D.Doughty Posted November 29, 2012 Share Posted November 29, 2012 Wow, I could probably take that.... how old do you have to be? I'm pretty sure you can do it whenever you want but they only offer the exam certain times of the year Link to comment Share on other sites More sharing options...
D.Doughty Posted November 29, 2012 Share Posted November 29, 2012 Wow B2 that's impressive I did the exam today and the lady there told me I will probably find out in 3 weeks. Did u just do B2? Or did u do an A level then upgrade to B2? I did A2 before I did B2. There's a major difference lol Link to comment Share on other sites More sharing options...
believe in blue forever Posted November 29, 2012 Share Posted November 29, 2012 I did A2 before I did B2. There's a major difference lol Yeah half of my class did B1 yesterday and they were saying their exam was super difficult where as those of us who went today thought it was pretty easy Link to comment Share on other sites More sharing options...
Peaches Posted November 29, 2012 Share Posted November 29, 2012 I can help in French if needed. Link to comment Share on other sites More sharing options...
TACIC Posted November 29, 2012 Share Posted November 29, 2012 I can do French too, I have done it my whole life Link to comment Share on other sites More sharing options...
KoreanHockeyFan Posted November 29, 2012 Share Posted November 29, 2012 First Year Calculus: I don't understand how they got to the f'(1) equation on the second part of the solution. Oh and on the first part of the solution, f'(x) is there because of implicit differentiation, correct? Thanks again. Link to comment Share on other sites More sharing options...
GodzillaDeuce Posted November 29, 2012 Share Posted November 29, 2012 I don't understand how they got to the f'(1) equation on the second part of the solution. Oh and on the first part of the solution, f'(x) is there because of implicit differentiation, correct? Thanks again. the f'(1) equation comes directly from the first equation you are given for f'(x), with the substitution x=1 when d/dx was applied to both sides, that was using implicit differentiation. as a result, you differentiated [f(x)]2 which needed the chain rule Link to comment Share on other sites More sharing options...
UMB Posted November 30, 2012 Author Share Posted November 30, 2012 Physics question - Electrostatics - is Brass a conductor or an insulator of charge? Is plastic a conductor or an insulator of charge? Link to comment Share on other sites More sharing options...
Peaches Posted November 30, 2012 Share Posted November 30, 2012 Is plastic a conductor or an insulator of charge? I'm pretty sure it's an insulator. Link to comment Share on other sites More sharing options...
SkeeterHansen Posted November 30, 2012 Share Posted November 30, 2012 I wonder what "Google search" thinks of this... Link to comment Share on other sites More sharing options...
Peaches Posted November 30, 2012 Share Posted November 30, 2012 @Yapierre I need an actual human to explain to me live Link to comment Share on other sites More sharing options...
KoreanHockeyFan Posted November 30, 2012 Share Posted November 30, 2012 the f'(1) equation comes directly from the first equation you are given for f'(x), with the substitution x=1 when d/dx was applied to both sides, that was using implicit differentiation. as a result, you differentiated [f(x)]2 which needed the chain rule Yeah I noticed that when I took a second look at it in the morning, doh! Thanks again though. Link to comment Share on other sites More sharing options...
CCF4E Posted November 30, 2012 Share Posted November 30, 2012 Multivariable Calculus, chapter of triple integrals Help please! Link to comment Share on other sites More sharing options...
GodzillaDeuce Posted November 30, 2012 Share Posted November 30, 2012 Multivariable Calculus, chapter of triple integrals Help please! you have two spheres in space, a small one on top of a large one where they overlap slightly (the smaller one sinks into the larger). the volume of each sphere alone is easy, but the volume of the entire snowman is the sum of each sphere minus the volume that they overlap. if you are using Cartesian coordinates, then what you want is to find the equation of the circle where the two spheres intersect (this is key). this will tell you the numbers for the bounds of integration in either x or y, and then the function that forms the bounds for the other variable. Once you have that, your bounds for z are easy, just integrate from the bottom of the "head" to the top of the "body". you can also switch to spherical or cylindrical coordinates (infact cylindrical might be best, theta does a full rotation, r goes from zero to the circle where the spheres intersect and z goes from the bottom to the top as before). Try that first, hopefully that's enough, but if you need the solution: the circle is x2+y2=3, so your x bounds would be +/- sqrt(3), y bounds are +/- sqrt(3-x2), and the bounds for z would be 4-sqrt(4-x2-y2) and sqrt(12-x2-y2) Link to comment Share on other sites More sharing options...
Mainly Mattias Posted November 30, 2012 Share Posted November 30, 2012 GD, I'm amazed you are still retaining this knowledge. Link to comment Share on other sites More sharing options...
GodzillaDeuce Posted November 30, 2012 Share Posted November 30, 2012 GD, I'm amazed you are still retaining this knowledge. hahaha, I'm not senile yet Link to comment Share on other sites More sharing options...
EoH Posted November 30, 2012 Share Posted November 30, 2012 has anyone read cormac mccarthy's book, outer dark? if you have much appreciate Link to comment Share on other sites More sharing options...
canucklax Posted December 1, 2012 Share Posted December 1, 2012 Can anyone give a good explanation on calculating limits? example: lim x->2 (x^(1/2)-2)/(x-4) Link to comment Share on other sites More sharing options...
GodzillaDeuce Posted December 1, 2012 Share Posted December 1, 2012 Can anyone give a good explanation on calculating limits? well sure, if you have about a month to talk about all of the different types of limits and the various different limit laws at your disposal. I'll say this, the most important type of limits in Calculus are difference quotients: limh->0 (f(x+h)-f(x))/h. The trick here is that you want to just be able to plug in the number zero for h, but doing so now would cause this function to become 0/0, which is undefined, so you must first do some simplification. For example, lets find limh->0 (f(x+h)-f(x))/h for the function f(x)=x2: limh->0 (f(x+h)-f(x))/h = limh->0 ((x+h)2-x2)/h =limh->0 (x2+2xh+h2-x2)/h =limh->0 (2xh+h2)/h =limh->0 (2x+h) =2x+0 =2x What is happening here is that the original expression ((x+h)2-x2)/h is a function of h that is undefined precisely where we are interested at looking: h=0. We get around that by saying "well if h=/=0, then this function looks exactly like 2x+h, and for this function at h=0 we get the expression 2x" example: lim x->2 (x^(1/2)-2)/(x-4) This function however is defined and continuous at the point x=2, so in this case we just plug that number in: limx->2 (sqrt(x)-2)/(x-4) = (sqrt(2)-2)/(2-4) =(sqrt(2)-2)/(-2) =1-1/sqrt(2) Link to comment Share on other sites More sharing options...
Mr. White Posted December 1, 2012 Share Posted December 1, 2012 I'm procrastinating my homework right now Link to comment Share on other sites More sharing options...
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