 February 23, 2020, 02:55:26 pm

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1193012 times) Tweet Share

0 Members and 35 Guests are viewing this topic.

#### ^^^111^^^

• MOTM: JULY 2019
• Forum Obsessive
•   • Posts: 292
• Respect: +24 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9585 on: January 15, 2020, 06:28:11 pm »
+4
But why is the acceleration negative?
This is because as Sine mentioned, we have to be consistent with vector quantities and as force is a vector, it has magnitude as well as direction. In the solutions they arbitrarily take the resistance force as positive. If the resistance force is positive, then the acceleration (which is in the opposing direction of the resistance force) is negative.

« Last Edit: January 15, 2020, 06:32:22 pm by ^^^111^^^ »

#### Evolio ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9586 on: January 15, 2020, 06:36:51 pm »
+2
This is because as Sine mentioned, we have to be consistent with vector quantities and as force is a vector, it has magnitude as well as direction. In the solutions they arbitrarily take the resistance force as positive. If the resistance force is positive, then the acceleration (which is in the opposing direction of the resistance force) is negative.

Sorry, I'm still a bit confused because don't we usually minus the smaller force from the bigger force.? And since it's sliding down, there is a bigger downward force than the friction force, thus it is able to slide down due to the imbalance?
I understand why the signs of the forces have switched because they are sliding down now (not being projected up)
2019: Biology , Mathematical Methods 
2020: Literature, Psychology, Specialist Mathematics, Chemistry

#### Sine

• Moderator
• ATAR Notes Legend
•     • • Posts: 3979
• ARTWORK 23
• Respect: +1134 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9587 on: January 15, 2020, 06:58:56 pm »
+4
Sorry, I'm still a bit confused because don't we usually minus the smaller force from the bigger force.? And since it's sliding down, there is a bigger downward force than the friction force, thus it is able to slide down due to the imbalance?
I understand why the signs of the forces have switched because they are sliding down now (not being projected up)
The friction force is one of the forces that is opposing motion (i.e. the object being pulled down). Also the "gravity" is the force pulling it down (mgsin(20)).

You don't always minus the smaller force from the bigger one. It should work either way if you use correct signs.

E.g. consider moving up the plane as positive then:
0.25R - mgsin(20) = ma
a = -1.05m/s^2
This is what the solutions say.

E.g. consider moving down the plane as positive then:
mgsin(20) - 0.25R = ma
mgsin(20) - 0.25(mgcos(20)) = ma
g(sin(20) - 0.25cos(20)) = a
a = 9.8 * ( sin(pi/9) - 0.25cos(pi/9))
a = 1.05m/s^2

However, since we denoted the initial velocity as 10 (i.e. positive 10) we should be using moving up the slope as positive.

So you just get that the object is accelerating down the slope whereas before you get the object is decelerating down the slope.

Always in these physics type question it is really useful to denote what you deem as positive before you start - even if it is just in your head.

#### ^^^111^^^

• MOTM: JULY 2019
• Forum Obsessive
•   • Posts: 292
• Respect: +24 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9588 on: January 15, 2020, 07:03:19 pm »
+4
Sorry, I'm still a bit confused because don't we usually minus the smaller force from the bigger force.? And since it's sliding down, there is a bigger downward force than the friction force, thus it is able to slide down due to the imbalance?
I understand why the signs of the forces have switched because they are sliding down now (not being projected up)
About taking away the smaller force from the bigger force, instead of thinking of it as smaller or bigger , think of it like this:
ΣF=ma
F1+F2=ma
F1 is a vector (a) and represents friction. F2 is a vector (b) and represents acceleration.  Instead of thinking of it as subtraction, to find the resultant vector (or the resultant force in this case), we add both vectors together (so F1 + F2). But here, the acceleration is negative (for previously mentioned reasons). Therefore F2 is negative. Now using vector addition, F1 + F2 becomes F1 + (-)F2 or F1 - F2.

Edit: Got there before me Sine. Use Sine's way, the way I showed you above was merely a try to make myself more clear.
« Last Edit: January 15, 2020, 07:07:19 pm by ^^^111^^^ »

#### Evolio ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9589 on: January 15, 2020, 08:35:37 pm »
+1
Oh, thank you so much Sine and ^^^111^^^!
I really appreciate all your help.
I understand it now, I had a wrong understanding of the concepts.
2019: Biology , Mathematical Methods 
2020: Literature, Psychology, Specialist Mathematics, Chemistry

#### TheEagle

• Forum Obsessive
•   • Posts: 211
• Respect: +2 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9590 on: January 16, 2020, 01:12:35 am »
0
Hey

I am curious to know why the equation of the asymptotes for the hyperbola are y=bx/a
Why don't we substitute infinity into bx/a?
I've attached a photo

#### ^^^111^^^

• MOTM: JULY 2019
• Forum Obsessive
•   • Posts: 292
• Respect: +24 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9591 on: January 16, 2020, 08:09:59 am »
+2
Hey

I am curious to know why the equation of the asymptotes for the hyperbola are y=bx/a
Why don't we substitute infinity into bx/a?
I've attached a photo
As x approaches infinity, y does not approach zero.

