{3}{2}(-125)^{2/3}\right] + \frac{1}{16} \left[ 48(512)^{4/3} + \tfrac{3}{2}(512)^{2/3}\right]$

$\displaystyle = -\frac{1}{16} \left[-48(-5)^4-\tfrac{3}{2}(-5)^2\right] + \frac{1}{16} \left[ 48(8)^4 + \tfrac{3}{2}(8)^2\right]$

$\displaystyle = -\frac{1}{16} \left[-48(625)-\tfrac{3}{2}(25)\right] + \frac{1}{16} \left[ 48(4096) + \tfrac{3}{2}(64)\right]$

Letting a calculator finish the work, I get that this equals

$\displaystyle =\frac{453483}{32} \approx 14171$

Assuming I didn't make any errors. :< - Mar 27th 2010, 11:29 AMs3a
You're right.

However, I don't get how you get from here:

$\displaystyle

= -\frac{1}{16} \int_{-125}^0 \left(64y^{1/3} + y^{-1/3}\right) dy + \frac{1}{16} \int_0^{512} \left(64y^{1/3} + y^{-1/3}\right) dy

$

to here:

$\displaystyle

= -\frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_{-125}^0 + \frac{1}{16} \left[ 48y^{4/3} + \tfrac{3}{2}y^{2/3}\right]_0^{512}

$ - Mar 27th 2010, 02:06 PMDeadstar
Since I'm browsing past I'll answer...

$\displaystyle -\frac{1}{16} \int_{-125}^0 \left( 64y^{1/3} + y^{-1/3} \right) dy = -\frac{1}{16} \left[ 64 \frac{y^{4/3}}{4/3} + \frac{y^{2/3}}{2/3} \right]_{-125}^0$ = $\displaystyle -\frac{1}{16} \left[ 64 y^{4/3} \cdot \tfrac{3}{4} + \tfrac{3}{2} y^{2/3} \right]_{-125}^0 = -\frac{1}{16} \left[ 48 y^{4/3} + \tfrac{3}{2}y^{2/3} \right]_{-125}^0$

Now you do the other integral. - Mar 27th 2010, 04:05 PMs3a
Oh my gosh! I'm so stupid, my eyes were thinking that the next step was still an integral! (It's kind of like a blind spot if you close one eye and look at a dot on a piece of graph paper; you won't see the dot because your brain fills in the dot's space with what it thinks it will be which is more graph paper.)