Tough (for me) arc length question

**s3a** · March 25th 2010, 10:50 AM

Question:
Find the length of the curve

= ?

Any help would be greatly appreciated!
Thanks in advance!

**General** · March 25th 2010, 11:33 AM

You have $x=f(y)$ and $a\leq y \leq b$

$L=\int_a^b \sqrt{ 1 + \left[ f'(y) \right]^2} \, dy$

**Deadstar** · March 25th 2010, 11:44 AM

You want to solve...

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

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

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

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

= $\int_{-125}^{512} \frac{1}{16} \frac{64y^{2/3} + 1}{y^{1/3}} dx$

**General** · March 25th 2010, 11:49 AM

Originally Posted by Deadstar

You want to solve...

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

The red one should be "+" not "-" ..

**Deadstar** · March 25th 2010, 12:44 PM

Originally Posted by General

The red one should be "+" not "-" ..

So it should be. It's just a typo, didn't carry it past the line after.

**s3a** · March 26th 2010, 08:54 AM

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.

**Deadstar** · March 26th 2010, 10:29 AM

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.

**drumist** · March 26th 2010, 10:30 AM

A correction is needed. This is correct:

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

However, this simplifies like so:

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

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

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

**s3a** · March 27th 2010, 05:55 AM

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.

**s3a** · March 27th 2010, 10:34 AM

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!

**drumist** · March 27th 2010, 11:05 AM

Reducing it first helps avoid the need for a limit. That is:

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

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

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

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

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

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

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

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

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

Assuming I didn't make any errors. :<

**s3a** · March 27th 2010, 11:29 AM

You're right.

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

$ = -\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:

$ = -\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} $

**Deadstar** · March 27th 2010, 02:06 PM

Originally Posted by s3a

You're right.

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

$ = -\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:

$ = -\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} $

Since I'm browsing past I'll answer...

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

Now you do the other integral.

**s3a** · March 27th 2010, 04:05 PM

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.)

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