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

What is the arclength of #(1/(1-e^t),2t)# on #t in [2,4]#?

Answer 1

Emma Aldridge

#approx 4.003334361#

We have

#x(t)=1/(1-e^t)# and by the power and chain rule we get

#x'(t)=-1*(1-e^(t))^(-2)(-e^t)#

#y(t)=2t#

then

#y'(t)=2#

so our integral is given by

#int_2^4sqrt((e^t/(1-e^t))^2+4)dt# by a numerical method we get

#approx 4.003334361#

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Answer 2

Cameron Alexander

To find the arc length of the curve ( (1/(1-e^t), 2t) ) on the interval ( t \in [2,4] ), you use the arc length formula:

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

where ( a = 2 ) and ( b = 4 ). First, find ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ) by differentiating the parametric equations:

[ \frac{dx}{dt} = \frac{d}{dt} \left(\frac{1}{1-e^t}\right) = \frac{e^t}{(e^t-1)^2} ] [ \frac{dy}{dt} = \frac{d}{dt} (2t) = 2 ]

Now, plug these into the arc length formula and integrate over the interval ( t \in [2,4] ):

[ L = \int_{2}^{4} \sqrt{\left(\frac{e^t}{(e^t-1)^2}\right)^2 + 2^2} , dt ]

Compute this integral to find the arc length ( L ).

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Answer from HIX Tutor

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