What is the arclength of #f(t) = (t-sqrt(t^2+2),t+te^(t-2))# on #t in [-1,1]#?

Question

Ezra Adams · Accepted Answer

Arc length #s=3.2352212144206" "#units

The formula to obtain the arc length #s# #s=int_(t_1)^(t_2) sqrt((dx/dt)^2+(dy/dt)^2)*dt# From the given #f(t)=(t-sqrt(t^2+2), t+t*e^(t-2))# We have #x=t-sqrt(t^2+2)" "#and #y= t+t*e^(t-2)# We need to obtain the derivatives #(dx)/dt# and #(dy)/dt# #(dx)/dt=1-t/sqrt(t^2+2)" "# and #(dy)/dt=1+t*e^(t-2)+e^(t-2)# We solve now the arclength #s# #s=int_(t_1)^(t_2) sqrt((dx/dt)^2+(dy/dt)^2)*dt# #s=int_(-1)^(1) sqrt((1-t/sqrt(t^2+2))^2+(1+t*e^(t-2)+e^(t-2))^2)*dt# The integral is complicated and requires the use of calculator or Simpson's Rule formula #s=int_(-1)^(1) sqrt((1-t/sqrt(t^2+2))^2+(1+t*e^(t-2)+e^(t-2))^2)*dt# #s=3.2352212144206" "#units God bless....I hope the explanation is useful.

Emilia Aguilar · Answer

undefined The arc length of the curve $ f(t) = (t - \sqrt{t^2 + 2}, t + te^{t - 2}) $ on the interval $[-1, 1]$ can be calculated using the following formula:

$$ L = \int_{-1}^{1} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$

Where $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $ are the derivatives of $ x $ and $ y $ with respect to $ t $, respectively.

So, first, compute $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $, then substitute them into the formula and integrate over the given interval $[-1, 1]$ to find the arc length.