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anonymous
 4 years ago
Calculus II  Arc Length Problem
Hello, I am having some serious problems calculating the Arc Length of a curve. Here is the problem;
(Putting in a new post to use equation calculator)
anonymous
 4 years ago
Calculus II  Arc Length Problem Hello, I am having some serious problems calculating the Arc Length of a curve. Here is the problem; (Putting in a new post to use equation calculator)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x = (y^3/33) + 11/(4y) \] on 3<= y <= 5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, I take the derivative. No problem; \[dy/dx = (y^2/11) + (11/(4y^2) \] Then Length = \[\sqrt(1 + (dy/dx)^2) \] So; \[Length = \sqrt(1 + (y^2/11 + 11/(4y))^2)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are off by a minus sign

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int_3^5\sqrt{1+(\frac{y^2}{11}\frac{11}{4y^2})^2}dy\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Holy god. If that's the problem I have been having this whole time I'm not sure whether I'm going to laugh or cry. Probably a bit of both.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh wow isn't that annoying!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0happens to me all the damn time, that is why it is good to get another pair of eyes

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int_3^5\sqrt{\frac{y^4}{121}+\frac{121}{16 y^4}+\frac{1}{2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, maybe you can tell me this  when I type ((y^2/11)  (11/(4y^2))^2 on my calculator it gives me; ((4y^4  121)^2)/1936y^4 Any idea why?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Same thing on Wolfram  if I type your above equation (everything under the root) and tell it to factor it gives me the same as above.
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