TrueTears question thread

[TrueTears]

i'm not sure what you're asking:


and

 if the stuff under the sqrt is negative.

< yeah that was what I was asking.

Thanks I get it now.

[TrueTears]

Also just another Q, what if the coefficients in front of the y" and y' are not reals. Hence the quadratic you end up with does not have complex conjugates as the roots, would the general solution to homogeneous 2nd order DE still work?

[TrueTears]

^^ I think it still works because even though the co-efficients are complex they still undergo the same algorithm as real numbers. ie i - i = 0 etc.

amirite?

[zzdfa]

i dont know, but just make up random complex coefficients for a b and c, get m, then differentiate and see if it works.

[TrueTears]

Thanks, just tried it now, general formula still holds.

[TrueTears]

Just a stupid question lol, are we allowed to use say work, energy [physics related formulas] for the dynamics part of spesh?

If I get the right answer would I get penalised?

[mark_alec]

Just a stupid question lol, are we allowed to use say work, energy [physics related formulas] for the dynamics part of spesh?

If you look at most dynamics questions, you will see that they cannot be solved by energy considerations, but require solving the second order differential equation, a = F/m. You should always use the methods in the course to solve problems given, not others, as you may not receive credit for solutions using them.

[TrueTears]

Thanks mark!

[TrueTears]

1. The graph of is shown below:



The values of a and b could be?

A.
B.

2.

Thank you !

[kamil9876]

1.)

You can see that at x=0, y=1.

meaning

1=cosec(-2b)

only option A solve the above.

2.)

C is a point, D and E are circles. only A and B seem plausible.

subbing in z=-i (which is part of the locus) we see:

B implies:

|-i|=|-i+1-i|
1=|1-2i|

But |1-2i|>1 hence a contradiction.

Leaving only A

==========================================================

Alternative/Proof that A is true (elimination is good so screw this on the exam but meh).

If you look at the origin and the point 1-i we are after the set of z such that the magnitude of the line connecting z to 1-i and z to 0 are equal. It can be justified that this line satisfies this condition by drawing a line from an arbitrary point on our locus going to 1-i  and another one going from this arbitrary point all the way to 0. Then draw a line from 1-i to 0. What we have is an iscoseles triangle because our locus is the perpendicular bisector of the line connecting 1-i and 0. Hence the magnitudes of the two other lines are equal as required.

[TrueTears]

Thanks kamil, I did Q 2 with elimination from the choices but I wanted to know a proof for it.

Thanks again.

[TrueTears]

Ok for Q2, just out of interest sake, how would you go "backwards" from the cartesian equation to the complex locus?

Ie, the equation of the line is y = x - 1, how do you get from that to the complex locus?

[Mao]

That would usually involve a bit of ingenuity. Though knowing the geometric interpretations would make it seem fairly easy. In this case, the line bisects the line from (0,0) to (1,-1). Hence |z| = |z-1+i|

There are many different possible expressions. An equivalent form is |z+1| = |z-1+2i|

[kamil9876]

A good systematic way of doing it is to write down the line as a vector, in our case it can be:



Then find any vector perpendicular to this, ie just assign the x component some arbitrary real number, a, and use the dot product:

where a is some real constant.

Now pick any point on the line in question, and 'add' to it the vector to end up at some point A and the vector to end up at some point  B. Now we can easily see that:

(where A and B are the two points but in complex number form)

[TrueTears]

Thanks guys.

Just another one:



I got the answer fairly easily, by elimination, A-D are all correct so I picked E. But I can not see how E is correct/incorrect? Can someone please explain why?

Thanks!