What I am doing wrong? arc length of parametric functions

So this seems to be a pretty straightforward question but I keep getting the arc length to be 0 and I redid this question many times..

Find the length of the parametrized curve given by

x(t) =[tex]t^{2}-8t + 24[/tex]
y(t) =[tex]t^{2}-8t -7 [/tex]

How many units of distance are covered by the point P(t) =(x(t), y(t)) between t=0, and t =8?

So my first step of course is to find dx/dt and dy/dt
[tex]\frac{dx}{dt}=2t-8[/tex]
[tex]\frac{dy}{dt}=2t-8[/tex]

Then set up the arc length equation
arc length = [tex]\int^{8}_{0}\sqrt{{\frac{dx}{dt}}^2+{\frac{dy}{dt}}^2}dt[/tex]
= [tex]\int^{8}_{0}\sqrt{{(2t-8)}^2+{(2t-8)}^2}dt[/tex]
= [tex]\int^{8}_{0}\sqrt{2{(2t-8)}^2}dt[/tex]
= [tex]\int^{8}_{0}\sqrt{2}(2t-8)dt[/tex]
=[tex]\sqrt{2}\int^{8}_{0}(2t-8)dt[/tex]
=[tex]\sqrt{2}\left[{t}^2-8t\right]^{8}_{0}dt[/tex]

Which give me an answer of zero when the answer is suppose to be 45.2548.

What I am doing wrong? Thanks.
 

HallsofIvy

Science Advisor
Homework Helper
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884
Have you forgotten that [itex]\sqrt{x^2}= |x|[/itex]?
 
But it still give me zero though. I don't see how it changes things. Could you elaborate a bit more? Thanks
 
107
1
you need [tex]\sqrt{2}[/tex][tex]\int[/tex][tex]^{8}_{0}[/tex]|2t-8|dt. to integrate this, notice that from 0 to 4, 2t - 8 is the negative of |2t - 8|. so we must multiply it by -1 to get 8 - 2t. from 4 to 8 2t - 8 = |2t - 8|, so w,e can leave it as is. so you must seperate this into two integrals, the 8 - 2t from 0 to 4, and 2t - 8 from 4 to 8, then add the results.
 
Last edited:
Oh I see. Thanks so much
 

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