Recent content by steel1

Homework Statement Not really a problem, more of a general question. When exactly can you use the Divergence test. Does it only work on both series and sequences?Homework Equations The series Diverges if lim ƩAn ≠ 0 The Attempt at a Solution If you take the lim of the series n^3/2n^3 ≠ 0 there...

Hmm, ok thanks for explaining. Also, 1 small question. why is the inner loop at .5 at 180 degrees, when cos(pi)+.5 is actually -.5?

Hmm. you would think A+B-B, the B's just cancel and your left with A. but i guess not though

So your saying that it takes longer than 2pi to complete before it starts tracing over itself again? and idea how i can find the time it takes a polar curve to complete 1 cycle?

Homework Statement Find the area inside the larger loop and outside the smaller loop of the limacon r=.5+cosθ Picture here http://www.wolframalpha.com/input/?i=r%3D.5%2Bcostheta Homework Equations Area = .5∫r^2The Attempt at a Solution To get the area of the outer loop, you just get the value...

Homework Statement Find the area enclosed by the graph r=2+sin(4θ) Homework Equations Area = .5∫r^2 The Attempt at a Solution Area = .5∫(2+sin(4θ))^2 =.5(4.5θ-1/16sin(8θ)-cos(4θ)) I can do the integration and all, but I am having trouble finding the limits of integration I...

yes, i was getting exactly half the right answer. forgot about the symmetry, thanks!

2.739

not sure what your trying to get at. :<

Homework Statement Find the area of the region that lies inside the curve r^2=8cos(2θ) and outside r=2Homework Equations area of polar curves = .5∫R^2(outside)-r^2(inside) dθThe Attempt at a Solution r^2=8cos(2θ) and r=2, so... 4=8cos(2θ) .5=cos(2θ) since .5 is positive, we need the angles in...

yes, i would prefer to solve this by u substitution, so i can see how its done.

Yeah, this is what i have so far U^2=t 2du=dt

If somebody could just do the entire thing via U substitution, that would be great. Then I can see how you did it. I'll be refreshing this page every few minutes.

derivative of u^2=t is 2u=1?

ok. If i set U=√t and Du=dt, where do i go from there? the du=dx was a typo.