# Thread: Tough (for me) arc length question

1. ## Tough (for me) arc length question

Question:
Find the length of the curve

= ?

Any help would be greatly appreciated!

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

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

3. You want to solve...

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

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

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

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

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

4. 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 "-" ..

5. 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.

6. 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.

7. 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.

8. 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:

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

9. 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.

10. 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!

11. 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. :<

12. 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}
$

13. 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.

14. 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.)