# Find the length of the curve

1. Oct 28, 2004

### Caldus

Can't seem to finish this problem:

Find the length of the curve:

$$x = 3y^{4/3}-((3/32)y^{2/3})$$

And -343 <= y <= 125.

I used the formula:

$$\int_{a}^{b}\ (1 + (x')^{2})^{1/2}dx$$
$$\int_{-343}^{125}\ (1+(4y^{1/3}-(1/16)y^{-1/2})^{2})^{1/2}dx$$

But how do you find this integral? Is this even the right integral to use? Thanks.

2. Oct 28, 2004

### Tide

First, you want $\left(y'\right)^2$ under ther radical. Then use the relationship between x and y to find y in terms of x from which you can find y'(x) and you should be able to complete the integration.

3. Oct 28, 2004

### Caldus

But isn't it just easier to do it the way I did (since it is given that y is between two values and not x)?

4. Oct 28, 2004

### Zurtex

Have you tried expanding it out?

5. Oct 28, 2004

### T@P

can i suggest a calculator?

6. Oct 28, 2004

### Caldus

What do you mean by expanding it out?

And the calculator will not give me an answer since 0 is not in the domain of y^(-1/3)...

7. Oct 28, 2004

### Zurtex

When I said expand out I mean like this:

$$\left(1+\left(4y^{\frac{1}{3}}-\frac{1}{16}y^{-\frac{1}{2}}\right)^{2}\right)^{\frac{1}{2}} = \left(1 + 16y^{\frac{2}{3}} - \frac{1}{2}y^{-\frac{1}{6}} + \frac{1}{128}y^{-1}\right)^{\frac{1}{2}}$$

Anyway I messed about with this and found it wasn't intergratable in terms of elementary functions sorry. Which is a bit of a shame as both your limits are cubes so their were some nice substitutions I tried 1st. I tried rearranging it as $y=f(x)$ and found that $1 + y'^2$ wasn't intergratable in terms of elementary functions as well.

Any way it occurred to me while I was coming up with a numerical solution that the function is not continuous at y=0 therefore the integral doesn't exist at y=0

8. Oct 28, 2004

### Caldus

So then how do I go about coming up with a numerical solution to it? Am I going to have to use limits?

9. Oct 29, 2004

There is no numerical solution, your function isn't continous for $y \leq 0[/tex] Have you tried getting to plot it out? You'll notice generally for y<0 your function isn't even real. 10. Oct 29, 2004 ### HallsofIvy Staff Emeritus It has been suggested repeatedly that you "multiply it out". Did you even try that? The whole point is this: [itex]\frac{dx}{dy}= 4y^{\frac{1}{3}}-\frac{1}{16}y^{\frac{-1}{3}}$. so $$$\frac{dx}{dy}$$^2= 16y^{\frac{2}{3}}- 1/2+ \frac{1}{256}y^{\frac{-2}{3}}$.

Now add 1 to that and the only change is that the "-1/2" in the middle becomes "+1/2". What's the square root of that?

11. Oct 30, 2004

### Caldus

I can't figure it out. What is the square root of it? Do you try to form a perfect square with it all or what?

12. Oct 30, 2004

### Caldus

Oh wait I think I know what you mean now. You just complete the square or whatever and then cancel out the square with the square root. So now I can do the integral. Problem is that I get a negative answer which doesn't make sense for the length of an arc. :\

Guess I'm still doing it wrong. Here's exactly what I did:

$$\int_{-343}^{125}\ (1+16y^{\frac{2}{3}}- 1/2+ \frac{1}{256}y^{\frac{-2}{3}})^{1/2}dx$$

$$\int_{-343}^{125}\ (16y^{\frac{2}{3}}+ 1/2+ \frac{1}{256}y^{\frac{-2}{3}})^{1/2}dx$$

$$\int_{-343}^{125}\ ((4y^{1/3}+(1/16)y^{-1/2})^{2})^{1/2}dx$$

$$\int_{-343}^{125}\ 4y^{1/3}+(1/16)y^{-1/2}dx$$

And then from there I got a negative answer and it's incorrect according to this program I'm using. :\

13. Oct 30, 2004

### HallsofIvy

Staff Emeritus
Well, there is one obvious mistake:

That "$y^{-\frac{1}{2}}$" should be "$y^{-\frac{1}{3}}$

Last edited: Oct 30, 2004
14. Oct 30, 2004

### Caldus

Yeah that was a typo. I actually used y^1/3 there. My final answer ended up being -5330.25 which makes no sense. :\

15. Oct 30, 2004

### Caldus

How can an arclength be negative? Is that even possible?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?