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#26 Re: Help Me ! » Convergent binomial expansions » 2007-07-04 09:58:34
I've looked at how to find the sum of an infinite geometric series, but that's about it. I guess i'll have to look up what you said about. Thanks.
#27 Re: Help Me ! » integration » 2007-07-04 09:56:27
Indeed, but since he gave no limits, I thought that much was obvious.
#28 Re: Help Me ! » integration » 2007-07-04 06:05:48
I assume by solve you mean find the integral? In that case it's:
x³ - 4x² + 6x + C
#29 Help Me ! » Convergent binomial expansions » 2007-07-03 06:37:21
- yonski
- Replies: 3
I'm learning about binomial expansions from my maths book but I don't really understand how you find the range of values of x for which the expansion is valid.
For example, take the expansion of √(1-3x). I calculate the first four terms to be:
1 - (3/2)x - (9/8)x² - (27/16)x³ + ...
Now because you're not raising the bracket to a positive integer, no coefficient will ever be equal to zero, so it goes on infinitely. I get that bit.
Then the book says that the expansion will only be convergent when modulus(x) < 1/3 . This bit i don't get.
Any help would be much appreciated!
#30 Re: Help Me ! » Differentiation- gradients » 2007-06-11 06:59:45
I thought that much was obvious.
#31 Re: Help Me ! » Differentiation- gradients » 2007-06-10 19:54:54
I think you're getting a bit confused with the differentiation. To differentiate a linear function like that you need to take the coefficient of x (in this case -1). Remember that the equation for a straight line is y = mx + c , where m is the gradient. When you differentiate you're looking for the sort of 'gradient function', so this should help to remind you.
Once you've done that, the second part of the question is easy. Since it's a straight line graph, the gradient will be the same at every point, i.e. -1.
Hope that helps.
#32 Re: Help Me ! » help? » 2007-06-07 11:15:43
No, think about the answer you're giving. How could the probability of them waiting more than 1 hour be 0.90332? This would mean that over 90% of people wait longer than an hour, but you've been told that the mean is 45 minutes!!! 
You need to go:
(X > 60) = P(Z > (60-45)/12)
= P(Z > 1.25) you were right up to this point
= 1 - P(Z < 1.25)
= 1 - 0.8944
= 0.1056
Always check that the answer is sensible lol.
#33 Re: Help Me ! » help? » 2007-06-07 08:36:51
Nah, you've got the right formula but you've used it wrongly (and put in the the wrong value for X). It's tough to really explain what the Z values represent without using diagrams, so I suggest you ask your teacher or whoever to go over it cos you're fumbling in the dark a bit at the moment (no offence).
Basically this is how you do the first question:
P(X < 20) = P(Z < (20-45)/12) using the formula you quoted
= P(Z < -2.08)
= 1 - P(Z < 2.08) using the fact the the normal distribution is symmetrical
= 1 - 0.9812 look up Z=2.08 in your table and find the corresponding value of theta(Z)
= 0.0188
and that's your answer 
#34 Re: Help Me ! » help? » 2007-06-07 06:22:25
We've just been doing this at school, and we always use a table of Z values to calculate this type of problem. Have you done anything with those? If you want me to go through it with that method i can do.
#35 Re: Help Me ! » Solutions to quadratic equations when graphs are given.. » 2007-06-07 04:56:19
I just think of the shape of the quadratic curve, so x^2 curves are 'valley shaped' and -(x)^2 curves are 'hill shaped'.
If you're looking to solve say x^2 + 5x + 6 < 0 , then you first factorise to give (x+3)(x+2) < 0 .
Now you know that the curve is valley shaped, and you're looking for the part of this that's below the x-axis. The factorised equation above gives the points that the curve crosses the x-axis, so the curve will be below the x-axis for all x values between these points. Therefore the solution is -3<x<-2 .
Say that instead of this you were looking for x^2 + 5x + 6 > 0 . Once again this factorises to give (x+3)(x+2) > 0 .
Now the inequality states that you're looking for the parts of the curve which are above the x-axis. So once again imagine the shape of the curve cutting the x-axis. This time it'll be > 0 when x<-3 or x>-2 .
Hopefully that helps somewhat, but the best way to learn is to keep doing lots of questions until it's drilled into your head!
#36 Help Me ! » Mechanics problem » 2007-06-07 03:31:11
- yonski
- Replies: 1
Hi there,
i had a mechanics exam this morning and i was struggling with this questions (the numbers were maybe a bit different, but it was basically the same). I'm worried that i've messed it up, so was wondering whether someone could go over my working please to put my mind at rest?
