timtamboy63 Posted December 14, 2011 Report Share Posted December 14, 2011 Try to search next time,http://www.ibsurvival.com/topic/8489-type-i-infinite-summation/ Reply Link to post Share on other sites More sharing options...
hraouf1 Posted December 14, 2011 Report Share Posted December 14, 2011 People i need help with my infinite summation IA , the part which is talking about Tn (a,x) , the validity , general statement , alot !!Help pleaseThere is a complete thread discussing this portfolio.Could you give me a link to it please , i can't find it Reply Link to post Share on other sites More sharing options...
opus92fan Posted December 15, 2011 Report Share Posted December 15, 2011 I'm having trouble with two questions on my homework for related rates. Could someone possibly explain the approach to the question without giving the answer?* A person 2 m tall walks away from a street light whose light is 5 m above the ground. If he walks at a speed of 1.5 m/s, at what rate does his shadow grow when he is 10 m away from the street light?* A book 15 cm long rests against a vertical bookshelf. If the bottom of the book slides away from the bookshelf at a speed of 3 cm/s, how quickly is the top of the book sliding down the bookshelf when the bottom of the book is 2 cm away from the bookshelf?Thank you! Reply Link to post Share on other sites More sharing options...
Drake Glau Posted December 15, 2011 Report Share Posted December 15, 2011 #1 I can't draw out in my head, sorry #2 is using Pythagorean Theory in rates. The book, floor, and book shelf form a right triangle where the book is the hypotenuse and the floor/bookshelf are the legs. The book doesn't change length so c=15. a2+b2=152 Now differentiate it all. 2a(da)+2b(db)=225 We're going to call the floor a. So a=2 and da=3 If a=2, c=15, b=sqrt(225-4)=14.866 So back to the differentiated one... 2(2)(3)+2(14.886)(db)=225 Solve for db. db=(225-12)/29.732 =7.16 Reply Link to post Share on other sites More sharing options...
Tianna2012 Posted December 21, 2011 Report Share Posted December 21, 2011 I'm studying for my final tommorow and I have no idea how to solve for the constant of an exponential growth or decay equation. for eample if I'm given 27200=32000r^1 How would I find r? Reply Link to post Share on other sites More sharing options...
rFumachi Posted December 21, 2011 Report Share Posted December 21, 2011 (edited) The answer to your example is r= 27200/32000, because you have r^1 = r.But if you have:a=br^Xa/b = r^XX root both sides|r| = X√(a/b) Edited December 21, 2011 by rFumachi Reply Link to post Share on other sites More sharing options...
AEdwy Posted December 27, 2011 Report Share Posted December 27, 2011 I'm stuck on a simultaneous equation and was wondering if someone would help me, WITH WORKING OUT! thanks3=p+qcos(o)-1=p+qcos(pi)Solve for p and q Reply Link to post Share on other sites More sharing options...
dessskris Posted December 27, 2011 Author Report Share Posted December 27, 2011 use elimination to get rid of p. you know the values of and , right? solve for q. then, solve for p using one of the equations. alternatively, you can treat p and q as x and y. express y in terms of x, and then plot those two equations on your GDC. then find the intersection. 1 Reply Link to post Share on other sites More sharing options...
Sarah Mahmoud Posted January 5, 2012 Report Share Posted January 5, 2012 does anyone know a good website to find ib practice questions?helppp SOS Reply Link to post Share on other sites More sharing options...
CkyBlue Posted January 5, 2012 Report Share Posted January 5, 2012 Doesn't your textbook designed specifically for IB? You needn't worry about it if you're in Pre-IB or the MYP right now. Just do your homework; if you're feeling shaky, purchase the course companion or study guide.Also get your hands on past papers by asking your teacher, or just googling them. IBS does not host past paper links. Reply Link to post Share on other sites More sharing options...
timtamboy63 Posted January 8, 2012 Report Share Posted January 8, 2012 (edited) (It's an induction question), I keep coming out with an answer that's off by a factor of two for Edited January 8, 2012 by timtamboy63 Reply Link to post Share on other sites More sharing options...
bomaha Posted January 8, 2012 Report Share Posted January 8, 2012 (edited) (It's an induction question), I keep coming out with an answer that's off by a factor of two for See this url: http://imageshack.us.../induction.png/ Hope this answers your question. Edited January 8, 2012 by bomaha 1 Reply Link to post Share on other sites More sharing options...
