- #1
motornoob101
- 45
- 0
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.
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.