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

leehuan

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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?
 

InteGrand

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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?
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.

Further reading: http://www2.onu.edu/~m-caragiu.1/bonus_files/HYPERSUB.pdf .
 
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leehuan

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Re: MATH1231/1241/1251 SOS Thread

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.

Further reading: http://www2.onu.edu/~m-caragiu.1/bonus_files/HYPERSUB.pdf .
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

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Re: MATH1231/1241/1251 SOS Thread

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...?


 
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RenegadeMx

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Re: MATH1231/1241/1251 SOS Thread

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
 

Paradoxica

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Re: MATH1231/1241/1251 SOS Thread

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.
 

Flop21

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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
 

InteGrand

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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
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

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Re: MATH1231/1241/1251 SOS Thread

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

 

InteGrand

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Re: MATH1231/1241/1251 SOS Thread

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

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Re: MATH1231/1241/1251 SOS Thread

Lol... :(

 

BobThaBuilda

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Re: MATH1231/1241/1251 SOS Thread

Hi,

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



Thanks!
 

InteGrand

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Re: MATH1231/1241/1251 SOS Thread

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

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Re: MATH1231/1241/1251 SOS Thread

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!
 

leehuan

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Re: MATH1231/1241/1251 SOS Thread

Getting kinda lost.



This bit was easy.



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



Clarification please?
 

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