What is the arclength of $f(t) = (sqrt(t-2),t^2)$ on $t in [2,3]$ ?

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Answer 1 · 2017-02-19T16:29:31

$L=int_2^3sqrt(1/(4(t-2))+4t^2)dtapprox5.1927$

The arc length of the parametric function $f(t)=(x(t),y(t))$ on $tin[a,b]$ is given through:

$L=int_a^bsqrt((x'(t))^2+(y'(t))^2)dt$

We see that:

$x(t)=sqrt(t-2)=(t-2)^(1/2)$

Then:

$x'(t)=1/2(t-2)^(-1/2)d/dt(t-2)=1/(2sqrt(t-2))$

And:

$y(t)=t^2" "=>" "y'(t)=2t$

Combining these, we see that:

$L=int_2^3sqrt((1/(2sqrt(t-2)))^2+(2t)^2)dt$

$L=int_2^3sqrt(1/(4(t-2))+4t^2)dtapprox5.1927$

This can't be integrated by hand.

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