What is the arclength of $(sqrt(3t-2),1/sqrt(t+3))$ on $t in [1,3]$ ?

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Answer 1 · 2017-05-13T04:58:36

$approx 1.64833$

The arc length for a parametric function is:
$L=int_a^b (sqrt((dx/dt)^2+(dy/dt)^2))dt$

In order to plug in values into this equation, we need to find $dx/dt$ and $dy/dt$ by differentiating the given function $(x(t),y(t))$ :

$x(t)=(3t-2)^(1/2)$

$dx/dt=1/2(3t-2)^(-1/2)(3)=3/2(3t-2)^(-1/2)=frac{3}{2sqrt(3t-2)}$

$y(t)=(t+3)^(-1/2)$

$dy/dt=(-1/2)(t+3)^(-3/2)=frac{-1}{2(t+3)^(3/2)}$

$L=int_1^3(sqrt((frac{3}{2sqrt(3t-2)})^2+(frac{-1}{2(t+3)^(3/2)})^2))dt$

$=int_1^3(sqrt(frac{9}{4(3t-2)}+frac{1}{4(t+3)^3}))dt$

$approx 1.64833$

What is the arclength of (sqrt(3t-2),1/sqrt(t+3))(sqrt(3t-2),1/sqrt(t+3)) on t in [1,3]t in [1,3]?