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

Question

Luke Alvarez · Accepted Answer

#~4.298# If we have a really small change in #t#, we will travel according to the following equation: #ds^2 = dx^2 + dy^2 = [((dx)/(dt))^2 +((dy)/(dt))^2]dt^2 # i.e. #ds = dt sqrt(((dx)/(dt))^2 +((dy)/(dt))^2) # The total length is therefore #int_(t_0)^(t_f) ds = int_(t_0)^(t_f) sqrt(((dx)/(dt))^2 +((dy)/(dt))^2) dt# Calculating the values, #(dx)/(dt) = 1 - e^t# #(dy)/(dt) = 1 + 2/t^2 # Therefore, plugging in the numbers, we have the equation #int_(1)^(2) sqrt((( 1 - e^t )^2 +(1 + 2/t^2)^2) dt # We can't solve this exactly (or at least not in any easy way if it is technically possible), so we plug this into a calculator such as WolframAlpha, getting the value around #4.298#.

Evelyn Agard · Answer

undefined To find the arc length of the curve represented by the parametric equations $f(t) = (t - e^t, \frac{t - 2}{t + 3})$ on the interval $[1, 2]$, we 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 = 2 $, and $ x(t) = t - e^t $ and $ y(t) = \frac{t - 2}{t + 3} $.

Now, let's find $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $:

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

Now, we plug these derivatives into the formula:

$$ L = \int_{1}^{2} \sqrt{(1 - e^t)^2 + \left(\frac{(t + 3)(1) - (t - 2)(1)}{(t + 3)^2}\right)^2} \, dt $$

$$ L = \int_{1}^{2} \sqrt{(1 - e^t)^2 + \left(\frac{5}{(t + 3)^2}\right)^2} \, dt $$

$$ L = \int_{1}^{2} \sqrt{(1 - e^t)^2 + \frac{25}{(t + 3)^4}} \, dt $$

This integral is not elementary and cannot be expressed in terms of elementary functions. Therefore, the arc length needs to be approximated using numerical methods such as Simpson's Rule or the Trapezoidal Rule.