What is the arc length of $f (t) = (\sqrt{t - 1}, 2 - 8 t)$ $f(t)=(sqrt(t-1),2-8t)$ over $t \in [1, 3]$ $t in [1,3]$ ?

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Answer 1 · 2018-03-12T08:58:49

$L = \frac{3}{2} \sqrt{114} + \frac{1}{32} ln (16 \sqrt{2} + 3 \sqrt{57})$ $L=3/2sqrt114+1/32ln(16sqrt2+3sqrt57)$ units.

$f (t) = (\sqrt{t - 1}, 2 - 8 t)$ $f(t)=(sqrt(t−1),2−8t)$

$f'(t)=(1/(2sqrt(t−1)),−8)$

Arc length is given by:

$L=int_1^3sqrt(1/(4(t−1))+64)dt$

Apply the substitution $t-1=u^2$ :

$L=int_0^sqrt2sqrt(1/(4u^2)+64)(2udu)$

Simplify:

$L=int_0^sqrt2sqrt(1+256u^2)du$

Apply the substitution $16u=tantheta$ :

$L=1/16intsec^3thetad theta$

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

$L=1/32[secthetatantheta+ln|sectheta+tantheta|]$

Reverse the last substitution:

$L=[1/2usqrt(1+256u^2)+1/32ln|16u+sqrt(1+256u^2)|]_0^sqrt2$

Insert the limits of integration:

$L=3/2sqrt114+1/32ln(16sqrt2+3sqrt57)$

What is the arc length of f(t)=(sqrt(t-1),2-8t) f(t)=(√t−1,2−8t)f(t)=(sqrt(t-1),2-8t) over t in [1,3]t∈[1,3]t in [1,3]?