Recent content by blake knight

"If x>0 or -1 less than or equal x <0."--->I think this statement has some problem.

Is your answer 0.06% derived from 152/23600?

I agree with tnutty, try differentiating it implicitly with respect to t(time).

Now assuming that both "x" and "a" are variables, can you try solving the problem? I have the solution already, I just want to see how you do it.

Lets try the first assumption, i.e., x is a variable and a as a constant: Sin(x+a) = SinxCosa + CosxSina--->differentiate both sides with respect to x. you'll get: cos(x+a) = [sinx*(0) + cosa*cosx] + [cosx*(0) + sina*(-sinx)]-->notice the derivative of cosa = 0 because the derivative of a...

My friends, first of all "x" & "a" can be both variable or can be a variable "x" and a constant "a" in the expression: Sin(x+a) = SinxCosa + CosxSina. To prove, try solving: Sin(30+5) = Sin30Cos5 + Cos30Sin5 and youll find out the formula works. Now, starting from the fact that they can both...

The only hint I can probably suggest is for you to focus on what is being asked, analyze the question very carefully, then gather the informations given above that "only" relates to the question.

Ok so then you'll agree that if v'=(x^-c)' then v=(x^1-c)/1-c<----which is what I suggested before...Only that after Integrating by parts youll end up having an Integrand susceptible for Integration by parts again, and to infinity. This case happens due to the fact that we have a variable as an...

You can substitute 0 now for x, you will end up getting: g'(0) = -sin g(x)<----this solves the problem.

Now, understanding what the problem really asks for, it is asking for g'(0), meaning what is the value of g'(x) when x=0.

You should obtain an equation in the form of: g'(x)=[2x-sin g(x)]/[1+xcos g(x)].

First of all there's something missing in this equation: the x in front of cos g(x).

Im a little bit confused too, I know we are going too far now...but its a little bit strange to me when you have a derivative sign for your v', like v'=(x^-c)' and have a dx at the end of the integral, Isnt it repeating? I've never seen an equation of this form before. I maybe wrong.

Oh I see what you are getting now : I thought there is a small 1 there in v'=(x^-c)^1 which of course is just equal to v'=(x^-c).

You will arrive at having: = [(x^1-c/1-c)*(a^2-x^2)^c-1] - [integral of (x^1-c/1-c)*(-2x)(c-1)(a^2-x^2)^c-2 dx]--->continue from here.