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

Question

Emily Ammerman · Accepted Answer

#11.927#

The arc length formula for a parametric equation is:

#A=int_a^bsqrt((dx/dt)^2+(dy/dt)^2#

We need to take the derivative of #(e^(2t)-t,t-t/e^(t-1))# which is in the form #(x(t),y(t))# in order to get #dx/dt# and #dy/dt#.

For #dx/dt#, use chain rule:

#dx/dt=2e^(2t)-1#

For #dy/dt#, use quotient rule. It may be easier to write #t/e^(t-1)# as #et/e^t# and then ignore the #e# since it's a constant:

#dy/dt=1-e(1(e^t)-(e^t)t)/e^(2t)=1-(1-t)e^(1-t)#

Now plug into the formula and solve:

#A=int_-1^1 sqrt((2e^(2t)-1)^2 +(1-(t-1)e^(t+1))^2)=11.927#

Samantha Adkison · Answer

undefined To find the arc length of the curve described by the parametric equations $ x(t) = e^{2t} - t $ and $ y(t) = t - \frac{t}{e^{t-1}} $ on the interval $ t $ in $[-1,1]$, we use the formula:

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

Where:
- $ \frac{dx}{dt} $ represents the derivative of $ x(t) $ with respect to $ t $
- $ \frac{dy}{dt} $ represents the derivative of $ y(t) $ with respect to $ t $
- $ a $ and $ b $ are the lower and upper limits of the parameter $ t $, respectively.

By calculating the derivatives, we get:

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

Now, we substitute these derivatives into the arc length formula and integrate over the interval $[-1,1]$:

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

This integral needs to be solved numerically, as it does not have a simple closed form solution. You can use numerical integration techniques such as the trapezoidal rule or Simpson's rule to approximate the value of the integral.