Question

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?

Answer 1

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

Answer 2

so for this one what do you get as y'?

Answer 3

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

Answer 4

no no

Answer 5

the whole thing inside has to be written as a square for you to do that

Answer 6

like so:
\[1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x^2}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x^2} \\ (3x^2+\frac{1}{12x^2})^2 \]

Answer 7

\[\sqrt{(3x^2+\frac{1}{12x^2})^2}=3x^2+\frac{1}{12x^2} \text{ for all } x \neq 0\]

Answer 8

which we don't have to worry about 0 since we are looking at [1/2,2]

Answer 9

so this is what you have:
\[\int\limits_{1/2}^{2} (3x^2+\frac{1}{12x^2}) dx\]

Answer 10

made a type-o

Answer 11

\[1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x\color{red}{^4}}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x\color{red}{^4}} \\ (3x^2+\frac{1}{12x^2})^2 \]

Answer 12

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

Answer 13

So in reality there is no square root for this problem? because the two integrals are equal to it?

Answer 14

yeah we got rid of the square root by writing previous integrand as
\[3x^2+\frac{1}{12x^2}\]

Answer 15

didn't you forget the plus 1?

Answer 16

remember this:
\[1+(3x^2-\frac{1}{12x^2})^2 \\ 1+(9x^4-\frac{1}{2}+\frac{1}{144x^4}) \\ 9x^4+\frac{1}{2}+\frac{1}{144x^4} \\ (3x^2+\frac{1}{12x^2})^2\]

Answer 17

those 4 expressions are equal expressions

Answer 18

\[\sqrt{1+(3x^2-\frac{1}{12x^2})^2} =\\ \sqrt{(3x^2+\frac{1}{12x^2})^2}\]

Answer 19

hey @Empty and @IrishBoy123 I have to leave
can you help @Greenwalrus further

Answer 20

ah yeah I see!

Answer 21

help :(

Answer 22

since I don't have a negative I can just ignore having two integrals and have one right