find the exact length of the curve.
x = tsint , y = tcost
i've come up with $$\int\limits_{0}^{1} \sqrt{2t^2+1}$$
after doing dx/dt and dy/dt and squaring under the square root symbols and using trig identities
how do i proceed to do this without a calculator now? or have i messed up along the way?

$$dx/dt = sint + tcost $$
$$dy/dt = cost-tsint$$
$$L = \int\limits_{0}^{1}\sqrt{(sint + tcost)^2 +(cost-tsint)^2 }$$

Question

find the exact length of the curve.
x = tsint , y = tcost
i've come up with $$\int\limits_{0}^{1} \sqrt{2t^2+1}$$
after doing dx/dt and dy/dt and squaring under the square root symbols and using trig identities
how do i proceed to do this without a calculator now? or have i messed up along the way?

$$dx/dt = sint + tcost $$
$$dy/dt = cost-tsint$$
$$L = \int\limits_{0}^{1}\sqrt{(sint + tcost)^2  +(cost-tsint)^2 }$$

shamil98 · Answer

i think i've found my mistake one sec

shamil98 · Answer

came up with this i i thought it was t^2(cos^2 + sin^2) by its this instead
$$L = \int\limits_{0}^{1} \sqrt{t^2(\cos^2t*\sin^2 t) + 1}$$

parthkohli · Answer

http://www.wolframalpha.com/input/?i=%5Cint_%7B0%7D%5E%7B1%7D+%5Csqrt%7B(%5Csin+t+%2B+t+%5Ccos+t)%5E2+%2B+(%5Ccos+t+-+t+%5Csin+t)%5E2+%7Ddt