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

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

Answer 1

Samantha Adkison

#approx 0.258144#

We have #x(t)=t/(1+t)^2# so #x'(t)=(t-1)^2/(1+t)^2#

#y(t)=-1/t# so #y'(t)=1/t^4# and we get the integral

#int_2^4sqrt(1/t^4+(t-1)^2/(1+t)^6)dt# by a numerical method we get

#approx 0.258144#

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Answer 2

Cameron Alexander

To find the arc length of the curve ( \left(\frac{t}{{(1+t)}^2}, -\frac{1}{t}\right) ) on the interval ( t \in [2,4] ), you can use the arc length formula:

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

In this case, ( a = 2 ) and ( b = 4 ), and the parametric equations are ( x(t) = \frac{t}{{(1+t)}^2} ) and ( y(t) = -\frac{1}{t} ). You'll need to find ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ), then substitute them into the arc length formula and integrate over the interval ( [2,4] ).

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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.