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

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

Answer 1

Isabella Ackerman

#L=1/8(3sqrt5+5sqrt13)+1/(8sqrt2)ln((sqrt10+3)/(sqrt26-5))# units.

#f(t)=(t^2-t,t^2-1)#

#f'(t)=(2t-1,2t)#

Arclength is given by:

#L=int_-1^1sqrt((2t-1)^2+4t^2)dt#

Expand the square:

#L=int_-1^1sqrt(8t^2-4t+1)dt#

Complete the square:

#L=1/sqrt2int_-1^1sqrt((4t-1)^2+1)dt#

Apply the substitution #4t-1=tantheta#:

#L=1/(4sqrt2)intsec^3thetad theta#

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

#L=1/(8sqrt2)[secthetatantheta+ln|sectheta+tantheta|]#

Reverse the substitution:

#L=1/(8sqrt2)[(4t-1)sqrt((4t-1)^2+1)+ln|4t-1+sqrt((4t-1)^2+1)|]_-1^1#

Insert the limits of integration:

#L=1/(8sqrt2)(3sqrt10+5sqrt26+ln((3+sqrt10)/(-5+sqrt26)))#

Hence:

#L=1/8(3sqrt5+5sqrt13)+1/(8sqrt2)ln((sqrt10+3)/(sqrt26-5))#

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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.