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

Question

Samantha Adkison · Accepted Answer

#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#

Charles Anctil · Answer

undefined To find the arc length of the curve $ \left(\sqrt{3t - 2}, \frac{1}{\sqrt{t + 3}}\right) $ on the interval $ t \in [1, 3] $, we'll use the formula for arc length:

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

where $ a = 1 $ and $ b = 3 $. First, find $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $. Then, substitute these derivatives into the formula and integrate over the given interval. After integration, you'll get the arc length.