Andiio's Questions Thread :)

[Andiio]

AHAHHA thanks for going to the effort of drawing that

Yeah I get it, so just basically solving for x and then adding on the period x n
I think that is basically TT's method.

But hm, I still can't get the right answer for my question; still not getting a -ve answer for the 2nd genearl solution

:S

[stonecold]

TT said it is still right.  they are just using a different X and Y starting points to you.

hence why I said it is not really testable.  as long as you still get the right solutions after subbing in random integers + 0, then you cannot be marked wrong.

to see if you are right, sub in -2,-1,0,1,2 etc. into yours and their answer.  odds are they are both right, but yours is just on a different scale. 

[Andiio]

Ahhh okay, awesome haha

So generally it's just:

1. Solve for x
2. add (period*n)
?

Never thought it was so easy! haha

[stonecold]

yepp.  maths textbooks are shit at explaining stuff... general solutions are actually really simple. 

more info here.

https://nrich.maths.org/discus/messages/145082/145446.html?1233671333

[Andiio]

Sweet haha, thanks heaps stonecold!

[kamil9876]

Thanks!

How do I find the general solutions of cos2(x+pi/3)=1/2? (Without applying the formula)

I used TT's method and ended up with:

x = n*pi - pi/6, x = pi/2 + n*pi, n E Z, but the answer states that x = n*pi - pi/6 (which is correct) and n*pi - pi/2 :S

Thanks!

To see they are both the same:



And clearly:

Stonecold is right: the "starting point" is just "shifted by 1"

in general, to see if two general solutions are equivalent just check if they both have the same period and their "starting points" differ by an integer multiple of this period.

[Andiio]

Hey guys,

Not sure what to do here for this part of the question.

What I did was:

A = 50sin(theta)
P = 20 + √[200-200cos(theta)]

Let A=P

Then I tried to solve it with my calc, but what would you put as the domain?

Or do you have to solve this through the use of identities?

Thanks!

[stonecold]

Try a domain between 0 and pi/2, as theta is an acute angle

[Andiio]

Try a domain between 0 and pi/2, as theta is an acute angle

Great idea! I completely disregarded and forgot about the diagram haha.

- Only obtained one solution, so hmmmmm :S Flawed question? I'll try a higher domain


EDIT: Got both solutions with a domain of [0,2pi]

[Andiio]

Hey guys, what's the most efficient method in using addition of ordinates to sketch a graph?

Is it to just plot points?

Thanks!

[brightsky]

Hey guys, what's the most efficient method in using addition of ordinates to sketch a graph?

Is it to just plot points?

Thanks!

Basically. Except you don't really need to draw in any points. If you've got the graphs for the two functions you're trying to add already, all you need to do is grab any two corresponding points with the same x-value, and then add their y-values. Do that for the whole length of the graph (or a good part of it so that you know what the general shape is) and then just sketch the whole graph. You can do all this essentially just by looking at the graph, no pencil-work needed except for the sketching part.

[Andiio]

When graphing a gradient function, (e.g. of a parabola, cubic etc), how would one determine the x-intercepts and other major points graphically?

An example is attached!

[Pixon]

I hope I've interpreted your question correctly...I'll edit it if I'm wrong.
If the question you've attached is the original function, you're just supposed to work out the gradients for each line.
So from [-3,0) the gradient will be 2/3. From (0,3] the gradient is -5/3. So for graphing the gradient function you would draw y=2/3, -3<x<0 and y=-5/3, 0<x<3
Note that the end points cannot be differentiated so mark them with open circles on your gradient graph.

For parabolas and cubics etc, they generally just want a reasonable estimate. x-ints of the gradient function are simply where stationary points are i.e. dy/dx=0, and be sure to check end points and discontinuity. Sometimes you can actually work out what the original function is from the graph with enough information, but usually they don't even care about that.

[Andiio]

I hope I've interpreted your question correctly...I'll edit it if I'm wrong.
If the question you've attached is the original function, you're just supposed to work out the gradients for each line.
So from [-3,0) the gradient will be 2/3. From (0,3] the gradient is -5/3. So for graphing the gradient function you would draw y=2/3, -3<x<0 and y=-5/3, 0<x<3
Note that the end points cannot be differentiated so mark them with open circles on your gradient graph.

For parabolas and cubics etc, they generally just want a reasonable estimate. x-ints of the gradient function are simply where stationary points are i.e. dy/dx=0, and be sure to check end points and discontinuity. Sometimes you can actually work out what the original function is from the graph with enough information, but usually they don't even care about that.

Thanks - why can't the endpoints of be differentiated though? I understand why dy/dx ≠ exist at x=0, but if the graph is continuous at the endpoints, then shouldn't it be a-okay? Not sure hm, I'm just being noob and missing something

[evaever]

End points are differentiable.

E.g. r = (t+1)^2 for t >= 0
Find velocity v(0).