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

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

Answer 1

#0.326586#

We have #x(t)=t/sqrt(t-1)# #x'(t)=1/sqrt(t-1)-t/(2*(t-1)^(3/2)# #y(t)=t/(t^2-1)# then #y'(t)=1/(t^2-1)-2*t^2/(t^2-1)^2# and our integral will be #int_2^3sqrt((t-2)^2/(4*(t-1)^3)+(t^2+1)^2/((t-1)^4*(t+1)^4))dt# which i can only solve by a numerical method.

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Answer 2

Emma Aldridge

To find the arc length of the curve ( f(t) = \left(\frac{t}{\sqrt{t-1}}, \frac{t}{t^2-1}\right) ) on the interval ([2,3]), you 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 ( \frac{dx}{dt} ) and ( \frac{dy}{dt} ) are the derivatives of ( x ) and ( y ) with respect to ( t ) respectively.

Taking derivatives:

[ \frac{dx}{dt} = \frac{\sqrt{t-1} - \frac{t}{2\sqrt{t-1}}}{t-1} ] [ \frac{dy}{dt} = \frac{t^2-1 - t(2t)}{(t^2-1)^2} ]

Substitute these into the arc length formula and integrate from 2 to 3:

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

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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.