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

Question

Scarlett Allison · Accepted Answer

#approx 0.431873# We have #x(t)=sqrt(t)# then #x'(t)=1/(2*sqrt(t))# #y(t)=1/sqrt(t^2+3)# then #y'(t)=-t/sqrt(t^2+3)^3# So we get the integral #int_1^2sqrt(1/(4*t)+t^2/(3+t^2)^3)dt# I have found only a numerical value: #approx 0.431873#

Ezra Adams · Answer

undefined To find the arc length of the curve described by the parametric equations $ x = \sqrt{t} $ and $ y = \frac{1}{\sqrt{t^2 + 3}} $ for $ t $ in the interval $[1, 2]$, we'll use the formula for arc length of a parametric curve:

$$ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$

First, compute $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $:

$$ \frac{dx}{dt} = \frac{1}{2\sqrt{t}} $$

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

Next, square and sum them:

$$ \left(\frac{dx}{dt}\right)^2 = \frac{1}{4t} $$

$$ \left(\frac{dy}{dt}\right)^2 = \frac{4t^2}{(t^2+3)^3} $$

$$ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = \frac{t^2 + 3}{t(t^2 + 3)^3} $$

Now, take the square root:

$$ \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \frac{\sqrt{t^2 + 3}}{t\sqrt{t^2 + 3}} = \frac{1}{t} $$

Finally, integrate from $ t = 1 $ to $ t = 2 $:

$$ L = \int_{1}^{2} \frac{1}{t} \, dt = \ln|t| \bigg|_{1}^{2} = \ln(2) - \ln(1) = \ln(2) $$

So, the arc length of the curve is $ \ln(2) $.