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

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

Don't mind me...just setting up some threads for more of my stupidity this upcoming semester.

1231/1241 Outline: https://www.maths.unsw.edu.au/sites/...41_s2_2016.pdf
1251 Outline: https://www.maths.unsw.edu.au/sites/...51_s1_2016.pdf

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

Is there a rule of thumb as to when you'd favour a hyperbolic substitution over a trigonometric substitution in integration?

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

Originally Posted by leehuan
Is there a rule of thumb as to when you'd favour a hyperbolic substitution over a trigonometric substitution in integration?
Often for a^2 + x^2, x = a*sinh(t) turns out to be neater than using tan. Just make sure if you're using hyperbolic ones, you're familiar with hyperbolic analogs to trig stuff, e.g. know how to integrate things like sech, simplify things like compositions of inverse hyperbolic functions with hyperbolic functions or expressing hyperbolic functions in terms of other ones, double angle identities for hyperbolic functions etc.

For x^2 - a^2, we can use x = a*cosh(t).

For a^2 - x^2, usually just a normal trig. substitution of x = a*sin(t) is neater.

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

Originally Posted by InteGrand
Often for a^2 + x^2, x = a*sinh(t) turns out to be neater than using tan. Just make sure if you're using hyperbolic ones, you're familiar with hyperbolic analogs to trig stuff, e.g. know how to integrate things like sech, simplify things like compositions of inverse hyperbolic functions with hyperbolic functions or expressing hyperbolic functions in terms of other ones, double angle identities for hyperbolic functions etc.

For x^2 - a^2, we can use x = a*cosh(t).

For a^2 - x^2, usually just a normal trig. substitution of x = a*sin(t) is neater.

Yeah I reckon they'll go over hyperbolics in the integration topic. It was in the outline from memory.

Also regarding hyperbolic identities, I was told that if you replaced sin with i.sinh and cos with cosh (in all your trigonometric identities) you get all the hyperbolic identities apparently...?

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

Originally Posted by leehuan
Yeah I reckon they'll go over hyperbolics in the integration topic. It was in the outline from memory.

Also regarding hyperbolic identities, I was told that if you replaced sin with i.sinh and cos with cosh (in all your trigonometric identities) you get all the hyperbolic identities apparently...?
$\noindent Yeah, it's based on the fact that \cos\left(ix\right) = \cosh x and \sin \left(ix\right) = i\sinh x (one way to see this is via Euler's Formula and the exponential definitions of the hyperbolic functions). (We can also show from these by replacing x with -ix that \cosh \left(ix\right) = \cos x and \sinh \left(ix\right) = i\sin x.)$

$\noindent We can use these to obtain analogous hyperbolic identities from trig. ones. (These explain why sometimes signs change or don't change, but otherwise are similar.) E.g. Consider the identity \cos 2\theta = 1-2\sin^2 \theta. Replacing \theta with ix, \cos \left(i\cdot 2x\right) = 1-2\sin^{2}\left(ix\right). Using the aforementioned identities, this tells us \cosh 2x = 1-2\left(i \sinh x\right)^2 = 1+2\sinh^2 x, and we've derived a double angle formula for cosh (and can see from where the sign change came about compared to the trig. one).$

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

Originally Posted by leehuan
Yeah I reckon they'll go over hyperbolics in the integration topic. It was in the outline from memory.

Also regarding hyperbolic identities, I was told that if you replaced sin with i.sinh and cos with cosh (in all your trigonometric identities) you get all the hyperbolic identities apparently...?
they do and they are easy to remember/do, from memory only hard parts could be having to do two substitutions within one question

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

Originally Posted by leehuan
Yeah I reckon they'll go over hyperbolics in the integration topic. It was in the outline from memory.

Also regarding hyperbolic identities, I was told that if you replaced sin with i.sinh and cos with cosh (in all your trigonometric identities) you get all the hyperbolic identities apparently...?
Given all the even identities, are exactly that, even, the squaring removes the existence of the imaginary unit. It works, because maths.

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

Originally Posted by Drsoccerball
Why do you think this thread will get closed?

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

https://youtu.be/fhjPMrfyVOM?t=1m26s

At the point linked... why does he use <2,0,1> or more so, HOW does he come up with that? He shows us he is using it since it's a vector of S, but never how he came up with it.

Thanks

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

Originally Posted by Flop21
https://youtu.be/fhjPMrfyVOM?t=1m26s

At the point linked... why does he use <2,0,1> or more so, HOW does he come up with that? He shows us he is using it since it's a vector of S, but never how he came up with it.

Thanks
By inspection. He was probably just looking for an easy vector in S, and the one he chose is easy to find (because we just make one of the terms 0 by making that component of the vector equal to 0 and then see what we need to do for the other entries. If we set the last entry equal to 1, it is easy to see we make the first one 2 to make the vector in S).

It is easy to confirm his choice of vector is in S by subbing it into the condition to be in S.

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

How on earth do you do this, my answers are wrong.

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

Originally Posted by Flop21
How on earth do you do this, my answers are wrong.

