Recent content by trap

Any help will be very gracious. If \text C_{1} , C_{2} , C_{3} are all non empty compact sets in \text R^n such that \text C_{k+1} \subset C_{k} for all k=1,2,3,..., then the set \text C = I_{k=1}^{\infty}C_{k} is also non-empty.

ok, i got it now, thanks for the help

can you help me with one more step? i really can't make two integrals with power of u..

yeah, i tried that, but i can't go further coz i can't get rid of the t^3

Anyone know how to solve this? \text 8\pi\int_{0}^{\infty}\frac{t^3}{(4+t^2)^\frac{5}{2}} dt

yeah, something about the parametric, cartesian, polar equations are what we are learning. But I still don't get how to find the 'maximum' and 'minimum' values of the curvature.

no...we are not learning vectors

Sorry...I don't really get what you just typed, coz I don't think I have learned those in my course. We are currently doing parametric equations and polar coordinates. Is there an other approach to the question?

so do I derive the curve and let it equal to zero to find the maximum? how about the minimum? also...I don't really get how do you find the curvature of the curve?

any clue? Determine maximum and minimum values of the curvature at points of the polar curve r = 3 + sin \theta .

I need help with this question which askes to prove it. Anyone has any idea? If \text C_{1} , C_{2} , C_{3} , … are all closed sets in \text R^n , then the set \text c = I_{k=1}^{\infty}c_{k} is also a closed set in \text R^n .

yes..the answer is supposed to be 2pi^2, unless it is a typo in the textbook

but i got the formula of surface area from my textbook, which is, \text{SA}=2\pi\int_{}^{}y\,\sqrt{\left(\frac{ dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}\, dt

I forgot to mention that the interval is t between (-pi/2, pi/2), how do i get the answer 2pi^2 with \text{SA}=2\pi\int_{-\pi/2}^{\pi/2}y\,\sqrt{\left(\frac{ dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}\, dt ?? so i have... dx = -cost dy = -sint here's what I've...

Can someone please help me with this question? x = 1-sint, y = 2+cost, rotate about y = 2 Find the surface area of the parametric curve. I don't know how to do it with y=2, I only know how if the question askes for rotating about the x-axis. The answer to the question is 2(pi)^2.