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Homeworksucks
 one year ago
Find the arc length of the graph x=(y^4/4)+1/(8y^2)
Homeworksucks
 one year ago
Find the arc length of the graph x=(y^4/4)+1/(8y^2)

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Homeworksucks
 one year ago
Best ResponseYou've already chosen the best response.0\[x=\frac{ y^4 }{ 4 }+\frac{ 1 }{ 8y^2 }\]

Homeworksucks
 one year ago
Best ResponseYou've already chosen the best response.0I've set up the integral as \[\int\limits_{1}^{2}\sqrt{1+(y^3\frac{ 1 }{ 4y^3 })^{2}}\] but I'm so lost on what to do from here

Homeworksucks
 one year ago
Best ResponseYou've already chosen the best response.0By the way, I forgot to specify an interval, the question states from y=1 to y=2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I apologize, I was typing my response and my laptop just decided to turn off :/

Homeworksucks
 one year ago
Best ResponseYou've already chosen the best response.0That's alright, take your time :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As a tip, though, often times when you have a derivative that ends up looking like yours. For example, derivatives like this: \[y  \frac{ 1 }{ 4y }\] \[\frac{ y^{2} }{ 3 }  \frac{ 1 }{ 4y^{2} }\] Basically derivatives with a minus sign in the middle and the negative fraction has a variable on bottom. In those case, the square root often simplifies to the derivative you have but with a plus sign instead of a minus. So I betcha that the root will simolify and youll have the integral of \[y^{3} + \frac{ 1 }{ 4y^{3} }\] Now of course I plan on showing this, but its always a nice idea to have in mind for these arc length problems. In reality, the integrals on these would be ridiculous in a normal context, so textbooks set them up so that they come out perfectly. The easiest set up is to have a function that has that sort of property where that arc length formula simplifies like that. But yeah, on to simplifying and showing what I said should be the result, lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I won't rewrite the integral over and over, let's just simplify the root \[\sqrt{1 + (y^{3}  \frac{ 1 }{ 4y^{3} })^{2}}\] \[\sqrt{1 + (\frac{ 4y^{6}  1 }{ 4y^{3} })^{2}}\] \[\sqrt{1 + \frac{ 16y^{12}  8y^{6} + 1 }{ 16y^{6} }}\] \[\sqrt{\frac{ 16y^{6} + 16y^{12}  8y^{6} + 1 }{ 16y^{6} }}\] \[\sqrt{\frac{ 16y^{12} + 8y^{6} +1 }{ 16y^{6} }}\] \[\sqrt{\frac{ (4y^{6}+1)^{2} }{ 16y^{6} }} = \frac{ 4y^{6} + 1 }{ 4y^{3} } = y^{3} + \frac{ 1 }{ 4y^{3} }\] Which is what I assumed would happen :) And that assumption comes from knowing that arc length problems are often set up that way and when your derivatives is in a form like yours, this is often the result. I didnt show every little detail, but can you follow what I did to simplify the root?

Homeworksucks
 one year ago
Best ResponseYou've already chosen the best response.0Yeah it's super clear now! Thanks!

Homeworksucks
 one year ago
Best ResponseYou've already chosen the best response.0My textbook just skips over that step and doesn't explain at all so I was totally lost

Homeworksucks
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep, no problem! ^_^
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