- #1
Kb1jij
- 19
- 0
I'm having trouble with the following:
The problem is to find the arc length of the following parametric function:
x=(e^-t)(cos t), y=(e^-t)(sin t) from 0 to [tex] \pi [/tex]
I found that
[tex] \frac{\partial y}{\partial t} = e^{-t}(\cos{t}-\sin{t}) [/tex],
[tex] \frac{\partial x}{\partial t} = -e^{-t}(\sin{t}+\cos{t})[/tex]
Then setting up the integral:
[tex]\int_{0}^{\pi} \sqrt{(-e^{-t}(\sin{t}+\cos{t}))^2+(e^{-t}(\cos{t}-\sin{t}))^2} dt [/tex]
I then simplified the square root to;
[tex] e^{-2t}(-4\cos{t}\sin{t}))=e^{-2t}*{-2\sin{2t}} [/tex]
This makes the integral:
[tex] \int_{0}^{\pi} e^{-t}\sqrt{-2\sin{2t} dt[/tex]
Can this integral even be solved?
I don't think this problem was ment to be that difficult, so I think I made a mistake somewhere, but I can figure what.
Thanks!
Tom
The problem is to find the arc length of the following parametric function:
x=(e^-t)(cos t), y=(e^-t)(sin t) from 0 to [tex] \pi [/tex]
I found that
[tex] \frac{\partial y}{\partial t} = e^{-t}(\cos{t}-\sin{t}) [/tex],
[tex] \frac{\partial x}{\partial t} = -e^{-t}(\sin{t}+\cos{t})[/tex]
Then setting up the integral:
[tex]\int_{0}^{\pi} \sqrt{(-e^{-t}(\sin{t}+\cos{t}))^2+(e^{-t}(\cos{t}-\sin{t}))^2} dt [/tex]
I then simplified the square root to;
[tex] e^{-2t}(-4\cos{t}\sin{t}))=e^{-2t}*{-2\sin{2t}} [/tex]
This makes the integral:
[tex] \int_{0}^{\pi} e^{-t}\sqrt{-2\sin{2t} dt[/tex]
Can this integral even be solved?
I don't think this problem was ment to be that difficult, so I think I made a mistake somewhere, but I can figure what.
Thanks!
Tom
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