Recent content by xstetsonx

yea thanks to mark i got an A on my exam!

yup yup everything makes lot senses now but i just want to know how he get the equation 1/cosx and 2cosx if i am doing from scratch how do i know it is that equation?

MARK you are once again the awesomest guy on here thanks alot

(2)^.5, 45 degrees?

em (cosx)^2+(sinx)^2?

i mean (1,1) that is where they intersect right?

1? cos^2+sin^2?

yup under the attachment above. well you said appropriate i am not clear. my teacher he said for d\theta is from 0 to pi/4 for the first one...etc i don't understand how you get this value according to that graph

thx phyzmatrix anyways

then how you suppose to figure out the boundaries? and can you expand on the description for r changes??

so i don't substitute t into dz/dx and dz/dy? and for the second i don't substitute v into dz/dx and dz/dy either? so my answers will contain x,y,t for the first one and x,y,v?

my teacher said you have to separate the area into to parts in order to solve it and i don't understand whysorry about my crappy drawing again

k so i am just going to do the first one since they are similar dx/dt=3t^2------------dy/dt=-sint and then do dz/dx and dz/dy? after that substituted the t into x and y? i am confuse

can someone explain to me why my teacher divided the area into two? I=\int1,0\int(2x-x^2)^.5,0 1/(x^2+y^2)^.5dydx ugggggh i tried to use the latex... anyway... he used the polar coordinate to do this. once he turned it into polar coordinate, he divided the area into 2 bounded by (pi/4 - o...

chain rule someone help please 1. let z=y^2-x^2cosy; x=t^3 y=cost, find dz/dt 2. let z=(x-y)^3;x=u+2v,y=2u-v,find dz/dvmy attempt: so i know the chain rule is (dz/dx)dx+(dz/dy)dy 1. should i substitute the x and y into t first or should i do the partial derivative first? 2. same thing what...