How do you find the arc length of the curve $y = x^{3}$ $y=x^3$ over the interval [0,2]?

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Answer 1 · 2018-03-09T06:17:01

A first-order approximation to the arc length gives $8 + {tan}^{- 1} (2 \sqrt{3})$ $8+tan^(-1)(2sqrt3)$ units.

$y = x^{3}$ $y=x^3$

$y'=3x^2$

Arc length is given by:

$L=int_0^2sqrt(1+9x^4)dx$

Apply the substitution $sqrt3x=u$ :

$L=1/sqrt3int_0^(2sqrt3)sqrt(1+u^4)du$

Complete the square in the square root:

$L=1/sqrt3int_0^(2sqrt3)sqrt((u^2+1)^2-2u^2)du$

Factor out the larger piece:

$L=1/sqrt3int_0^(2sqrt3)(u^2+1)sqrt(1-(2u^2)/(u^2+1)^2)du$

For $x in [0,2sqrt3]$ , $(2u^2)/(u^2+1)^2<1$ . Take the series expansion of the square root:

$L=1/sqrt3int_0^(2sqrt3)(u^2+1){sum_(n=0)^oo((1/2),(n))(-(2u^2)/(u^2+1)^2)^n}du$

Isolate the $n=0$ and $n=1$ cases:

$L=1/sqrt3int_0^(2sqrt3){(u^2+1)-u^2/(u^2+1)}du+1/sqrt3sum_(n=2)^oo((1/2),(n))(-2)^nint_0^(2sqrt3)u^(2n)/(u^2+1)^(2n-1)du$

Hence

$L=1/sqrt3int_0^(2sqrt3)(u^2+1/(u^2+1))du+1/sqrt3sum_(n=2)^oo((1/2),(n))(-2)^nint_0^(2sqrt3)u^(2n)/(u^2+1)^(2n-1)du$

Ignoring the higher-order terms, a first-order approximation gives:

$L~~1/sqrt3[1/3u^3+tan^(-1)u]_ 0^(2sqrt3)$

Insert the limits of integration:

$L~~8+tan^(-1)(2sqrt3)$

How do you find the arc length of the curve y=x^3y=x3y=x^3 over the interval [0,2]?