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Find the exact length of the curve.

$ x = \displaystyle \frac{y^4}{8} + \displaystyle \frac{1}{4y^2} $ , $ 1 \le y \le 2 $

$L=\frac{33}{16}$

Applications of Integration

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Campbell University

Oregon State University

Harvey Mudd College

Boston College

and this problem we're gonna be using arc length formula, um, and arc length formula goes by this, it's l is equal to the integral from we'll call it a to B of the square root of one class x D y. Well, uh, Andi, with this, we know that wise from 1 to 2, we know that the lower bound is going to be one and Anthony too as a result of that. And then we know that x were given X So we wanna find dx dy y If we take dx dy y so differentiate x with respect to y will end up getting Yeah, why fourth eight plus lied to the negative too Owner for on dso When we differentiate that that's our X value. So we differentiate and we end up getting, um, that dx dy y is why cube? So we can just plug this in here? Why cubed over to minus Why? To the negative third over. Yeah, that right there is our our cling formula. And when we square this, we can get other, um values, but simplifying it, what will ultimately get underneath the radical with its squared is going to be wide to the sixth or minus one half. Plus, why do the negative six, Sheriff, you were getting the same answer Arc length, then, um, what we end up getting when we, um, calculate this when we add one. Um, this is what we get. So underneath the radical all this will end up being Why cubed over to that's why to a negative third to Goodbye, Andi. It's gonna be getting rid of this right here. Um and then once we integrate this, we are just integrate with respect to y and plug in our values to the fundamental form of calculus. What we end up getting is that l is equal to everyone over 16 plus 1/8, which is what we got before on. Ultimately, this is going to be a 33/16.

California Baptist University

Applications of Integration