anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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freckles
  • freckles
you 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
  • freckles
so for this one what do you get as y'?
anonymous
  • anonymous
I get \[\sqrt{1+((3x^2-(1/12x^2))^2}\] so could I basically just cancel out the square root

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freckles
  • freckles
no no
freckles
  • freckles
the whole thing inside has to be written as a square for you to do that
freckles
  • freckles
like 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
  • freckles
\[\sqrt{(3x^2+\frac{1}{12x^2})^2}=3x^2+\frac{1}{12x^2} \text{ for all } x \neq 0\]
freckles
  • freckles
which we don't have to worry about 0 since we are looking at [1/2,2]
freckles
  • freckles
so this is what you have: \[\int\limits_{1/2}^{2} (3x^2+\frac{1}{12x^2}) dx\]
freckles
  • freckles
made a type-o
freckles
  • freckles
\[1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x\color{red}{^4}}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x\color{red}{^4}} \\ (3x^2+\frac{1}{12x^2})^2 \]
freckles
  • freckles
by 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
  • anonymous
So in reality there is no square root for this problem? because the two integrals are equal to it?
freckles
  • freckles
yeah we got rid of the square root by writing previous integrand as \[3x^2+\frac{1}{12x^2}\]
anonymous
  • anonymous
didn't you forget the plus 1?
freckles
  • freckles
remember 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
  • freckles
those 4 expressions are equal expressions
freckles
  • freckles
\[\sqrt{1+(3x^2-\frac{1}{12x^2})^2} =\\ \sqrt{(3x^2+\frac{1}{12x^2})^2}\]
freckles
  • freckles
hey @Empty and @IrishBoy123 I have to leave can you help @Greenwalrus further
anonymous
  • anonymous
ah yeah I see!
anonymous
  • anonymous
help :(
anonymous
  • anonymous
since I don't have a negative I can just ignore having two integrals and have one right

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