# arc length difficulty

• Apr 5th 2009, 10:22 AM
Stonehambey
arc length difficulty
Not sure if I'm having another brain freeze here or something, but I'm having trouble solving the following types of problems. Find the arc length of the following curves between the points specified

1) $\displaystyle y = x^{\frac{1}{2}} - \frac{1}{3}x^{\frac{3}{2}}$, between $\displaystyle x = 1$ and $\displaystyle x = 4$

2) $\displaystyle y = \frac{1}{3}x^3 + \frac{1}{4}x^{-1}$, between $\displaystyle x=1$ and $\displaystyle x=2$

Both times I apply the formula for arc length I end up having to find the root of a rather unpleasant expression. If someone could tell me the best way to go about these problems I'd be very grateful :)

Stonehambey
• Apr 5th 2009, 10:28 AM
Jhevon
Quote:

Originally Posted by Stonehambey
Not sure if I'm having another brain freeze here or something, but I'm having trouble solving the following types of problems. Find the arc length of the following curves between the points specified

1) $\displaystyle y = x^{\frac{1}{2}} - \frac{1}{3}x^{\frac{3}{2}}$, between $\displaystyle x = 1$ and $\displaystyle x = 4$

2) $\displaystyle y = \frac{1}{3}x^3 + \frac{1}{4}x^{-1}$, between $\displaystyle x=1$ and $\displaystyle x=2$

Both times I apply the formula for arc length I end up having to find the root of a rather unpleasant expression. If someone could tell me the best way to go about these problems I'd be very grateful :)

Stonehambey

first tell us what expressions you are getting for us to see if you have a problem setting up the formula correctly. if you are good there, then we can move forward
• Apr 5th 2009, 10:43 AM
Stonehambey
Quote:

Originally Posted by Jhevon
first tell us what expressions you are getting for us to see if you have a problem setting up the formula correctly. if you are good there, then we can move forward

Sure :)

For the first one

$\displaystyle y = x^{\frac{1}{2}} - \frac{1}{3}x^{\frac{3}{2}}$

$\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{\frac{1}{2}}$

$\displaystyle [y']^{2} = \frac{1}{4}x^{-1} - \frac{1}{2} + \frac{1}{4}x$

$\displaystyle s = \int^{4}_{1} \sqrt{\frac{1}{4}x^{-1} + \frac{1}{2} + \frac{1}{4}x} \, \, dx$

and this is where I get stuck, although I'm sure I'm missing something obvious. I tried pulling out the quarter and messing with it a bit like that. But at the end of the day it's those powers of x which are confusing me
• Apr 5th 2009, 10:50 AM
Jhevon
Quote:

Originally Posted by Stonehambey
Sure :)

For the first one

$\displaystyle y = x^{\frac{1}{2}} - \frac{1}{3}x^{\frac{3}{2}}$

$\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{2}x^{\frac{1}{2}}$

$\displaystyle [y']^{2} = \frac{1}{4}x^{-1} - \frac{1}{2} + \frac{1}{4}x$

$\displaystyle s = \int^{4}_{1} \sqrt{\frac{1}{4}x^{-1} + \frac{1}{2} + \frac{1}{4}x} \, \, dx$

and this is where I get stuck, although I'm sure I'm missing something obvious. I tried pulling out the quarter and messing with it a bit like that. But at the end of the day it's those powers of x which are confusing me

taking our cue in going from the $\displaystyle y'$ line to the $\displaystyle [y']^2$ line, we notice that $\displaystyle \frac 14x^{-1} + \frac 12 + \frac 14x = \left( \frac 12x^{1/2} + \frac 12 x^{-1/2}\right)^2$ (Wink)
• Apr 5th 2009, 11:19 AM
Stonehambey
Haha, why didn't I spot that? :P

Thanks man, I see the second one is very similar now :)