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#1 Re: Help Me ! » help please am sooooooo confused :(( » 2011-05-17 22:08:55

No worries bob:) I was just getting concerned for a second there haha smile Glad to help smile

#2 Re: Help Me ! » help please am sooooooo confused :(( » 2011-05-17 20:57:39

Sorry to interrupt but wouldn't the graph be y=4?
it passes through (-3,4) and (2,4) if this means (x,y).
As the y coordinate is repeated for 2 different x-values and we know it is a line why is it y=2?

#3 Re: Help Me ! » Why doesn't n-tuple notation work for curvilinear coordinates? » 2011-05-16 23:12:43

I think I get it now, is it because that notation would be ambiguous as to which coordinate system we were using if we allowed spherical and cylindrical coordinates to use the same notation?

#4 Help Me ! » Why doesn't n-tuple notation work for curvilinear coordinates? » 2011-05-16 20:43:30

Replies: 2

By N-Tuple notation I mean representing the standard basis vectors i,j,k as (1,1,1) for R=i+j+k for example.

#5 Re: Help Me ! » Why is it that {R:r=1} where R is a position vector resembles a circle » 2011-05-16 19:50:08

Thanks so much for the reply Bob,
So if we had {R:r<=1}
We would have position vectors where the end of each vector resembles a point inside the circle?

i.e. the lengths don't count as points only the ends of the vectors give us points?

#6 Help Me ! » Why is it that {R:r=1} where R is a position vector resembles a circle » 2011-05-16 15:26:28

Replies: 3

The position vector meaning the tail of the vector is at the origin. And this is in R^2 with plane polar coordinates.

I thought that the sketch of this set would resemble a set of position vectors with magnitude 1 extending out to a radius of 1 which trace out to reach a point which would lie on a circle of radius 1, but because of the lengths of each of the position vectors, wouldn't the region be a filled in circle(a disc)? Why do we neglect the lengths?

Any help would be much appreciated, this idea has confused me for a while.

#7 Re: Help Me ! » non-linear second order DE vs linear second order DE » 2010-09-04 14:03:36

Hi Bobbym,
I've been quite well, I'm in second semester of 1st year uni now, the workload is quite tough but I'm enjoying it nonetheless.
How are you?

And thanks for the reply, I can see that the expression can be simplified to a linear nature, however, how about


Would that be classified as non-linear? It seems to me that it would be.

#8 Help Me ! » non-linear second order DE vs linear second order DE » 2010-09-04 12:48:36

Replies: 3

Hi all,
Hopefully a quick question.
From my understanding a DE is linear, of any order, as long as the independent variable and all its derivatives are linear.
So does this mean that

(------)       = 1

Would be classifed as non-linear? As y's second derivative is of a non-linear nature?

#9 Help Me ! » Seperable Differential Equations Conceptual Help » 2010-08-13 13:30:46

Replies: 2

Hey all,
I'm having difficulty understanding several concepts with regards to solutions of seperable differential equations.

The S.D.E I am talking about is of the form f'(x)=f(y)*f(x).

From my understanding, a solution of an initial value problem must satisfy BOTH the D.E and the initial conditions given in the stated problem, and this is where my confusion begins.

Apparently, a constant solution satisfies f(y)=0 and if the initial conditions y-value matches the constant solution found, that is your solution to your S.D.E initial value problem.

Ok so for an example f'(x)=3*x*y-6*x with y=2 when x=1.
f'(x) = 3*x(y-2)
so then f(y) = y-2
which means that f(y)=0 for y=2.

now, I understand that when x=1, y=2 will be satisfied by y=2 but what I don't get is how on earth y=2 satisfies the D.E? as f'(x)=0 not 3*x(y-2).
How can it be called a constant 'solution' if it only satisfied the initial conditions?

This follows onto my next question, if I changed the initial conditions of the previous question to say y=1 when x=1
why can we then say that the solution must lie beneath the line y=2? I've heard that solutions can't cross, but if by solutions they mean solutions to IVP's well then, how can y=2 be a constant solution in this case? it doesn't satisfy the initial conditions OR the D.E.

I know also, that dividing by zero comes into play when trying to seperate the D.E. because if y satisfies the initial condition while also the constant solution, you would be dividing by zero. But I'm still quite unclear with all this, any help would be greatly appreciated.


#10 Re: Help Me ! » trig identity issue » 2009-09-13 19:12:17

Thanks for the reply bobbym:)
thats interesting your example with x^2-2x=0
because technically you ca divide both sides by zero which will make x=2.
So why is it that this results in missing solutions though? You've got me interested with this now.

#11 Help Me ! » trig identity issue » 2009-09-12 14:23:21

Replies: 3

Hi all just one hopefully quick question which is really bothering me

Solve sin2x=sinx over [0,2pi]

sin2x=2sinxcosx from double angle formula

divide both sides by 2sinx
x= pi/3 or 5pi/3

this is wrong however, what is it that is incorrect?

#12 Dark Discussions at Cafe Infinity » VCE mid-year results » 2009-08-02 22:29:21

Replies: 3

Anyone else get there mid-year exam results?
Got an A for physics VERY HAPPY!!!:D

#13 Re: Help Me ! » Integration Application: Volume generated by area bounded by two curve » 2009-07-18 17:14:31

Hi cathelyn13,
I haven't encountered this specific type of integral volumes/areas stuff before but I'm assuming it works similar to the way I would go about it.
I would firstly work out the areas between the two curves
so using this formula;
Area∫ [f(x)]² – [g(x)]² dx
Where f(x) is the greater area.
and then find the volume of this
Volume rotated about y=pi ∫ x² dy

#14 Re: Help Me ! » A Kinematics Question » 2009-07-18 17:08:53

Just starting physics? it's great stuff, you'll enjoy it

#15 Re: Help Me ! » c constant with applications of differential equations » 2009-07-18 17:02:32

luca-deltodesco wrote:

the two equations are exactly the same in the end; both C and K are both constants, so the product CK is also a constant.

thus e^k(t-c) is essentialy exactly the same as e^(tk-c), only that c has a different value because it is not multiplied by k

You could say that your teachers solution is simplified.


