## anonymous 5 years ago Y^2 + 2y = 2x + 1 set up an integral for the length of the curve

1. anonymous

The kind of integral you end up with will depend on whether you parametrize, or use one of the given variables to integrate over. If you consider your curve, it's a parabola turned on its right side. It isn't the case that this thing is some function of x since a vertical line drawn through the graph cuts the graph at more than one point...BUT...if you draw a line horizontally, the graph IS only cut at one point, so you have something that is a function of y, namely, $x(y)=\frac{1}{2}y^2+y-\frac{1}{2}$Now, consider a little bit of length on your graph, and call it$\delta s$By Pythagoras' Theorem, you have$(\delta s)^2=(\delta x)^2+(\delta y)^2$ (i.e. the distance between two very close points. Then$(\frac{\delta s}{\delta y})^2=1+(\frac{\delta x}{\delta y})^2 \rightarrow \frac{\delta s}{\delta y}=\sqrt{1+\left( \frac{\delta x}{\delta y} \right)^2}$Multiply through by delta y and take the limit as delta y approaches zero to get$\delta s = \sqrt{1+\left( \frac{\delta x}{\delta y} \right)^2}\delta y \rightarrow ds= \sqrt{1+\left( \frac{d x}{d y} \right)^2}dy$The length of your curve then, between two limits defined by y will be$s=\int\limits_{y_1}^{y_2}\sqrt{1+\left( \frac{d x}{d y} \right)^2}dy$

2. anonymous

Now, you have$x=\frac{1}{2}y^2+y-\frac{1}{2} \rightarrow \frac{dx}{dy}=y+1$Substitute this into the formula,$s=\int\limits_{y_1}^{y_2}\sqrt{1+(y+1)^2}dy$and integrate.

3. anonymous

You can start with a substitution of u=y+1 and take it from there. It's too much to go through here, but if you need help, plug the integral into Wolfram Alpha and then 'steps' in the solution if you need assistance.

4. anonymous

The answer in my book says that before you integrate you get sqr root of 1+1/(2x+2) not sqr root 1+(y+1^2)?

5. anonymous

Hi LoK

6. anonymous

Like I said in the beginning, the integral you get will depend on what path you want to take. The actual number you get in the end should be the same.

7. anonymous

Hi dichalao

8. anonymous

Okay thank you!

9. anonymous

welcome

10. anonymous

feel free to give me a fan point...that took forever! ;p

11. anonymous

dichalao, did you have another look at your problem?

12. anonymous

Yea, but i still don't get it.

13. anonymous

blm, if you need more help, let me know.

14. anonymous

dichalao, go back to the old thread.

15. anonymous

LOL how do i go back to the old thread

16. anonymous

I will go back there, you go to Group Home and click on what I'm viewing.