Here's the question you clicked on:
liliy
find the length: x=e^tcos t y=e^tsin t. t is betw 0 and 2
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?
to find the curve you do sqaure root of(dx/dy)^2+ (dy/dx)^2)
are you looking for distance between y values?
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:)
@liliy :o i did :P and no one answered it and theres only 4 questions but there not bath there career education .-.
fine. im a nice person so ill help: but its really not fair what u did. A,A,A,D
;p I'm sorry ill delete mine
im just saying having it in ur own question is better.
(also you can legit just google the questions)
i did lmfao and I couldn't find any thing thats why I'm here
alright:). well i hope i helped
i just gave u the answers : 1.A 2.A 3.A 4.D
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 \[\int\limits_{0}^{2}\sqrt{(\frac{dx}{dt})^{2} +(\frac{dy}{dt})^{2}} dt\] \[\frac{dx}{dt} = e^{t}(\cos t -\sin t)\] \[\frac{dy}{dt} = e^{t} (\sin t + \cos t)\] \[\rightarrow \sqrt{2}\int\limits_{0}^{2}e^{t} dt\]
wait, i dont undestand ur last step. wat happend to the sin and cos?
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
dx/dt is subtracting betw the cos and sin, so it doesnt cancel out!