Is there a rule of thumb as to when you'd favour a hyperbolic substitution over a trigonometric substitution in integration?
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
Last edited by leehuan; 7 Aug 2016 at 3:05 PM.
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 .
Last edited by InteGrand; 23 Jul 2016 at 7:27 PM.
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...?
If I am a conic section, then my e = ∞
Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.
In b4 thread closes
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.
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.)
Lol...
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).)
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?
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.
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).
How did this get simplified to this??
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