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

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

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Answer 1 · 2018-05-28T12:41:28

$\approx 16.0071$ $approx 16.0071$

Explanation:

We hat
$x (t) = {(t^{2} + t)}^{- \frac{1}{2}}$ $x(t)=(t^2+t)^(-1/2)$
$x'(t)=-1/2*(2*t+1)/(t^2+t)^(3/2)$
$y(t)=2-8t$
$y'(t)=-8$
So we have to solve
$int_1^3sqrt(64+(-1/2*(2*t+1)/(t^2+t)^(3/2))^2)dt$
I have got only a numerical value.

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

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What is the arc length of f(t)=(1/sqrt(t+t^2),2-8t) f(t)=(1√t+t2,2−8t)f(t)=(1/sqrt(t+t^2),2-8t) over t in [1,3]t∈[1,3]t in [1,3]?

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Explanation:

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