# First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (2 Viewers)

#### leehuan

##### Well-Known Member

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

#### InteGrand

##### Well-Known Member

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.

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#### leehuan

##### Well-Known Member

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...?

#### InteGrand

##### Well-Known Member

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...?
$\bg_white \noindent Yeah, it's based on the fact that \cos\left(ix\right) = \cosh x and \sin \left(ix\right) = i\sinh x (one way to see this is via Euler's Formula and the exponential definitions of the hyperbolic functions). (We can also show from these by replacing x with -ix that \cosh \left(ix\right) = \cos x and \sinh \left(ix\right) = i\sin x.)$

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

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##### Kosovo is Serbian

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

##### -insert title here-

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.

#### InteGrand

##### Well-Known Member

Why do you think this thread will get closed?

#### Flop21

##### Well-Known Member

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

#### InteGrand

##### Well-Known Member

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.

#### Flop21

##### Well-Known Member

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

#### InteGrand

##### Well-Known Member

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.)

#### leehuan

##### Well-Known Member

Lol...

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

#### InteGrand

##### Well-Known Member

Lol...

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

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

#### BobThaBuilda

##### New Member

Hi,

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

Thanks!

#### InteGrand

##### Well-Known Member

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).)

#### BobThaBuilda

##### New Member

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!

##### -insert title here-

Lol...

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

#### leehuan

##### Well-Known Member

Getting kinda lost.

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

This bit was easy.

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

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

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