Edit: Sorry, I will explain a bit more. For y=bx/a, x can be anything. If you were to substitute ∞ for x, you will get y --> ∞ as well. I mean, sure if you want you can substitute infinity for x, but there isn't a need to.

Is this what you meant, or did you mean substituting ∞ for bx/a?
If this is what you meant, we don't substitute ∞ for b/xa, because for y=bx/a the function of the hyperbola should not be defined. If we took bx/a as ∞, then the function will be defined, and this contradicts the definition of an asymptote.
« Last Edit: January 16, 2020, 08:24:26 am by ^^^111^^^ »

#### TheEagle

• Forum Obsessive
•   • Posts: 211
• Respect: +2 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9592 on: January 16, 2020, 12:02:18 pm »
+1
As x approaches infinity, y does not approach zero.

Edit: Sorry, I will explain a bit more. For y=bx/a, x can be anything. If you were to substitute ∞ for x, you will get y --> ∞ as well. I mean, sure if you want you can substitute infinity for x, but there isn't a need to.

Is this what you meant, or did you mean substituting ∞ for bx/a?
If this is what you meant, we don't substitute ∞ for b/xa, because for y=bx/a the function of the hyperbola should not be defined. If we took bx/a as ∞, then the function will be defined, and this contradicts the definition of an asymptote.

I kinda see...
What I meant was:

As x approaches infinity,

Y= (b*infinity)/a  *  sqroot (1 - a^2/infinity^2)   = infinity
However the video I was watching had considered the sqroot as 1 when x approaches infinity, which I understand. But they didnt substitute it in the outside (bx/a)

#### ^^^111^^^

• MOTM: JULY 2019
• Forum Obsessive
•   • Posts: 292
• Respect: +24 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9593 on: January 16, 2020, 03:09:54 pm »
+4
I kinda see...
What I meant was:

As x approaches infinity,

Y= (b*infinity)/a  *  sqroot (1 - a^2/infinity^2)   = infinity
However the video I was watching had considered the sqroot as 1 when x approaches infinity, which I understand. But they didnt substitute it in the outside (bx/a)
Solving for y, you get to y= +/- (b/a)x * square root of(x^2 - a^2)
If x is approaching infinity, then the a^2 is insignificant. Therefore:
y = +/-(b/a) * square root of((x^2)).

Remember that x is not infinity itself, it's x approaches infinity. You should be clear from what I typed before, but if you wish, you can view the proof below:
The limit as x approaches infinity of (b/a square-root-of(x^2 - a^2) - bx/a)= 0
Here is a proof of this that I found from a textbook:
lim{x approaches infinity}(b/a (sqrt(x^2 - a^2) - x))

= b/a lim{x approaches infinity}(sqrt(x^2 - a^2) - x)
= b/a lim{x approaches infinity} (sqrt(x^2 - a^2) - x)(sqrt(x^2 - a^2) + x) divided by (sqrt(x^2 - a^2) + x
= b/a lim{x approaches infinity}  (x^2 - a^2)^2 - x^2 divided by  sqrt(x^2 - a^2) + x
= b/a lim{x approaches infinity} x((1 - a^2/x^2)^2 - 1) divided by x(sqrt(1 - a^2/x^2) + 1
= b/a lim[x approaches infinity] (1 - a^2/x^2)^2 - 1 divided by sqrt(1 - a^2/x^2) + 1
= b/a lim{x approaches infinity}(1)^2 - 1 divided by  sqrt(1) + 1
=0/2
=0
Edit: Sorry I just realized that I added too much spacing « Last Edit: January 16, 2020, 03:19:03 pm by ^^^111^^^ »

#### TheEagle

• Forum Obsessive
•   • Posts: 211
• Respect: +2 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9594 on: January 16, 2020, 09:17:00 pm »
+1
Solving for y, you get to y= +/- (b/a)x * square root of(x^2 - a^2)
If x is approaching infinity, then the a^2 is insignificant. Therefore:
y = +/-(b/a) * square root of((x^2)).

