find the length of the curve C from t=0 to t pi/2 where x(t)=2sin^3t and y(t)=2cos^3t. set up the correct integral and simplify the integrand.
i think im supposed to use the formula L= sqrt[(dx/dt)^2 + (dy/dt)^2]
for x' i got 36sin^4tcos^2t
y' = -36sin^2tcos^4t
how am i supposed to add(or subtract) them so that it =6sintcost ? that was the answer given for the part inside the integrand. so before rooting im supposed to get 36sin^2tcos^2t but how?
i think i might have screwed up on my math typing again...i wanted to say x=2sin^3(t) and y=2cos^3(t). just t not 3t so "sin and cos cubed t" o.o if not then i dont know how to get 3t...
whats the technique to deferentiate those trigs with cubes. can you go
2cos^3(t)
2(cos(t))^3 u=cos(t)
6(u^2)u'
-6cos^2(t)sin(t)
or does the sign switch both times you differentiate? it becomes negative then postive again? or the sign only switches when you differentiate the last cos(t)?