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anonymous
 one year ago
FInd the length of curve y=x^3+(1/(12x)) on interval [1/2,2]
I get that I take the derivative and square it.
and plug it in to sqrt(1+y'^2)
but what confuses me is do I do the inverse of derivative to equal out.
also can I just take the square root of everything inside and then take the integral?
anonymous
 one year ago
FInd the length of curve y=x^3+(1/(12x)) on interval [1/2,2] I get that I take the derivative and square it. and plug it in to sqrt(1+y'^2) but what confuses me is do I do the inverse of derivative to equal out. also can I just take the square root of everything inside and then take the integral?

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freckles
 one year ago
Best ResponseYou've already chosen the best response.2you can take the square root of the thing inside if the inside itself can be written as a square but depending on the limits you might have to adjust your sign of the integrand

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so for this one what do you get as y'?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get \[\sqrt{1+((3x^2(1/12x^2))^2}\] so could I basically just cancel out the square root

freckles
 one year ago
Best ResponseYou've already chosen the best response.2the whole thing inside has to be written as a square for you to do that

freckles
 one year ago
Best ResponseYou've already chosen the best response.2like so: \[1+(3x^2\frac{1}{12x^2})^2 \\ 1+(9x^4\frac{1}{2}+\frac{1}{144x^2}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x^2} \\ (3x^2+\frac{1}{12x^2})^2 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{(3x^2+\frac{1}{12x^2})^2}=3x^2+\frac{1}{12x^2} \text{ for all } x \neq 0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2which we don't have to worry about 0 since we are looking at [1/2,2]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so this is what you have: \[\int\limits_{1/2}^{2} (3x^2+\frac{1}{12x^2}) dx\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[1+(3x^2\frac{1}{12x^2})^2 \\ 1+(9x^4\frac{1}{2}+\frac{1}{144x\color{red}{^4}}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x\color{red}{^4}} \\ (3x^2+\frac{1}{12x^2})^2 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2by the way the whole sign thing I mentioned earlier...pretend we have: \[\int\limits_{1}^2 \sqrt{x^2} dx \\ \text{ well } \sqrt{x^2}=x \text{ if } x \ge 0 \\ \text{ and } \sqrt{x^2}=x \text{ if } x<0 \\ \text{ so } \\ \int\limits_{1}^2 \sqrt{x^2} dx=\int\limits_{1}^0 (x )dx+\int\limits_0^2 x dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So in reality there is no square root for this problem? because the two integrals are equal to it?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yeah we got rid of the square root by writing previous integrand as \[3x^2+\frac{1}{12x^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0didn't you forget the plus 1?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2remember this: \[1+(3x^2\frac{1}{12x^2})^2 \\ 1+(9x^4\frac{1}{2}+\frac{1}{144x^4}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x^4} \\ (3x^2+\frac{1}{12x^2})^2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2those 4 expressions are equal expressions

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{1+(3x^2\frac{1}{12x^2})^2} =\\ \sqrt{(3x^2+\frac{1}{12x^2})^2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2hey @Empty and @IrishBoy123 I have to leave can you help @Greenwalrus further

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since I don't have a negative I can just ignore having two integrals and have one right
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