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

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

Originally Posted by InteGrand
Compute the partial derivatives (which they've done) and sub. in the given (x,y) point.

$\noindent E.g. They found F_x \left(x,y\right) = \pi y^2 \cos \left(\pi xy^2\right). So at (x,y) = (2,-1), F_x (2,-1) = \pi \cdot (-1)^2 \cos \left(\pi \cdot 2\cdot (-1)^2\right) = \pi \cos 2\pi = \pi, since \cos 2\pi = 1.$
Ohh okay! They subbed the points in, explains things. Thanks!

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

Another tangent to surface question, how do I do this one: S: z^2+x^2+y^2 = 1., x0 = (1/3, 1/2, root(23)/6)

Find normal vector and equation of the tangent plane to the surface S at the point x0.

What is confusing me is the z. So do I move everything but the z to the RHS? Then solve?

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

Originally Posted by Flop21
Another tangent to surface question, how do I do this one: S: z^2+x^2+y^2 = 1., x0 = (1/3, 1/2, root(23)/6)

Find normal vector and equation of the tangent plane to the surface S at the point x0.

What is confusing me is the z. So do I move everything but the z to the RHS? Then solve?
$\noindent There's an easier way to do this particular one. This surface is a sphere centred at the origin, so the normal to the surface at any point \left(x,y,z\right) on the surface can geometrically be seen to just be that same vector \left(x,y,z\right). This means a normal to S at that point \mathbf{x}_0 is \mathbf{n} = \begin{bmatrix}1/3 \\ 1/2 \\ \sqrt{23}/6 \end{bmatrix} (i.e. same vector as \mathbf{x}_0). Note that \mathbf{n}\cdot \mathbf{x}_0 = \mathbf{x}_0 \cdot \mathbf{x}_0 = \left \|\mathbf{x}_0\right \|^2 = 1^2 = 1 (as \mathbf{x}_0 lies on the unit sphere). So an equation of the tangent plane is$

\begin{align*} \mathbf{n}\cdot \mathbf{x} &= \mathbf{n}\cdot \mathbf{x}_0 \\ \iff \frac{1}{3} x +\frac{1}{2}y +\frac{\sqrt{23}}{6}z &= 1. \end{align*}

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

Originally Posted by Flop21
Another tangent to surface question, how do I do this one: S: z^2+x^2+y^2 = 1., x0 = (1/3, 1/2, root(23)/6)

Find normal vector and equation of the tangent plane to the surface S at the point x0.

What is confusing me is the z. So do I move everything but the z to the RHS? Then solve?
$\noindent Here's the other method for doing these types of Q's that'll work even if the surface isn't something like a sphere where we can easily see a normal. We don't need to try and solve for z. We just need to make use of the following fact. If a smooth surface S is defined implicitly by the equation F\left(x,y,z\right) = C (where C is a constant), then a normal to S at the point \mathbf{x}_0 on the surface is given by \nabla F evaluated at the point \mathbf{x}_0. (Recall that \nabla F is the gradient vector \left(F_x, F_y, F_z\right).)$

$\noindent For your particular question, the surface is F\left(x,y,z\right)=1, where F\left(x,y,z\right) = x^2 + y^2 + z^2. So \nabla F = \left(2x,2y,2z\right) is a normal to the surface at any point \left(x,y,z\right) on the surface. So at the given point \mathbf{x}_0 =\left(\frac{1}{3}, \frac{1}{2}, \frac{\sqrt{23}}{6}\right) (which we can check is indeed on the surface by substituting it into the surface's equation and seeing it holds), a normal to the surface here is \left(2\times \frac{1}{3}, 2\times \frac{1}{2}, 2\times \frac{\sqrt{23}}{6}\right). From this, we can simplify things and since we have a normal and we know the point, we can find an equation for the tangent plane.$

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

how do I do this one

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

Originally Posted by Flop21

how do I do this one
Row-reduce that augmented matrix (get it into row-echelon form) and you should find a zero row at the bottom, with some linear expression involving x,y and z in this row in the right-hand augmented part. There will be solutions, i.e. we will have v be in S, if and only if this expression equals 0.