Procrastination Posted January 13, 2012 Report Share Posted January 13, 2012 Hey guys could you please help me with this: Solve the following inequality (Solve for x): x +(x-1/a+1) > (x+1/a+1) - ax I already developed it and got "x > -ax" but I don't think it is correct. For what value(s) of p does |(3x/2) - 7| </= p -3 have no solutions? Help would be appreciated Reply Link to post Share on other sites More sharing options...
CkyBlue Posted January 13, 2012 Report Share Posted January 13, 2012 recheck your work. 1) x +[(x-1)/(a+1)] > [(x+1)/(a+1)] - ax x+[(x-1)/(a+1)]-[(x+1)/(a+1)]>-ax x+(-2/a+1)>-ax (-2/a+1)>-ax-x 2/(a+1)>ax+x 2/(a+1)>x(a+1) 2/(a+1)^2>x If that's right, for all values of a, that equation should hold true. Sometimes that's the simplest it gets; you can't always have a numerical answer 2) This is what you do, I'm pretty sure... (for the sake of convenience I'll just use <) |(3x/2)-7|<p-3 |1.5x-4|<p Draw this graph, and figure out the range. In this case, p cannot be less than 0. For reference, the cusp is at x=2+2/3 Someone check me? 1 Reply Link to post Share on other sites More sharing options...
Procrastination Posted January 13, 2012 Report Share Posted January 13, 2012 Ohh I just realized I didn't changed one sign in my calculations, that's why I got it all wrong. Anyways, thank you Reply Link to post Share on other sites More sharing options...
itty.bitty.kitty Posted February 2, 2012 Report Share Posted February 2, 2012 consider the equation: 2xy^2=x^2y+3find y when x=1 and y>0im using implicit derivative but i cant seem to find the answer...if someone were to answer by tonight it would be much appreciated Reply Link to post Share on other sites More sharing options...
kiwi.at.heart Posted February 2, 2012 Report Share Posted February 2, 2012 (edited) consider the equation: 2xy^2=x^2y+3find y when x=1 and y>0im using implicit derivative but i cant seem to find the answer...if someone were to answer by tonight it would be much appreciatedI might be looking at this way to simply but to me it looks like a simple quadratic equation.Sub in x=1 gets you:2y^2=y+3 --> 2y^2-y-3=0When that is factorised you end up with (2y-3)(y+1)=0Solved you get 1.5 and -1 and since y is greater than 0 the answer is 1.5 Edited February 2, 2012 by kiwi.at.heart Reply Link to post Share on other sites More sharing options...
itty.bitty.kitty Posted February 2, 2012 Report Share Posted February 2, 2012 oh wow thank you....i dont know why i didnt see that and im sorry i made a mistake its y<0 so the answers -1 but thank you Reply Link to post Share on other sites More sharing options...
itty.bitty.kitty Posted February 17, 2012 Report Share Posted February 17, 2012 I think i am the only one who has questions.......the shaded region is bounded by f(x)=x1/2, x=a and the x-axisthe shaded region is revolved around the x-axis through 360 degrees and the volume formed is .845 pi find the value of a...sorry i cant show the graph i have no idea how to show it Reply Link to post Share on other sites More sharing options...
Drake Glau Posted February 17, 2012 Report Share Posted February 17, 2012 (edited) Ok, so this makes a little bullet like shape.Now picture this as an "infinite" amount of thin discs forming the whole.Each disc's "volume" would be the area of the circle where f(x)=radius. so you have the SUM of an infinite amount of pi(x1/2)2 which is just pi*xNow to sum all these discs together is just the integral of x(pi) dx which is x2pi/2The bounds are from 0 to aThis means that [a2pi/2]-[02pi/2]=0.845piThat second part is just zero so,a2pi/2=0.845pia2pi=1.690pia2=1.690a=sqrt(1.690)=1.3In general terms, when you are given a function rotating around the x axis it turns out to be the integral of [f(x)]2pi with whatever bounds you need. This is also if there is only one function, if there is two the little "discs" now have holes in them and complicates the integral part a little more. Edited February 17, 2012 by Drake Glau Reply Link to post Share on other sites More sharing options...
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