Recent content by Galadirith

Hi tiny-tim, thanks so much for the reply, Yes that was one of my chains of taught, at sort of the end area :D, well actually I was treating T1 and T2 as vectors, not absolute value so I would have T1 = -T2, the reason being is the motivation for this is to be able to implement this type of...

Hey Sky.Anthony, Sorry, i did mean the answer you worked out was correct, the way you did it is perfectly fine, using a substitution certainly makes the differentiating the function easier somewhat as one doesn't have to deal with long complex expression, but without using a substitution as you...

Hey Sky.Anthony :D, Yeh that looks good to me, now I am not sure how you did \frac{d}{dx}(x^x) Im quessing as you let a=xx you did ln(a) = xln(x) and differentiated implicitly. However another way of doing this is considering x^x = e^{ln(x^x)} = e^{xlnx} although you may have done it that...

Hi guys, I was thinking of a problem the other day; consider a group of particles, connected to each other by light elastic strings, there is no restriction on which particles can be connected to which, then can one determine the position of a particle at time t. So I set about trying to...

Hi ganondorf29, I think actually your way over complicating this :D. I suppose much better than under complicating it. So let's start from a good point you got to: \frac{1}{2}\int sin^3(u) \ du = \frac{1}{2}\int sin^2(u)sin(u) \ du (careful you wrote x in the final sin in yours...

Hi amb1989, That was an interesting way of trying to solve the problem, using the idea of the trig functions in triangles, never tried that myself. So your wanting to evaluate: \int_1^2 \frac{\sqrt{x^2-1}}{x} \ dx Ill assume that the lower bound was in fact 1 and not 2 as the order...

Oh ok thename1000, sure think I just assumed since the answer was there. So question #1, you need to first find k. Now you don't need to integrate anything to find k for the first question as we are given the force and the length of the spring from which we can determine the extension, so we...

Hi thename1000, So you've solved 1.). Now for two I think the best way is to have two integrals as you have suggested. I would say yes this is done always, but then this is quite a specific question, there's really only this way to solve it. So required equation: E_{el} = \int_{x_0}^{x_1}...

Hi KillerZ, Look perfect to me except the very end. You have: arcsin(u) - ln|x| = ln|c| the next step isn't quite right, it should go: arcsin(u) = ln|x| + ln|c| arcsin\left(\frac{y}{x}\right) = ln|cx| e^{arcsin\left(\frac{y}{x}\right)} = cx but other than that its perfect, I...

Hi icosane, Well first ill assume that f(x,y) = z. Perhaps this is something that you forgot to include, or perhaps it inst specified in the question, which might then be where all the confusion lies. So considering the two equations: \textbf{r}\ = \ <t+(1/t),t-(1/t),8> \ \ (i)...

Ok sure thing neutron. I was assuming that you had already covered basic calculus skills, but that's kl. So let's first look at the case where its in the form ax. So a represents and integer, just like the example you asked about 4x. Now let's we say the f(x) = 10x. Now let's look at the...

Ah rite neutron star, now I think I can give you a bit more help. Rite I think you are reading the notation wrong. Now you wrote fl(3) however that is not what the question will be it will be f'(3), where that little "dash" in-between the f and the (3) bit is just that, a dash, read...

Hi neutron star. Now its slightly ambiguous a question as fr(x), could mean raising the function to a power of r at x, ie [f(x)]r and I wouldn't be completely sure without know the context the question is asked in. However generally fr(x), means the rth derivative of f(x) (this is know as...

Hey ksm100, that's no problem. Rite well I believe I can see why your not completely happy with it. So all I have done is rearrange (i), so because of that if I can prove that the final inequality I presented is true, then any rearrangement of that inequality must also be true. However, Ill...

Hi ksm100, Well this was interesting I wasn't sure where to go from where you left off, I just learned about induction in my A-levels last year. I think everything you have done up to that point is perfect. So as you said, if we can show that...