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

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?

**glenn101**- 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.

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?

**glenn101**- 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.

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

cos(y'')=1?

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

**glenn101**- 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

(d^2y)^2

(------) = 1

(dx^2)

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

**glenn101**- 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.

Thanks,

Glenn101.

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.

**glenn101**- 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

2sinxcosx=sinx

divide both sides by 2sinx

cosx=sinx/2sinx

cosx=1/2

x= pi/3 or 5pi/3

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

**glenn101**- Replies: 3

Anyone else get there mid-year exam results?

Got an A for physics VERY HAPPY!!!:D

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

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

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

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

x=-2

is your asymptote:)

**glenn101**- 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.

dP/dt=kP

dt/dP=1/kP

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?

tk-c=loge|P|

e^(tk-c)=P

Alternatively I tried this approach;

t=1/k∫(1/P) dP

t=1/kloge|P|+c

t-c=1/kloge|P|

k(t-c)=loge|P|

e^k(t-c)=P

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.

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

Fscales=750N

W=686N

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

F=ma

750=70a

a=10.71ms^-2

oh and MASS NEVER CHANGES!

sin(2x)

------------- = tan(x)?

cos(2x) +1

**glenn101**- 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;

-2sin(2x)

------------- =-tan(x)?

2+2sin(2x)

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?:|

oh my god, I'm sorry I integrated!

I've been doing integration all day, sorry!

I also did Volumes of solids of revolutions before, fascinating stuff there.

I'm doing Differential Equations tomorrow

Hi Gercel and Bobbym,

I believe she wants to know how to evaluate

y=2x^2-3x-1/x^3

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)

---------------------

v^2

= -3x^2

----------

x^6

= -3/x^4

and then you just combine that and the other terms using the power rule and you get;

4x-3-3/x^4

there we go, again sorry for that

check by the power rule:

= 4x-3-3/x^4

interesting method there bobbym

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.