Okay, so you have 2 particles attached to a taut string which passes over a smooth pulley, as in the diagram below:
Particle A rests on the slope, and the coefficient of friction between the slope and the particle is 4/7 . Also, tan(x) = 3/4 .
(a) The system is released from rest and the two particles move a distance h (A does not reach the pulley and B does not reach the floor). Calculate the potential energy lost by the system as a whole when the particles have moved a distance h.
(b) When the particles have moved distance d, their speed is v ms/s, where v^2 = kgh . Work out the value of k.
My solutions:
(a) In moving a distance h up the slope, A gains potential energy (3/5)mgh joules.
In moving downwards a distance h, B loses potential energy 2mgh joules.
Therefore net energy loss is: 2mgh - (3/5)mgh = (7/5)mgh joules.
(b) Some of this potential energy lost goes towards work done against friction, and some gets converted into kinetic energy.
Work done against friction = (4/7)mghcos(x) = (16/35)mgh joules.
Therefore kinetic energy = (7/5)mgh - (16/35)mgh = (33/35)mgh joules.
Hence the speed of the particles is v, where
0.5*3mv^2 = (33/35)mgh (This is the bit i'm unsure about - do you take the mass as the sum of the two?)
(3/2)v^2 = (33/35)gh
v^2 = (22/35)gh .
Therefore k = 22/35.
Is that okay, or have i made some mistakes? Any help appreciated
#37 Help Me ! » Collisions problem » 2007-03-27 06:40:17
- yonski
- Replies: 4
Hi there,
having a bit of trouble with this question:
Two particles A and B move on a smooth horizontal table. The mass of A is m, and the mass of B is 4m. Initially A is moving with speed u when it collides with B, which is at rest on the table. As a result of the collision, the direction of motion of A is reversed. The coefficient of restitution between the particles is e.
(a) Find an expression for the speed of A and the speed of B immediately after the collision.
In the subsequent motion, B strikes a smooth vertical wall and rebounds. The wall is perpendicular to the direction of motion of B. The coefficient of restitution between the wall and B is 4/5. Given that there is a second collision between A and B,
(b) show that 1/4 < e < 9/16 .
Given that e = 1/2 ,
(c) find the kinetic energy lost in the collision between A and B.
Okay, so that's the question.
I've done part (a) and found that the speeds are
Va = (u/5)(4e-1)
Vb = (u/5)(1+e)
and then done part (b) too.
For part (c) though, I don't see how you can find the total energy lost, since you don't know the initial speed or the speeds after the collision. I found the percentage energy loss to be 60% but that's all I could figure out.
Any help appreciated.
Cheers, Jon.
#38 Help Me ! » Remainder Theorem » 2007-02-19 09:12:57
- yonski
- Replies: 1
Hi there,
how would i go about expressing the following improper fraction in 'mixed' number form, using the remainder theorem?...
(2x^2 + 4x + 5) / (X^2 - 1)
Would it be correct to write it as:
A(x^2 - 1) + (Bx + C)/(x^2 - 1)
And then find the values of A, B and C? I know it's something like that.
Thanks for any help!
Jon.
#39 Help Me ! » Logarithms » 2007-02-07 11:06:39
- yonski
- Replies: 1
How do i calculate the value of y for which 2log(base3)y - log(base3)(y+4) = 2 ?
It's the log(y+4) bit that confuses me... i remember my teacher telling me what to do with those, and i remember thinking 'oh yeah, why didn't i think of that', but i've forgotten and i still can't think ofi it!
Thanks, Jon.
#40 Re: Help Me ! » Electricity question » 2006-11-27 08:57:38
Hmm yes, i see your problem.
Having looked at it again, I think the way to look at it is using symmetry and some PD logic. If you divide that bit of the circuit up symmetrically as follows
Now the PD at each of the identically coloured junctions is going to be the same, so as far as electricity is concerned they are the same point. This means that you can look at the circuit as 4 sections, each with resistors in parallel, and these 4 sections are themselves connected in series. Hence we have
R1 = (1/10 + 1/10)-¹ = 5
R2 = (1/10 + 1/10 + 1/10 + 1/10)-¹ = 5/2
R3 = R2 = 5/2
R4 = R1 = 5
So the total resistance is 5 + 5/2 + 5/2 + 5 = 15 ohms. And that gives the current as 50/15 = 3.333 amps.
