What is the arc length of the curve given by $r (t) = (9 \sqrt{2}, e^{9 t}, e^{- 9 t})$ $r(t)= (9sqrt(2),e^(9t),e^(-9t))$ on $t \in [3, 4]$ $t in [3,4]$ ?

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Answer 1 · 2018-03-12T08:44:55

A zeroth-order approximation gives $2 sinh (36) - 2 sinh (27)$ $2sinh(36)-2sinh(27)$ units.

$r (t) = (9 \sqrt{2}, e^{9 t}, e^{- 9 t})$ $r(t)=(9sqrt2, e^(9t), e^(−9t))$

$r'(t)=(0, 9e^(9t), -9e^(−9t))$

Arc length is given by:

$L=int_3^4sqrt(0+81e^(18t)+81e^(-18t))dt$

Simplify:

$L=9int_3^4sqrt(e^(18t)+e^(-18t))dt$

Apply the identity $coshx=1/2(e^x+e^-x)$ :

$L=9int_3^4sqrt(2cosh(18t))dt$

Apply the identity $cosh2x=2cosh^2x-1$ :

$L=9sqrt2int_3^4sqrt(2cosh^2(9t)-1)dt$

Factor out the larger piece:

$L=18int_3^4cosh(9t)sqrt(1-1/2sech^2(9t))dt$

For $t in [3, 4]$ , $1/2sech^2(9t)<1$ . Take the series expansion of the square root:

$L=18int_3^4cosh(9t){sum_(n=0)^oo((1/2),(n))(-1/2sech^2(9t))^n}dt$

Isolate the $n=0$ term:

$L=18int_3^4cosh(9t)dt+18sum_(n=1)^oo((1/2),(n))(-1/2)^nint_3^4sech^(2n-1)(9t)dt$

A zeroth-order approximation gives:

$L~~2[sinh(9t)]_ 3^4$

Hence:

$L~~2sinh(36)-2sinh(27)$

What is the arc length of the curve given by r(t)= (9sqrt(2),e^(9t),e^(-9t))r(t)=(9√2,e9t,e−9t)r(t)= (9sqrt(2),e^(9t),e^(-9t)) on t in [3,4]t∈[3,4] t in [3,4]?