find the length: x=e^tcos t
y=e^tsin t. t is betw 0 and 2

Question

find the length: x=e^tcos t  
y=e^tsin t. t is betw 0 and 2

anonymous · Answer

hahha. if i were the real bieber do u think i would calll myself liliy and would i actually be smart enough to find this site and do math qs?

anonymous · Answer

to find the curve you do sqaure root of(dx/dy)^2+ (dy/dx)^2)

anonymous · Answer

ya?

australopithecus · Answer

are you looking for distance between y values?

anonymous · Answer

holy crap. DO nOT!!! come into MY QUESTION and write a whole long question like this. please open ur open question and i will be glad to help u out with these:)

anonymous · Answer

@liliy :o i did :P and no one answered it and theres only 4 questions but there not bath there career education .-.

anonymous · Answer

fine. im a nice person so ill help: but its really not fair what u did. A,A,A,D

anonymous · Answer

;p I'm sorry ill delete mine

anonymous · Answer

relax, its fine

anonymous · Answer

im just saying having it in ur own question is better.

anonymous · Answer

(also you can legit just google the questions)

anonymous · Answer

i did lmfao and I couldn't find any thing thats why I'm here

anonymous · Answer

alright:). well i hope i helped

anonymous · Answer

with what o.O?

anonymous · Answer

i just gave u the answers : 1.A 2.A 3.A 4.D

anonymous · Answer

:O

anonymous · Answer

@liliy thanks :O

dumbcow · Answer

to find length of the parametric curve
you need to integrate over the arc length of the curve
$$\int\limits_{}^{}\sqrt{dx^{2} +dy^{2}}$$
to factor in variable t, multiply by dt/dt
$$\int\limits_{0}^{2}\sqrt{\frac{dx^{2} +dy^{2}}{dt^{2}}} dt$$
which can be written as
$$\int\limits_{0}^{2}\sqrt{(\frac{dx}{dt})^{2} +(\frac{dy}{dt})^{2}} dt$$
$$\frac{dx}{dt} = e^{t}(\cos t -\sin t)$$
$$\frac{dy}{dt} = e^{t} (\sin t + \cos t)$$
$$\rightarrow \sqrt{2}\int\limits_{0}^{2}e^{t} dt$$

anonymous · Answer

wait, i dont undestand ur last step. wat happend to the sin and cos?

dumbcow · Answer

plug in the expressions for dx/dt and dy/dt , square them and combine terms
when you do that you notice that sin^2 + cos^2 = 1 , and the rest of the sin and cos terms cancel
so it simplifies to that nice easy integral i have above

anonymous · Answer

dx/dt is subtracting betw the cos and sin, so it doesnt cancel out!