kingjewel1

hmm

If i remember rightly, it was that it doesn't have any calculable area or something like that. If you actually try to integrate it: [math]y=x^x[/math] [math]logy=xlogx[/math] therefore y=e^(xlogx) If we used integration by parts... [math]integrate 1.e^(xlogx)[/math] u=e^(xlogx) dv=1 It just gets more and more complicated... ie du=x^x(logx+1)dx; v=x [math]x.e^(xlogx)-integrate e^(xlogx)(logx+1)dx[/math] Thats not going anywhere! so x^(x+1)-int x^x.(LNx+1) just as bad.... sorry for the poor use of LaTeX, as i'm just figuring it out

Why is it not intergratable though? If you drew it, it would have a meaningful area between x=0 and x=5? would it not? I do remember a thread on x^x a few weeks back, but i don't remember what was said exactly. Cheers guys.

Hi there. I'm having problems integrating [math]y=x^x[/math] Any help would be appreciated.

Ah that looks good! I did AB.OP=0, P=x,y,z (-4*x)+(-4*y)+(8*z)=0 -24+96t=0 t=1/4 if D is twice away as P then t=1/2 therefore 3i-3j+3k Thank you Atheist

Yes my mistake, i took the equation from |AD| i should have done [math]x=5-4t[/math] [math]y=-1-4t[/math] [math]z=-1-8t[/math] this gives me t=1/2 ^t=0 thanks For the next method, I used [math]A (5i-j-k) B(i-5j+7k)[/math] // a_par to AB [math]r=5i-j-k+t(i-5j+7k)[/math] a_perp A.(x,y,z)=0 B.(x,y,z)=0 if z=1 then [math]r=1/2i+2/3j+k[/math] would these be correct? if sqrt27=a_perp-a_par how do i solve if r is in terms of t?

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Thank you very much Atheist. It helped me greatly. so i did [math]|OA|=sqrt(5^2+1^2+1^2)[/math] since |OD|=|OA| from above x,y,z [math]|OD|^2=(x-5)^2+(y+1)^2+(z+1)^2[/math] [math]27=16t^2+16t^2+64t^2[/math] [math]t=sqrt(9/32)[/math] [math]5-4t=2.87[/math] [math]-1-4t=-3.12[/math] [math]-1+8t=3.24[/math] so therefore [math]d=3i-3j+3k[/math] i must have made a mistake because i didn't think they were supposed to be decimals :S Your second method is interesting. i'll post my workings. Cheers

Hi guys I'm on my vectors again. Last time you really helped me understand what i was doing! Refered to a fixed origin A=(5i-j-k) B=(i-5j+7k) Find the position vector of the point D, where Dis not A on AB such that |OD|=|OA| i did OD= xi+yj+zk from r=a+t(B-A) x=5-4t y=-1-4t z=-1+8t |AD|^2=(x-5)^2+(y+1)^2+(z+1)^2 |AD|^2=(5-4t-5)^2+(-1-4t+1)^2+(-1+8t+1)^2 d(|AD|^2)/dt=(32+32+128)t am i using the right method? i also did Cos(OAOD) but got silly decimals. for the vector eq. Thanks in advance!

Okay I understand. Thank you I hadn't thought of it in that way, that's why. Vectors are very badly explained in all the books i have. Funnily enough, my exam paper will have 8 questions on 6 topics, and invariably vectors is the one holding the most points. Now that you have enlightened me i see how the methods are really quite simple. It's just knowing how to apply them to other questions. Would you suggest a sourse ie book with similar vector methods for me ? Thanks again

How did you come about substituting 0 and 1 into C(x)?

for d) I did |AP|:|PB| sqrt3^2+5^2+4^2=sqrt50 sqrt6^2+10^2+8^2=sqrt200 1:2

Okay. I used 1/2|AxB|= det u=|i, j, k| |1,- 5,- 7| |10, 10, 5| 1/2sqrt(45^2+75^2+60^2)=53.0 u^2 correct or A.B=|A||B|COSf 10-50-35=(sqrt75)(sqrt225)COSf f=54.77 1/2sqrt75.15.sinf =53.03 correct I like the cross product, it seems much cleaner and more efficient

Thank you for clarifying that Atheist. For the area of the triangle i can use two methods? a) cross product of modulus AOxOB b)cosine rule Which one would you prefer to use?

Hi there guys, I'm 18 and have ve been accepted by Imperial in the UK, the Hague, Amsterdam, and Utrecht to do business and economics. My dilema is this: my heart lies in maths, but I can't apply to any UK universities to do maths now, until next year. What should I do? Stay another year and reapply then? I speak german, french n spanish well enough to do maths in any of those languages, so maybe that's an option? How easy is it to switch courses at university (ie immediately)? Thanks, i'd appreciate any advise.

Thank you! I didn't know that it was possible to add together the different x,y,z terms of the column matrix:s ie (9).(1+9t)=0 (15).(-5+15t)=0 (12).(-7+12t)=0 i had previously worked out t for each row, hence getting the wrong answer. so we added ALL like terms and worked out the t. we then substitute it into the initial equation for AB and get 4i+0j-3k Thank you guys!!! I didn't realise it was actually an intersection problem.

So now that we have t=-1 as you said, i should now substitute it into the r equation i-5j-7k-(9i+15j+12k) =-8i-20j-19k ? 4i-3k is the right answer according to the back of the book

Ah thanks for spotting the error

Thank you both very much for your help. although i am familiar with the dot product (which is what i tried to apply), i fail to see how it yields an appropriate answer. ie. (B-A)*P=0 since (9,-15,-12).(1-9t,-5-15t,-7-12t)=0 thus giving t=-1 for i, and other answers for j and k. namely 1/3 and 7/12. Oder habe ich etwas nicht kapiert? Thanks

Hi, i'm having problems with this question and would appreciate it very much if someone would help me out. I've got my C4 exam in a month and none of my teachers has a clue about vectors... Referred to the origin we have 2 points points A (i-5j-7k) and B (10i+10j+5k). P is a point on AB a) Find a vectr for the line passing through AB(i got i-5j-7k+t(9i-15j-12k)) b)Find the position vector of P such that OP is perpendicular to AB. I really don't know how to do b). c) find area of triangle OAB ( i can do when i have b) d) find ratio in which P divides AB (not sure) Can anyone tell me where i can get a good guide to vectors (book or internet, etc)? It's my weakest topic Cheers guys, Phi

thanks very much for your help!

thanks for your help!

cheers! thanks a lot! my problem was that i didn't know if the way i was trying to solve the function was the correct one. thanks again

Hi there guys! this one's interesting but got me stumped. How would i show that the values of x at which sinxe^(-sinx) has stationary points form an arithmetic sequence? Thanks!

Ok i understand now. so if we were to work it through in the same manner: 1+2sin^2x/cos^2x=-cosx 2sin^2x+cos^2x=-cos^3x 2(-cos^2x+1)=+cos^2x=-cos^3x 2-2cos^2x=-cos^3x 2-cos^2x=-cos^3x and where do i go from here?