# 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  Reply With Quote

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?  Reply With Quote

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.

Further reading: http://www2.onu.edu/~m-caragiu.1/bon...s/HYPERSUB.pdf .  Reply With Quote

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.

Further reading: http://www2.onu.edu/~m-caragiu.1/bon...s/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...?  Reply With Quote

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...?  Reply With Quote

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  Reply With Quote

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.  Reply With Quote

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

In b4 thread closes  Reply With Quote

9. ## Re: MATH1231/1241/1251 SOS Thread Originally Posted by Drsoccerball In b4 thread closes
Why do you think this thread will get closed?  Reply With Quote

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  Reply With Quote

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.  Reply With Quote

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

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

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.)  Reply With Quote

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

Lol...   Reply With Quote

15. ## Re: MATH1231/1241/1251 SOS Thread Originally Posted by leehuan Lol...   Reply With Quote

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!  Reply With Quote

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).)  Reply With Quote

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!  Reply With Quote

19. ## Re: MATH1231/1241/1251 SOS Thread Originally Posted by leehuan Lol...   Reply With Quote

20. ## 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?  Reply With Quote

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.  Reply With Quote

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  Reply With Quote

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).  Reply With Quote

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

How did this get simplified to this??   Reply With Quote

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.  Reply With Quote

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