Recent content by A_I_

I need to proove the Minkowski's inequality for integrals. I am taking a course in analysis. [ int(f+g)^2 ] ^(1/2) =< [int(f^2)]^(1/2) + [int(g^2)]^(1/2) now we are given that both f and g are Riemann integrable on the interval. So by the properties of Riemann integrals, so is f^2,g^2...

yes F = kQ1Q2/r where K is the coulomb constant :) But Still i didnt find the formula :( any hint? :)

the problem is that i don't know the formula of the potential energy when we are dealing with charges :( i know E=mgh where mg is the force and h is the distance. If we put F (coulomb force) to be the force in this problem, and r? to be the distance, and then we integrate it??it still...

show that the potential energy due to electrostatic forces of uniformly charged sphere of radius R and total charge Q is: 3Q^2/5R i don't even know how to start?? i really don't want to hate physics:frown: :frown: :frown: i was reading through in my textbook and they only gave...

A 14.4 KeV photon from 57 Fe is red shifted as it rises from a sourceat ground level to an absorber placed at the top of a tower of a height of 20 m because it has to expend energy to climb the gravitational potential. Derive an expression for thered shift as a fraction of the energy of the...

I get what you've said. Thanks for the explanation

for the first part, that's what i was trying to tell him, that we only use the chain rule when we have a multivariable function, which is not the case. As for the second part, i am not sure about it, because i know you can only use the chain rule with multivariable function. Did u consider x...

here you go

what do you exactly want to know? do you want an example of the chain rule? what you wrote: f(x)= g(r(x)) is (gor)(x) and is different from the chain rule. do you have an im?

the chain rule is used when u have: f(x,y) and x=g(t) and y=r(t) but since your f and your u are both functions of x, thus you can not use the chain rule and you can not say: f'(x) = (dy/du)*(du/dx) ok?

your mistake was taking partial derivatives u write f'(x) = dy/dx = (dy/du)/(du/dx) since y and u are both functions of x u can not apply the partial derivative (the chain rule formula)

hehe good my method was simple: f'(2x) = x^2 --> f(2x) = x^3 / 3 we multiply the denominator and numerator by 8 --> f(2x) = 8x^3 / 24 ---> f(2x) = (2x)^3 / 24 ---> f(x) = x^3 / 24 --> f'(x) = x^2 / 8

let me know what your answer is to see if you did it right

u have f'(2x) = x^2 first find f(2x) and try to write it as a function of (2x) and then after u do that substitute x for 2x you therefore find f(x) finally derive it and you obtain f'(x) does it make any sense?

g(x) is a function of x that we do not know its form. y(t) =(1/2) integral 0-->t [ sin(2t-2x)*g(x) ]dx i tried to use integration by substitution and by parts but the problem is that g(x) has an unknown form. the actual problem is that y" +cy' +dy = g(t) y(0) = y0 y'(0) = y'0...