#41 Re: Help Me ! » Electricity question » 2006-11-21 11:07:48
Yeah, i was kind of looking at it the same way but couldn't figure out how to combine them like you have. It all seems logical now though. We kept answering all the worksheets really quickly in class so my physics teacher was like 'right, this one'll sort you out' lol. Tomorrow i'm going in with a big grin on my face Thanks mathsyperson.
P.S. You are a little bit awesome, maybe.
#42 Help Me ! » Electricity question » 2006-11-21 08:52:19
- yonski
- Replies: 6
Okay, so this is definitely not maths, but it's really bugging me and i thought someone round here might be able to help :-) I've been given the following diagram and been asked to find the total resistance and the current in the circuit...
Apologies for the rubbish drawing (the jagged lines are supposed to be resistors btw lol). Each resistor has a resistance of 10 ohms. I think the idea is to simplify the circuit into something that can be worked with more easily, but i'm not quite sure how.
Thank you!
#43 Re: Help Me ! » Equation of a straight line in vector form » 2006-10-27 08:59:41
The book says r = a + tb.
But i thought i was looking for the equation of the line which goes through point A and is parallel to b? But r does neither of these? That's what i'm confused about.
#44 Help Me ! » Equation of a straight line in vector form » 2006-10-26 03:27:10
- yonski
- Replies: 3
I've read through the section of my text book on this but i don't seem to understand it.
I'm looking for the equation of the line which goes through point A, with position vector a, and is parallel to the given vector b, right?
Since AP is parallel to b, AP = tb, where t is a scalar. Okay, i get that.
The vector b is the direction vector of the line. Right, okay.
Therefore the position vector r can be written as a+tb. Yep i see that.
Then it makes out as if r is the answer. But how is that the equation of the line which goes through point A and is parallel to b? It does neither!
Can anyone help explain it to me? 
Thanks!
#45 Re: Help Me ! » Differentiation Problem » 2006-10-22 07:34:03
What happens if Tom loves Maria, but Maria doesn't love him back?
#46 Re: Help Me ! » Differentiation Problem » 2006-10-20 07:09:06
Thanks, that's cleared those up. I was almost there with em, just required a little more thinking outside the box with the first one. So if the gradient is infinite/undefined, there's likely to be a division by zero in there?
#47 Help Me ! » Differentiation Problem » 2006-10-19 21:30:42
- yonski
- Replies: 6
Hi there,
just going through some differentiation exercises but i've found one or two i can't figure out...
1. (a) If (1+x)(2+y) = x^2 + y^2 , find dy/dx in terms of y and x.
(b) Show that there are two points at which the tangents to the curve are parallel to the y-axis.
Now i've done part (a) and found dy/dx = (2x-2-y)/(1+x-2y) . I think that's right, but how on earth do you do part (b)?!
And this is the second question...
2. Find the coordinates of the turning point on the curve y^3 + 3xy^2 - x^3 = 3 .
I thought it looked relatively simple - find dy/dx then set it as equal to zero. But when i do this i end up with
(x^2 - y^2)/(y+2x) = 0 , and i don't know where to go from there lol.
Thanks for any help!
Jon.
#48 Help Me ! » Partial Fractions » 2006-10-06 10:18:31
- yonski
- Replies: 2
Hi there,
what's the best way to convert improper fractions to partial fractions? I'm using this sort of method at the moment...
(4x^3 + 10x + 4) / x(2x + 1) = Ax + B + C/x + D/(2x + 1)
and then working out A, B, C, D etc by putting in values for x, or equating the coefficients. This usually works fine, until I came across this one
(x^3 + 1) / (x^2 + 1)
and suddenly my method seems to collapse. Is this a good method to use or is there a better one?
Thanks.
#49 Re: Help Me ! » A little motion problem » 2006-09-27 07:31:24
Okay, yeah, it makes sense when you put it like that. That is the right answer cos it's in the back of the book. I just couldn't understand why.
Thanks.
#50 Help Me ! » A little motion problem » 2006-09-27 06:37:16
- yonski
- Replies: 3
Hi there,
this is more of a physics question, but maybe someone out there can help.
"A body (not a dead body i hope) hangs from a spring-balance which is suspended from the ceiling of a lift. What is the mass of the body if the balance registers a reading of 70N when the lift has an upward acceleration of 4m/s^2? (g=10m/s^2)"
This has been bugging me all day but i can't seem to get the right answer. If anyone could run me through it i'd be very happy!
Thanks, Jon.