What is the arclength of $f (t) = (- 2 {(t + 3)}^{3}, {(t - 2)}^{2})$ $f(t) = (-2(t+3)^3,(t-2)^2)$ on $t \in [1, 2]$ $t in [1,2]$ ?

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Answer 1 · 2018-06-03T10:24:46

$\approx 122.006$ $approx 122.006$

We have
$x (t) = - 2 {(t + 3)}^{3}$ $x(t)=-2(t+3)^3$
so
$x'(t)=-6)t+3)^2$
$y(t)=(t-2)^2$
so

$y'(t)=2(t-2)$

so our integral is given by

$int_1^2sqrt((-6(t+3)^2)^2+(2(t-2))^2)dt$
A numerical method gives us $approx 122.006$

What is the arclength of f(t) = (-2(t+3)^3,(t-2)^2)f(t)=(−2(t+3)3,(t−2)2)f(t) = (-2(t+3)^3,(t-2)^2) on t in [1,2]t∈[1,2]t in [1,2]?