Recent content by Callisto

Hi all, I was wondering what some of you thought about 'Information Theoretic Process Physics'? The topic seems fascinating, however somewhat difficult to come to grips with. I attached a link for those who might be interested. There have been papers published which are surly worth a...

If the emissivity of a flat surface varies with zenith angle according to e=E*cos(theta) where E is the emissivity at zenith. Would this surface radiate isotropically? I think that because the emissivity varies then the emitted radiation varies accordingly so the energy measured from...

If the emissivity e varies with zenith angle according to e = E*cos(theta) where E is the emissivity normal to the surface. Would this surface be an isotropic source of radiation? Well, since e varies with angle then the flux density must vary accordingly so the surface would radiate...

I see that we get L = int:dtheta*sqrt((dr/dtheta)^2+r^2) which is the length of the polar curve r=f(theta) how do i decide what is a worthy substitution for dtheta?

Hi all, I seeking some advice about the calculus of variations. I am an undergraduate and i am enrolled in a topic of the above mentioned. After successfully completing the requirments for the topic, 3 weeks after commencement i am feeling way out of my depth. I understand that the calculus...

Cheers James R!

Taylor series for 1/(1-x) = (1-x)^-1 = 1 + x^2 + x^3 + x^4 +...= SUMi x^i Then sub in exp(-1) for x That's it?

Hi How do you expand (1-exp(-1))^-1 as Taylor series Callisto

thanks for your replies' from 0 = 2*lambda*x+ 4*rho*x = 2*x(lambda +2*rho) = 0 when x = 0 if x = 0 then y^2 = z^2 so z = y = +/-sqrt(15) so the min value is -sqrt(15), that must mean the max value is sqrt(15) is this right?

This problem was given in my calc class during the semester, "Find the lowest point on the intersection of the sphere x^2+y^2 +z^2 = 30 and the cone 2*x^2 +y^2 = c^2". I don't know how to solve this problem with lagrange multipliers. How is it done? Thanks! Callisto

Thanks 2f must be minimum since if the object p is at infinity then 1/p=0 so q = f for an object at infinity.

1/0+1/0=1/0, 1/0 doesn't = 1/f is this why 0's are excluded?

1/0+1/0=0 , 0 doesn't = 1/f therefore 0's are excluded from the problem, correct? Callisto

solving 1/p+1/p=1/f for p i get -qf/(f-q) now i substitute this into p+q, then i get -q^2/(f-q). now if i use calculus to find the minimum value i differentiate -q^2/(f-q) and let it equal zero so d/dq= -q(2f-q)/(f-q)^2 =0 when q = 0 or 2f Repeating this process for p then p = 0 or...

Hi Peaple :!) How do you determine the minimum distance p+q between the object and image for given focal length given 1/p + 1/q = 1/f this may seem trivial, but i can't figure it out. any tips please Callisto