What I am doing wrong? arc length of parametric functions

1. Mar 22, 2008

motornoob101

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) =$$t^{2}-8t + 24$$
y(t) =$$t^{2}-8t -7$$

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
$$\frac{dx}{dt}=2t-8$$
$$\frac{dy}{dt}=2t-8$$

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

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

What I am doing wrong? Thanks.

2. Mar 22, 2008

HallsofIvy

Staff Emeritus
Have you forgotten that $\sqrt{x^2}= |x|$?

3. Mar 22, 2008

motornoob101

But it still give me zero though. I don't see how it changes things. Could you elaborate a bit more? Thanks

4. Mar 22, 2008

matticus

you need $$\sqrt{2}$$$$\int$$$$^{8}_{0}$$|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: Mar 22, 2008
5. Mar 22, 2008

motornoob101

Oh I see. Thanks so much