# Thread: First Year Mathematics B (Integration, Series, Discrete Maths & Modelling)

1. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by leehuan
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!

2. ## Re: MATH1231/1241/1251 SOS Thread

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

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3. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by leehuan

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

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Yeah.

4. ## Re: MATH1231/1241/1251 SOS Thread

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?

5. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by Flop21
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).

6. ## Re: MATH1231/1241/1251 SOS Thread

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

7. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by Flop21
I'm not sure what they want me to enter here. All results I get decimals, after finding the derivative of that integral.

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

8. ## Re: MATH1231/1241/1251 SOS Thread

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.

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

9. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by leehuan
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.

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

10. ## Re: MATH1231/1241/1251 SOS Thread

Lol they made us solve this differential equation by considering how it is an exact ODE. But the solution y=Ax just bugs me... is there a shorter way to doing this one haha

$\frac{x\frac{dy}{dx}-y}{x^2}=0$

11. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by leehuan
Lol they made us solve this differential equation by considering how it is an exact ODE. But the solution y=Ax just bugs me... is there a shorter way to doing this one haha

$\frac{x\frac{dy}{dx}-y}{x^2}=0$
Note the LHS is d/dx (y/x).

12. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by InteGrand
Note the LHS is d/dx (y/x).
Oops..........................

That moment when I saw it for this harder one but not the easier one
$e^x\sin y + e^x\cos y \frac{dy}{dx}=0$

13. ## Re: MATH1231/1241/1251 SOS Thread

divide by e^x cos(y) a bit of rearrange you should get -cot(y)dy=dx

14. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by leehuan
Oops..........................

That moment when I saw it for this harder one but not the easier one
$e^x\sin y + e^x\cos y \frac{dy}{dx}=0$
Reverse product?

15. ## Re: MATH1231/1241/1251 SOS Thread

divide by e^x cos(y) a bit of rearrange you should get -cot(y)dy=dx
I guess that works too.

16. ## Re: MATH1231/1241/1251 SOS Thread

Reverse product?
Yeah, I'm saying I missed reverse product for the first but saw it for the e^(xy) one
divide by e^x cos(y) a bit of rearrange you should get -cot(y)dy=dx
Final answer ends up with exp(xy)cos(x)=C though

17. ## Re: MATH1231/1241/1251 SOS Thread

This is boring and tiring. What is the giveaway as to what values of p and q I am supposed to pick.

$\\Show that the equation \\ (2y^2+4x^2y)dx+(4xy+3x^3)dy=0\\ can be turned into an exact equation by multiplying throughout by a function of the form x^py^q, where p and q are constants. Hence solve the equation.$

18. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by leehuan
This is boring and tiring. What is the giveaway as to what values of p and q I am supposed to pick.

$\\Show that the equation \\ (2y^2+4x^2y)dx+(4xy+3x^3)dy=0\\ can be turned into an exact equation by multiplying throughout by a function of the form x^py^q, where p and q are constants. Hence solve the equation.$
Remember that d/dz z^n = nz^(n-1), I think that should help you...

In this case, xy² ???

19. ## Re: MATH1231/1241/1251 SOS Thread

Remember that d/dz z^n = nz^(n-1), I think that should help you...
It's hard to juggle that when I have to consider two constants at a time and not just one

20. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by leehuan
This is boring and tiring. What is the giveaway as to what values of p and q I am supposed to pick.

$\\Show that the equation \\ (2y^2+4x^2y)dx+(4xy+3x^3)dy=0\\ can be turned into an exact equation by multiplying throughout by a function of the form x^py^q, where p and q are constants. Hence solve the equation.$
$\noindent Multiply through by x^p y^q to obtain \underbrace{\left(2x^p y^{2+q} + 4x^{2+p}y^{1+q}\right)}_{A(x,y)}\, \mathrm{d}x +\underbrace{\left(4x^{1+p}y^{1+q} + 3x^{3+p}y^q\right)}_{B(x,y)}\, \mathrm{d}y = 0.$

$\noindent Now, \frac{\partial A}{\partial y} = 2(2+q)x^p y^{1+q} + 4(1+q)x^{2+p} y^q and \frac{\partial B}{\partial x} = 4(1+p)x^{p} y^{1+q} + 3(3+p) x^{2+p}y^q. We want these two things to be equal for the ODE to be exact, so equating coefficients of like terms, we clearly will take 2(2+q) = 4(1+p) and 4(1+q) = 3(3+p). Solve this system of simultaneous equations and we'll have the required values of p and q. Once we have these values, the rest is just the usual tedium.$

21. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by leehuan
It's hard to juggle that when I have to consider two constants at a time and not just one
y is on the right, and there is a 3, so you want 3y². x is on the left and you have 4x² so you want 4x³. so xy² is a possibility...

22. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by InteGrand
$\noindent Multiply through by x^p y^q to obtain \underbrace{\left(2x^p y^{2+q} + 4x^{2+p}y^{1+q}\right)}_{A(x,y)}\, \mathrm{d}x +\underbrace{\left(4x^{1+p}y^{1+q} + 3x^{3+p}y^q\right)}_{B(x,y)}\, \mathrm{d}y = 0.$

$\noindent Now, \frac{\partial A}{\partial y} = 2(2+q)x^p y^{1+q} + 4(1+q)x^{2+p} y^q and \frac{\partial B}{\partial x} = 4(1+p)x^{p} y^{1+q} + 3(3+p) x^{2+p}y^q. We want these two things to be equal for the ODE to be exact, so equating coefficients of like terms, we clearly will take 2(2+q) = 4(1+p) and 4(1+q) = 3(3+p). Solve this system of simultaneous equations and we'll have the required values of p and q. Once we have these values, the rest is just the usual tedium.$
Andddd this is where intuition collides head on with mechanical rigour...

23. ## Re: MATH1231/1241/1251 SOS Thread

I can't identify the type of this one...

$x^2\frac{dy}{dx}-xy=y^2$

...unless I did my partial derivatives wrong

24. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by leehuan
I can't identify the type of this one...

$x^2\frac{dy}{dx}-xy=y^2$

...unless I did my partial derivatives wrong
make dy/dx the subject and you'll see.

substitution y(x)=x.v(x)

25. ## Re: MATH1231/1241/1251 SOS Thread

Originally Posted by laters
make dy/dx the subject and you'll see.

substitution y(x)=x.v(x)
Oh crap I forgot about that g(y/x) because they said they'd give us the substitution in the exam

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