- #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 \pi
I found that
\frac{\partial y}{\partial t} = e^{-t}(\cos{t}-\sin{t}),
\frac{\partial x}{\partial t} = -e^{-t}(\sin{t}+\cos{t})
Then setting up the integral:
\int_{0}^{\pi} \sqrt{(-e^{-t}(\sin{t}+\cos{t}))^2+(e^{-t}(\cos{t}-\sin{t}))^2} dt
I then simplified the square root to;
e^{-2t}(-4\cos{t}\sin{t}))=e^{-2t}*{-2\sin{2t}}
This makes the integral:
\int_{0}^{\pi} e^{-t}\sqrt{-2\sin{2t} dt
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 \pi
I found that
\frac{\partial y}{\partial t} = e^{-t}(\cos{t}-\sin{t}),
\frac{\partial x}{\partial t} = -e^{-t}(\sin{t}+\cos{t})
Then setting up the integral:
\int_{0}^{\pi} \sqrt{(-e^{-t}(\sin{t}+\cos{t}))^2+(e^{-t}(\cos{t}-\sin{t}))^2} dt
I then simplified the square root to;
e^{-2t}(-4\cos{t}\sin{t}))=e^{-2t}*{-2\sin{2t}}
This makes the integral:
\int_{0}^{\pi} e^{-t}\sqrt{-2\sin{2t} dt
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
Last edited:
