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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?

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  1. freckles
    • one year ago
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    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

  2. freckles
    • one year ago
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    so for this one what do you get as y'?

  3. anonymous
    • one year ago
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    I get \[\sqrt{1+((3x^2-(1/12x^2))^2}\] so could I basically just cancel out the square root

  4. freckles
    • one year ago
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    no no

  5. freckles
    • one year ago
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    the whole thing inside has to be written as a square for you to do that

  6. freckles
    • one year ago
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    like so: \[1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x^2}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x^2} \\ (3x^2+\frac{1}{12x^2})^2 \]

  7. freckles
    • one year ago
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    \[\sqrt{(3x^2+\frac{1}{12x^2})^2}=3x^2+\frac{1}{12x^2} \text{ for all } x \neq 0\]

  8. freckles
    • one year ago
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    which we don't have to worry about 0 since we are looking at [1/2,2]

  9. freckles
    • one year ago
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    so this is what you have: \[\int\limits_{1/2}^{2} (3x^2+\frac{1}{12x^2}) dx\]

  10. freckles
    • one year ago
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    made a type-o

  11. freckles
    • one year ago
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    \[1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x\color{red}{^4}}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x\color{red}{^4}} \\ (3x^2+\frac{1}{12x^2})^2 \]

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

  13. anonymous
    • one year ago
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    So in reality there is no square root for this problem? because the two integrals are equal to it?

  14. freckles
    • one year ago
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    yeah we got rid of the square root by writing previous integrand as \[3x^2+\frac{1}{12x^2}\]

  15. anonymous
    • one year ago
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    didn't you forget the plus 1?

  16. freckles
    • one year ago
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    remember this: \[1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x^4}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x^4} \\ (3x^2+\frac{1}{12x^2})^2\]

  17. freckles
    • one year ago
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    those 4 expressions are equal expressions

  18. freckles
    • one year ago
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    \[\sqrt{1+(3x^2-\frac{1}{12x^2})^2} =\\ \sqrt{(3x^2+\frac{1}{12x^2})^2}\]

  19. freckles
    • one year ago
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    hey @Empty and @IrishBoy123 I have to leave can you help @Greenwalrus further

  20. anonymous
    • one year ago
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    ah yeah I see!

  21. anonymous
    • one year ago
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    help :(

  22. anonymous
    • one year ago
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    since I don't have a negative I can just ignore having two integrals and have one right

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