To find where it meets a certain axis, we set the other two variables to 0 and solve for the last variable.

E.g. to find where it hits the y-axis, we would let x = 0 and z = 0 and solve for y in the plane equation. The point where it meets the y-axis would be of the form (0, y*, 0), where y* is the y-intercept. In other words, both other entries are 0 (similarly for the other axial intercepts). (After all, points are on the y-axis iff they are of the form (0, y, 0) for some real number y, and similarly for the other axes.)

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

Lol...

$\text{If }\alpha \in \mathbb{C}\text{ satisfies }|\alpha|=1\text{, show that }|\alpha+1|+|\alpha-1|\le 2\sqrt{2}$

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

Originally Posted by leehuan
Lol...

$\text{If }\alpha \in \mathbb{C}\text{ satisfies }|\alpha|=1\text{, show that }|\alpha+1|+|\alpha-1|\le 2\sqrt{2}$
$\noindent We can write \alpha = \cos t + i\sin t, for some t\in \mathbb{R}. So |\alpha +1| +|\alpha -1|= \sqrt{\left(\cos t+1\right)^2 + \sin^2 t} + \sqrt{\left(\cos t -1\right)^2 + \sin^2 t} = \sqrt{2 +2\cos t} + \sqrt{2-2\cos t} = \sqrt{2}\left(\sqrt{1 + \cos t} + \sqrt{1-\cos t}\right). So we just need to show \sqrt{1+\cos t}+\sqrt{1-\cos t}\leq 2.$

$\noindent Substitute u =\cos t, then we just need to show that \sqrt{1+u}+\sqrt{1-u}\leq 2 for u\in \left[-1,1\right]. But this is true since the L.H.S. here is non-negative and LHS^2 = 1+u + 2\sqrt{1-u^2} + 1-u = 2+ 2\sqrt{1-u^2} \leq 2+2 = 4 =RHS^2 (as 0\leq \sqrt{1-u^2}\leq 1). This completes the proof.$

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

Hi,

I'm not sure how to do this question because of the z^2 - Part D

Thanks!

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

Originally Posted by BobThaBuilda
Hi,

I'm not sure how to do this question because of the z^2 - Part D

Thanks!
When we have a smooth surface given by F(x,y,z) = const., a normal vector is grad F. So since here F(x,y,z) = x^2 + y^2 + z^2, grad F = (2x, 2y, 2z). So a normal at a point (x, y, z) on the surface is (2x, 2y, 2z).

(Or just note in this case it's a sphere centred at the origin, so an obvious normal is (x, y, z).)

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

Originally Posted by InteGrand
When we have a smooth surface given by F(x,y,z) = const., a normal vector is grad F. So since here F(x,y,z) = x^2 + y^2 + z^2, grad F = (2x, 2y, 2z). So a normal at a point (x, y, z) on the surface is (2x, 2y, 2z).

(Or just note in this case it's a sphere centred at the origin, so an obvious normal is (x, y, z).)
Thanks InteGrand!

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

Originally Posted by leehuan
Lol...

$\text{If }\alpha \in \mathbb{C}\text{ satisfies }|\alpha|=1\text{, show that }|\alpha+1|+|\alpha-1|\le 2\sqrt{2}$
$\noindent Let one of the lengths be x and the other one be y. \\ Geometrically, the condition is that the two numbers are lengths of a right angled triangle with hypotenuse 2. \\ So that means x^2 + y^2 = 4, and the aim is to maximise the sum of the lengths. There are approximately \infty ways to do this so here's two:\\1. Parametrise the lengths as x=2\cos{\phi}, y=2\sin{\phi}, then use the auxiliary transformation and elementary trigonometric inequalities. \\ 2. Use the Arithmetic-Mean Quadratic-Mean Inequality: \sqrt{\frac{x^2+y^2}{2}} \geq \frac{x+y}{2} \\ The answer appears immediately.$

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

Getting kinda lost.

$\\ \text{Try to find the general solution for }\frac{dy}{dx}=3y^{\frac{2}{3}}\\ \text{Your answer will probably be }y=(x+C)^3$

This bit was easy.

$\\ \text{Observe }y=0\text{ is also a solution and it cannot be expressed as }y=(x+C)^3\text{ for any value of }C\\ \text{How do you account for this? Are there any other solutions?}$

Yeah, I don't know how to account for it. And according to the back of the book the solutions are:

$y=\begin{cases}(x-a)^3 &\text{if }xb;\text{ where }a

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

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.

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

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

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

Originally Posted by leehuan
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).

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

How did this get simplified to this??

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

Originally Posted by Flop21
How did this get simplified to this??

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

$\noindent E.g. They found F_x \left(x,y\right) = \pi y^2 \cos \left(\pi xy^2\right). So at (x,y) = (2,-1), F_x (2,-1) = \pi \cdot (-1)^2 \cos \left(\pi \cdot 2\cdot (-1)^2\right) = \pi \cos 2\pi = \pi, since \cos 2\pi = 1.$

Page 1 of 16 12311 ... Last

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•