Recent content by raincheck

hmmm, I asked someone else and they got 8pi too... Well I had one more question... if I were finding the area between r = sqrt[cos2(theta)] and r = 2cos(theta) do I basically do the same thing as finding the area between two regular curves? Each time I try it, I come up with zeros...

Homework Statement "Find the area of the region described: The region that is enclosed by the rose r=4cos3[theta]" Homework Equations A= [integral] (1/2)r^2 d[theta] The Attempt at a Solution I'll use Q as [theta].. I'm not really sure, but I set up (1/2) [integral] (16(cos^2)3Q) dQ ...

so .001 = (x^4)/4! ? Would I solve for x? Thanks so much for helping me! :]

OH ok " |Rn(x)| is less than or equal to (M/(n+1)!)|x-xo|^(n+1) " so (1/4!)|-xo|^(4) ? I already know xo right? I still don't know what it is though..

" |Rn(x)| is less than or equal to (M/(n+1)!)|x-xo|^(n+1) " so am I using 3 as n? Sorry for completely not getting this!

Ohh ok awesome! So R3(x) would be f(x) - p3(x) since its the remainder? maybe... .008?

Hahah ummmmm... the p3?

So: let xo = 0? f(x)= sinx, f(0)=0 po(x)=0 f'(x)= cosx, f'(0)=1 p1(x)=x f''(x)= -sinx, f''(0)=0 p2(x)= x f'''(x)= -cosx, f'''(0)=-1 p3(x)=x - (1/3!)x^3

OH! Hmm..all that's on my question page is.. 0?

Hmm.. I don't know isn't just infinity? Or does it get a value?

Isnt it the nth+1 derivative of ...sinx? Hahah I don't feel like I'm getting it at all :[

Right.. M is the upper bound.. And I'm trying to find an interval with x=0 in it, so could the upper bound be pi? or 1? Ok, I'm not sure.

I guess I would say xo is 0? And M is |F^(n+1)(x)| ?

Oh.. is it Rn(x) = f(x) - pn(x) = f(x) - [sigma] (f^(k)*(xo))/k! * (x-xo)^k ? That's the only other equation I can find..

Hmm well that makes sense, but how do I find the remainder term? I don't understand that..