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

#### InteGrand

##### Well-Known Member

When you divided by y in separating variables, you had to assume y =/= 0, making you lose that solution of y = 0.

And what did you want clarification with for the solution from the back of the book? You can check that it is differentiable everywhere and satisfies the given ODE.

• leehuan

#### leehuan

##### Well-Known Member

Whoops... clumsily forgot about the y≠0 problem...

Yeah that makes sense but, I have no idea where they got a≤x≤b from. Can't figure out the significance of using two different constants a and b to replace C

#### InteGrand

##### Well-Known Member

Whoops... clumsily forgot about the y≠0 problem...

Yeah that makes sense but, I have no idea where they got a≤x≤b from. Can't figure out the significance of using two different constants a and b to replace C
The solution they gave is essentially a more 'general' sort of solution than (x – c)^3, which is a shifted cubic. The more 'general' solution given is essentially taking this cubic but 'widening out' the horizontal point of the cubic to any arbitrary length. This preserves differentiability of the function because the derivatives at these 'joining points' x = a and x = b are still 0, as the derivative approaches 0 from either side and the function is still continuous at these points (and the function is differentiable everywhere else too clearly).

• leehuan

#### Flop21

##### Well-Known Member

How did this get simplified to this?? #### InteGrand

##### Well-Known Member

How did this get simplified to this?? Compute the partial derivatives (which they've done) and sub. in the given (x,y) point.

• Flop21

#### Flop21

##### Well-Known Member

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

Ohh okay! They subbed the points in, explains things. Thanks!

• InteGrand

#### Flop21

##### Well-Known Member

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?

#### InteGrand

##### Well-Known Member

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?

• Flop21

#### InteGrand

##### Well-Known Member

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?

• Flop21

#### Flop21

##### Well-Known Member how do I do this one

• eyeseeyou

#### InteGrand

##### Well-Known Member 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.

• Flop21

#### leehuan

##### Well-Known Member

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

##### -insert title here-

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

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

• leehuan

#### leehuan

##### Well-Known Member

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

##### -insert title here-

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.

#### leehuan

##### Well-Known Member

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

##### -insert title here-

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.

#### InteGrand

##### Well-Known Member

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

• kawaiipotato and leehuan

#### Flop21

##### Well-Known Member

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??? 