Question:
Find the length of the curve
= ?
Any help would be greatly appreciated!
Thanks in advance!
You want to solve...
$\displaystyle \int_{-125}^{512} \sqrt{1 + (f'(y))^2} dx$
= $\displaystyle \int_{-125}^{512} \sqrt{1 + \bigg{(}4y^{1/3} - \frac{1}{16} y^{-1/3}\bigg{)}^2} dx$
= $\displaystyle \int_{-125}^{512} \sqrt{1 + 16y^{2/3} - \frac{1}{2} + \frac{1}{256 y^{2/3}} } dx$
= $\displaystyle \int_{-125}^{512} \sqrt{\frac{1}{256} \frac{(64y^{2/3} + 1)^2}{y^{2/3}}} dx$
= $\displaystyle \int_{-125}^{512} \frac{1}{16} \frac{64y^{2/3} + 1}{y^{1/3}} dx$
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 $
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.
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!
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. :<
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}
$
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.
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.)