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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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\[s=\int\limits_{y_1}^{y_2}\sqrt{1+\left( \frac{d x}{d y} \right)^2}dy\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Like 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0feel free to give me a fan point...that took forever! ;p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dichalao, did you have another look at your problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yea, but i still don't get it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0blm, if you need more help, let me know.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dichalao, go back to the old thread.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LOL how do i go back to the old thread

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I will go back there, you go to Group Home and click on what I'm viewing.
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