# First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

##### Kosovo is Serbian

why are you still doing matricies
lol what they did in 1131 was just adding/subtracting multiplying them, here they learn vector spaces, basis etc and eigenvalues/vectors

#### Flop21

##### Well-Known Member

why are you still doing matricies
I believe this topic is called "linear combinations and spans" or the overall topic is "vector spaces".

#### turntaker

##### Well-Known Member

are you doing things like intersection of lines, planes etc

#### Flop21

##### Well-Known Member

are you doing things like intersection of lines, planes etc
we're doing things like vector spaces (e.g. show that the set is vector space), subspaces (e.g. show that the line segment defined by blah is not a subspace of R3 or find distinct members of the set blah). And I guess we are coming to a point involving matrices in sets or subspaces or whatever.

Soz don't really know wtf I'm talking about at this point lol.

#### turntaker

##### Well-Known Member

we're doing things like vector spaces (e.g. show that the set is vector space), subspaces (e.g. show that the line segment defined by blah is not a subspace of R3 or find distinct members of the set blah). And I guess we are coming to a point involving matrices in sets or subspaces or whatever.

Soz don't really know wtf I'm talking about at this point lol.
Nvm I was thinking about vectors not matricies. But the two are connected somehow.

#### InteGrand

##### Well-Known Member

I'm stuck on a similar one

How do I find this vector???

You essentially have two conditions:

x -(1/4)y + 0z = 0

and

0x -(3/4)y + z = 0

So the vector (x,y,z) is in the column space of A iff it satisfies those two conditions.

You can turn those conditions into a matrix and get it into row-echelon form (actually in this case, it already is in row-echelon form).

Then do the usual procedure of setting a non-leading column's variable to a free parameter (in this case, that variable is z, so set z = lambda say), then use back substitution as usual to get x and y in terms of lambda.

This will mean you'll end up with x, y, z in terms of lambda, which means you can get a vector b as desired. For the sake of example, if you ended up with x = 2lambda, y = -lambda, z = lambda, we'd have (x,y,z) = (2lambda, -lambda, lambda) = lambda (2,-1,1), and thus a vector we could choose for b would be (2,-1,1).

#### leehuan

##### Well-Known Member

Nvm I was thinking about vectors not matricies. But the two are connected somehow.
In a way a matrix is just a ton of vectors smacked side by side

#### leehuan

##### Well-Known Member

Am I doing something wrong with algebra or do I actually need to go for partial fractions? (Please don't complete the question)

$\bg_white Let y=xv \implies \frac{dy}{dx}=x\frac{dv}{dx}+v$

\bg_white \begin{align*}\frac{dy}{dx}+\frac{2xy}{x^2+y^2}&=0\\ x\frac{dv}{dx}+v+\frac{2x^2v}{x^2+x^2v^2}&=0\\ x\frac{dv}{dx}&=-v-\frac{2v}{1+v^2}\\ &=v\left(\frac{3+v^2}{1+v^2}\right)\\ \frac{dv(1+v^2)}{v(3+v^2)}&=-\frac{dx}{x}\end{align*}

#### InteGrand

##### Well-Known Member

Am I doing something wrong with algebra or do I actually need to go for partial fractions? (Please don't complete the question)

$\bg_white Let y=xv \implies \frac{dy}{dx}=x\frac{dv}{dx}+v$

\bg_white \begin{align*}\frac{dy}{dx}+\frac{2xy}{x^2+y^2}&=0\\ x\frac{dv}{dx}+v+\frac{2x^2v}{x^2+x^2v^2}&=0\\ x\frac{dv}{dx}&=-v-\frac{2v}{1+v^2}\\ &=v\left(\frac{3+v^2}{1+v^2}\right)\\ \frac{dv(1+v^2)}{v(3+v^2)}&=-\frac{dx}{x}\end{align*}
I think you made a simplification error in getting the second last line.

#### leehuan

##### Well-Known Member

I think you made a simplification error in getting the second last line.
Oh my bad I dropped a negative in that line which I later reintroduced. But wait

Factoring -v out I have

1 + 2/(1+v^2) = (1+v^2+2)/(1+v^2) = (3+v^2)/(1+v^2) right?

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#### InteGrand

##### Well-Known Member

Oh my bad I dropped a negative in that line which I later reintroduced. But wait

Factoring -v out I have

1 + 2/(1+v^2) = (1+v^2+2)/(1+v^2) = (3+v^2)/(1+v^2) right?

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Correct!

#### leehuan

##### Well-Known Member

But then.. Damn so I do have to p.f. it don't I haha

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#### InteGrand

##### Well-Known Member

But then.. Damn so I do have to p.f. it don't I haha

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Yeah.

#### Flop21

##### Well-Known Member

I've got a formula here for the normal vector to the surface:

<Fx(xo,yo), Fy(xo,yo), -1>

but then somewhere else I see them using this:

<-Fx(xo,yo), -Fy(xo,yo), 1>

Which one is correct?

#### InteGrand

##### Well-Known Member

I've got a formula here for the normal vector to the surface:

<Fx(xo,yo), Fy(xo,yo), -1>

but then somewhere else I see them using this:

<-Fx(xo,yo), -Fy(xo,yo), 1>

Which one is correct?
We can use either for the normal, because those two vectors you've written are just negatives of each other, which means they are both valid normals (remember, two vectors that are just multiples of each other will both be valid direction vectors for a line).

#### Flop21

##### Well-Known Member

I'm not sure what they want me to enter here. All results I get decimals, after finding the derivative of that integral.

#### leehuan

##### Well-Known Member

I'm not sure what they want me to enter here. All results I get decimals, after finding the derivative of that integral.

$\bg_white \text{Si}^\prime(x)=\frac{\sin x}{x}\text{ by FTC1}\\ \text{So let }x=3.\\ \text{Have you tried typing it in exact?}$

#### leehuan

##### Well-Known Member

A bit lost in how to incorporate the nominal rate here. Need help with setting up the relevant ODE. I have a formula to find the effective rate for continuous compounding but it's taught in ACTL and not MATH.

$\bg_white A savings account is opened with a deposit of A dollars. At any time t years thereafter, money is being continuously deposited into the account at a rate of (C + Dt) dollars per year. If interest is being paid into the account at a nominal rate of 100R\% per year, compounded continuously, find the balance B(t) dollars in the account after t years.$

#### InteGrand

##### Well-Known Member

A bit lost in how to incorporate the nominal rate here. Need help with setting up the relevant ODE. I have a formula to find the effective rate for continuous compounding but it's taught in ACTL and not MATH.

$\bg_white A savings account is opened with a deposit of A dollars. At any time t years thereafter, money is being continuously deposited into the account at a rate of (C + Dt) dollars per year. If interest is being paid into the account at a nominal rate of 100R\% per year, compounded continuously, find the balance B(t) dollars in the account after t years.$
$\bg_white \noindent At any time t, the change in B comes from the instantaneous deposit rate of C+Dt, and from the continuous compounding, which means instantaneous rate of increase in B due to interest added as RB(t). So the balance satisfies B^\prime (t) = C +Dt + RB(t) (since these are the two contributions to the instantaneous rate of increase). This is the ODE to solve. This is a linear first order ODE and can be solved via integrating factors. The initial condition given is B(0) = A, which will allow us to find the arbitrary constant that will come up.$

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#### leehuan

##### Well-Known Member
$\bg_white \frac{x\frac{dy}{dx}-y}{x^2}=0$