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

A hint on part c) please before I succumb to being 100% stuck. (Aside from conjug(x)=x)

$For n>1, let \omega_1, \omega_2, \dots, \omega_n be the n distinct nth roots of 1 and let A_k be the point on the Argand diagram which represents \omega_k. Let P represent any point z on the unit circle, and let PA_k denote the distance from P to A_k$

$a) Prove that (PA_k)^2=(z-w_k)(\overline{z}-\overline{w_k})$

$b) Deduce that \sum_{k=1}^{n}(PA_k)^2=2n$

$c) Now let P represent the point x on the real axis, -1

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

Originally Posted by leehuan
A hint on part c) please before I succumb to being 100% stuck. (Aside from conjug(x)=x)

$For n>1, let \omega_1, \omega_2, \dots, \omega_n be the n distinct nth roots of 1 and let A_k be the point on the Argand diagram which represents \omega_k. Let P represent any point z on the unit circle, and let PA_k denote the distance from P to A_k$

$a) Prove that (PA_k)^2=(z-w_k)(\overline{z}-\overline{w_k})$

$b) Deduce that \sum_{k=1}^{n}(PA_k)^2=2n$

$c) Now let P represent the point x on the real axis, -1
lol steven was doing this the other day

The point P is on the real axis so the conjugate of P is itself.

So, the conjugate distances from above will be (x-ω)(x-ω*) = x² - 2xcosθ +1

On the other hand, the product of all the conjugate pairs form all the irreducible quadratic factors of the degree n polynomial of unity.

Throw in the factor of (x+1) based on the parity of n.

Lastly, chuck in the 1-x factor which appears for all values of n.

This is equal to 1-x^n

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

Originally Posted by Paradoxica
lol steven was doing this the other day

The point P is on the real axis so the conjugate of P is itself.

So, the conjugate distances from above will be (x-ω)(x-ω*) = x² - 2xcosθ +1

On the other hand, the product of all the conjugate pairs form all the irreducible quadratic factors of the degree n polynomial of unity.

Throw in the factor of (x+1) based on the parity of n.

Lastly, chuck in the 1-x factor which appears for all values of n.

This is equal to 1-x^n
That was a bit too rushed. I had the idea of the quadratic factors but I don't see how they transform into 1-x^n

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

Originally Posted by leehuan
That was a bit too rushed. I don't get how the quadratic factors transform into 1-x^n
TL;DR

Factorise the nth polynomial of unity into it's complex factors and use the knowledge that x is inside the unit circle to obtain the distances you want.

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

This isn't helping sorry. Too rushed and you TLDRd it further. I don't see it....

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

Originally Posted by leehuan
This isn't helping sorry. Too rushed and you TLDRd it further. I don't see it....
...

x-ω

ω is one of the nth roots of unity

do I have to say more.

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

Originally Posted by leehuan
This isn't helping sorry. Too rushed and you TLDRd it further. I don't see it....
Essentially, here is a sketch.

$\noindent Let the product in question equal \rho. Recall from Paradoxica's earlier remarks that PA_k ^2 = \left(x -\omega_k \right)\left(x -\overline{\omega}_k\right). Take the product of both sides from k=1 to n: \prod \limits _{k=1}^{n} PA_k ^2 = \prod \limits _{k=1} ^{n} \left(x- \omega_k\right)\left(x -\overline{\omega_k}\right) \Rightarrow \rho^2 \equiv \left(\prod \limits _{k=1}^{n} PA_k\right)^2 = \prod _{k=1}^{n}\left(x -\omega_k\right) \cdor \prod \limits _{k=1}^{n}\left(x- \overline{\omega_k}\right).$

$\noindent You should be able to convince yourself that \left\{\omega_k \right\}_{k=1}^{n} and \left\{\overline{\omega_k}\right\}_{k=1}^{n} are exactly the same sets. This means the two products on the R.H.S. of the last line are the same, so \rho^2 = \left(\prod \limits _{k=1}^{n}\left(x-\omega_k\right)\right)^2. Since this right-hand product is x^n-1 (factorise the polynomial p(z) = z^n -1 and sub. in z=x), we have \rho^2 = \left(x^n -1\right)^2. Since \rho is a product of distances, it is positive, so \rho = 1-x^n (as -1

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

Originally Posted by InteGrand
Row-reduce that augmented matrix (get it into row-echelon form) and you should find a zero row at the bottom, with some linear expression involving x,y and z in this row in the right-hand augmented part. There will be solutions, i.e. we will have v be in S, if and only if this expression equals 0.
I'm stuck on a similar one

How do I find this vector???

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

why are you still doing matricies

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

Originally Posted by turntaker
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

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

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

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

are you doing things like intersection of lines, planes etc

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

Originally Posted by turntaker
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.

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

Originally Posted by Flop21
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.

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

Originally Posted by Flop21
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

(assuming your answer was right, I didn't check it).

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).

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

Originally Posted by turntaker
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

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

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

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

\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*}

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

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

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

\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.

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

Originally Posted by InteGrand
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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