Recent content by vesu

Oh right, that makes sense. Thanks!

Ah, I think I've got it. I ended up with 2\sinh^{-1}(\sqrt{x}) + C. Now we're supposed to let y = 2\sinh^{-1}(\sqrt{x}) + C, find the inverse, then "use implicit differentiation to prove that the fundamental theorem of calculus holds, i.e. prove that \frac{dy}{dx} = \frac{1}{\sqrt{(x+x^2)}}". I...

Homework Statement \int \frac{1}{\sqrt{x + x^2}} dx We have been told to use the substitution x = \sinh^2{t}.Homework Equations \int \frac{1}{\sqrt{a^2 + x^2}}dx = \sinh^{-1}(\frac{x}{a}) + C Maybe?The Attempt at a Solution I'm not really sure where to start, we haven't done any questions...

Thanks PhanthomJay, I figured it and the rest of the questions out. Also my apologies, after all my stubbornness it was actually 9kN. I don't know how I managed to repeatedly misread that every time.

They have given us the value of 9N. This is a very introductory class so I think they are just trying to teach us the basics with examples that are not necessarily practical or realistically possible. The weight of the members hasn't been mentioned at any point in anything we've covered...

Woops sorry, the wording in the question seems to alternate between using "beam" and "member" to describe them. I'll give that a shot, thanks! Somehow I completely missed that when looking through my lecture notes.

Hi, sorry the beam is in compression I forgot to mention that. It is definitely 9N.

We have a truss with seven beams in total, this part of the question is asking for the minimum area of the square cross-section of an individual beam so that the structure will not fail (i.e. so that any individual beam will not fail). There is one beam that has the greatest length and force (6m...

Homework Statement A beam (note: part of a truss, I chose to use the beam in the truss with the most force and length for my calculations, which I assume is the correct thing to do) of length 6m with a force of 9N is being constructed out of a material with a Young's Modulus of E = 70 \times...

so zw = 0 re^{i\theta} \times qe^{i\phi} = 0 rqe^{i(\theta + \phi)} = 0 e^{i(\theta + \phi)} \not= 0 \frac{rqe^{i(\theta + \phi)}}{e^{i(\theta + \phi)}} = \frac{0}{e^{i(\theta + \phi)}} rq = 0 \therefore r=0 or q=0 \therefore z = 0 or w = 0 Um, I can't find anything at all in our course...

Is it? I was just using |zw|=|z||w|, I don't think that's what we're trying to prove. :S

Sorry, I'm not really sure what to do with your suggestion and I'm still trying to figure out if either of the proofs I have is correct! Is this proof okay if the other one isn't? (even though this doesn't involve exponential polar form, that was only a hint, not a requirement) let z = a +...

Uh... I understand what you're saying about what's wrong with my proof but I really don't know what to do even with the hint. :confused: I mean I know r = |z| and q = |w| so rq=|zw| but I don't know where that gets me.

This has confused me even more. :S also I cringe every time I re-read this: I was very very tired. :frown:

I'm a little confused, could you explain how this is incorrect? zw = 0 zw = re^{i\theta} \times qe^{i\phi} zw = rqe^{i(\theta + \phi)} e^{i(\theta + \phi)} \not= 0 \therefore r=0 or q=0 \therefore z = 0 or w = 0 I guess you could also do it this way to eliminate i altogether so that you're...