Home > Calculus > Determining the Length of a Parametric Curve (Parametric Form)

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

Answer 1

#approx 10.37224180#

We have

#x(t)=e^(-2t)-t^2# then #x'(t)=-2^(-2t)-2t#+

#y(t)=t-t/e^(t-1)# then

#y'(t)=1-e^(1-t)+e^(1-t)t# so we get the integral #int_(-1)^1sqrt((-2e^(-2t)-2t)^2+(1-e^(1-t)+e^(1-t)t)^2)dt# By a numerical method we get

#approx10.37224180#

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Answer 2

Luke Alvarez

To find the arc length of the parametric curve ( (e^{-2t} - t^2, t - \frac{t}{e^{t-1}}) ) on the interval ([-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} ) and ( \frac{dy}{dt} ) are the derivatives of the parametric equations with respect to ( t ).

Compute ( \frac{dx}{dt} = \frac{d(e^{-2t} - t^2)}{dt} ) and ( \frac{dy}{dt} = \frac{d(t - \frac{t}{e^{t-1}})}{dt} ).
Substitute these derivatives into the formula.
Integrate the resulting expression over the given interval ([-1, 1]).

After integrating, you'll have the arc length of the curve on the specified interval.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.