Question

Y^2 + 2y = 2x + 1 set up an integral for the length of the curve

Answer 1

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\[s=\int\limits_{y_1}^{y_2}\sqrt{1+\left( \frac{d x}{d y} \right)^2}dy\]

Answer 2

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.

Answer 3

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.

Answer 4

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)?

Answer 5

Hi LoK

Answer 6

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.

Answer 7

Hi dichalao

Answer 8

Okay thank you!

Answer 9

welcome

Answer 10

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

Answer 11

dichalao, did you have another look at your problem?

Answer 12

Yea, but i still don't get it.

Answer 13

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

Answer 14

dichalao, go back to the old thread.

Answer 15

LOL how do i go back to the old thread

Answer 16

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