Remember that x is not infinity itself, it's x approaches infinity. You should be clear from what I typed before, but if you wish, you can view the proof below:
The limit as x approaches infinity of (b/a square-root-of(x^2 - a^2) - bx/a)= 0
Here is a proof of this that I found from a textbook:
lim{x approaches infinity}(b/a (sqrt(x^2 - a^2) - x))

= b/a lim{x approaches infinity}(sqrt(x^2 - a^2) - x)
= b/a lim{x approaches infinity} (sqrt(x^2 - a^2) - x)(sqrt(x^2 - a^2) + x) divided by (sqrt(x^2 - a^2) + x
= b/a lim{x approaches infinity}  (x^2 - a^2)^2 - x^2 divided by  sqrt(x^2 - a^2) + x
= b/a lim{x approaches infinity} x((1 - a^2/x^2)^2 - 1) divided by x(sqrt(1 - a^2/x^2) + 1
= b/a lim[x approaches infinity] (1 - a^2/x^2)^2 - 1 divided by sqrt(1 - a^2/x^2) + 1
= b/a lim{x approaches infinity}(1)^2 - 1 divided by  sqrt(1) + 1
=0/2
=0
Edit: Sorry I just realized that I added too much spacing Thank you so much! You are very clear at explaining! Appreciate your help! #### TheEagle

• Forum Obsessive
•   • Posts: 211
• Respect: +2 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9595 on: January 21, 2020, 07:59:29 pm »
0
Hello!

can someone please give me an explanation for the reflection explanation, I don't quite understand

#### ^^^111^^^

• MOTM: JULY 2019
• Forum Obsessive
•   • Posts: 292
• Respect: +24 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9596 on: January 21, 2020, 09:45:36 pm »
+4
Hello!

can someone please give me an explanation for the reflection explanation, I don't quite understand
To be quite honest I wouldn't really call it a reflection, but what I think they mean is that:
(ax,by) ==> (bx,ay).
That is, the dilations of the equations have interchanged.
OR
The major axis becomes a minor axis and vice versa(as seen from my attachment).

As for the rest of the explanation, we can use substitution to solve.
x2/4 + y2/9 - 1= x2/9 + y2/4 -1
===>>> x2/4 + y2/9=x2/9 + y2/4
===>>>9x2+4y2=9y2 + 4x2
===>>>9x2-4x2=9y2 -  4y2
===>>>5x2=5y2
===>>>y=+ or (-) x
Then we can substitute y as +/- x to the equations and, therefore, prove the intersections as the vertices of a square
Honestly to show the points of intersection as vertices square, I don't think it is necessary to take the reflection points to substitute to the equations, although it would be the orthodox method in this case.

#### TheEagle

• Forum Obsessive
•   • Posts: 211
• Respect: +2 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9597 on: January 23, 2020, 12:16:32 am »
+1
To be quite honest I wouldn't really call it a reflection, but what I think they mean is that:
(ax,by) ==> (bx,ay).
That is, the dilations of the equations have interchanged.
OR
The major axis becomes a minor axis and vice versa(as seen from my attachment).

As for the rest of the explanation, we can use substitution to solve.
x2/4 + y2/9 - 1= x2/9 + y2/4 -1
===>>> x2/4 + y2/9=x2/9 + y2/4
===>>>9x2+4y2=9y2 + 4x2
===>>>9x2-4x2=9y2 -  4y2
===>>>5x2=5y2
===>>>y=+ or (-) x
Then we can substitute y as +/- x to the equations and, therefore, prove the intersections as the vertices of a square
Honestly to show the points of intersection as vertices square, I don't think it is necessary to take the reflection points to substitute to the equations, although it would be the orthodox method in this case.

Thanks so much #### TheEagle

• Forum Obsessive
•   • Posts: 211
• Respect: +2 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9598 on: January 24, 2020, 10:02:54 pm »
0
Hello all

I am having trouble finding the range of x over the domain of (pi/2, 3pi/2). I keep getting negative infinity to negative a, but the textbook says negative infinity to positive a?

#### ^^^111^^^

• MOTM: JULY 2019
• Forum Obsessive
•   • Posts: 292
• Respect: +24 ##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9599 on: January 25, 2020, 09:16:32 am »
+1
Hello all

I am having trouble finding the range of x over the domain of (pi/2, 3pi/2). I keep getting negative infinity to negative a, but the textbook says negative infinity to positive a?
Edit: Sorry I realized I made a mistake, tbh I am also not sure why it is positive a, so let's just wait till help comes.
« Last Edit: January 25, 2020, 09:27:19 am by ^^^111^^^ »