In the same way, you might end up with an integration where when you manipulate it you end with a constant like:

I(x) = f(x) + 4.5C

which ofcourse, you can equally just write as

I(x) + f(x) + C'

the C is written with the apostrophe here just to explicitly show that it is not the same constant as the first equation, but both are correct

Thanks for the reply luca,
it seems you are right that they end to be the same, however, I discovered that plugging values into e^(tk-c) and e^k(t-c) gave different results, they only came out to be the same when I solved for P after finding values for k and c. Any ideas?

for e^k(t-c) I have found that k= 1/2loge(11/10) and c= -6loge(10)/loge(11/10)
for e^(tk-c) I have found k=1/2loge(11/10) and c=-3loge(10)
actually never mind your absolutely right.
Thanks ever so much:D

#16 Re: Help Me ! » Help fiding Vertical Asymptote » 2009-07-18 16:23:19

makaros wrote:

Hi, I need help in finding vertical asymptote in particular question:

f(x)= 2x²-5/x+2, x≠ -2
a. Show that x= -2 is a verical asymptote.

Please explain how to find it..

Thank you

Hi Makaros,

vertical asymptotes appear when the denominator of a fraction = 0
so x+2=0
is your asymptote:)

#17 Help Me ! » c constant with applications of differential equations » 2009-07-17 20:34:06

Replies: 2

Hi all, I'm getting frustrated with dealing with applications of differential equations.
With regards to when to add in the c constant, and how you are to go about manipulating it.
Take for example this applications question;

A city with population P, at time t years after a certain date, has a population which increases at a rate proportional to the population at that time.
a) i) Set up a DE to describe this situation;
Let P= Population
    t= time (years)
dP/dt proportional to p
hence Dp/dt=kP where k>0.

ii) Solve to obtain a general solution.
t=∫1/kP dP
this is where the problem arises;I want to solve the integral for P which will help with later questions.
My teacher goes about it this way.

t= 1/k integral(1/P) dP
tk= loge|P|+c      oh and if you solve for t here would you get; t=1/kloge|P|+c/k?

Alternatively I tried this approach;
t=1/k∫(1/P) dP
which is correct? and why?

I have solved for P because the next question asks to find the population after x years given other data.

#18 Re: Help Me ! » Physics » 2009-07-08 13:22:07

Hi careless25,
the way I go about motion questions is to first list my data; Oh and is this on Earth? or we don't know?
m= 70kg
Now for weight we I don't like to write F=ma
I instead write W=mg
where W= weight of object
          g= gravitational field strength at his location
and hence W=70(9.8) or (10)
                   = 686N and that is his actual weight.
The scales read differently as he "jumped" on it however 686N is still his weight.
and so;
Weight = 686N
a should be= 9.8ms^-2
but since the scales read greater than his weight, his acceleration read by the scales as you have recalled would be


#19 Re: Help Me ! » How does the identity sinx/cosx work? » 2009-07-08 12:57:34

oh sorry, your quite right, cos was supposed to be on the denominator. And so how is it, that this identity works
------------- = tan(x)?
cos(2x) +1

#20 Help Me ! » How does the identity sinx/cosx work? » 2009-07-04 14:09:04

Replies: 4

Ok I understand that sinx/cosx=tanx
and hence sin2x/cos2x=tan2x
but I can't understand how something I came across the other day works with this identity, somehow;

         ------------- =-tan(x)?
I get where the negative is coming from, that's obvious, but I would have thought it would have been something like tan(2x)?

Can someone please explain?:|

#21 Re: Help Me ! » HELP Find a Derivative of a Square Root Function » 2009-07-03 19:12:56

oh my god, I'm sorry I integrated!
I've been doing integration all day, sorry!

#22 Re: Help Me ! » Integration » 2009-07-03 19:09:47

Haha, I might:P
I also did Volumes of solids of revolutions before, fascinating stuff there.
I'm doing Differential Equations tomorrowup

#23 Re: Help Me ! » HELP Find a Derivative of a Square Root Function » 2009-07-03 18:56:24

Hi Gercel and Bobbym,
I believe she wants to know how to evaluate
Using both the product and quotient rule methods.
I'll show you the quotient rule way;

Ok so we see some terms where we can apply the power rule, oh and by the way this all can be done by using the power rule on each term, anyway as for the quotient.

For the term 1/x^3
Let u =1, v=x^3
     du/dx= 0,   dv/dx= 3x^2.
dy/dx= v(du/dx)-u(dv/dx)
         = -3x^2
        =     -3/x^4
and then you just combine that and the other terms using the power rule and you get;
there we go, again sorry for that

check by the power rule:
= 4x-3-3/x^4

#24 Re: Help Me ! » log help » 2009-07-03 18:45:49

interesting method there bobbym

#25 Re: Help Me ! » Integration » 2009-07-03 18:41:23

Ah, I see it now, thank for showing that.
And I'm glad you noticed me trying some of those substitutions, I still need the practise, so I thought I'd see if there are any on this site. And don't worry I've been thoroughly practising these with my textbook too.

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