August 09, 2020, 10:15:33 am

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1301687 times) Tweet Share

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#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9660 on: July 06, 2020, 04:16:23 pm »
+3
KeltingMeith's method is the one usually taught for finding square roots of a complex number, without using polar form. Here is an alternative method which I also encourage students to learn, because the algebra is often easier.

Suppose we want to solve $z^2 = a+bi$ where $a$ and $b$ are real. We let $z = x+yi$, where $x$ and $y$ are both real, and hence our equation is $(x+yi)^2=a+bi$.

Expanding out the LHS and equating real and imaginary parts, we get $x^2 - y^2 = a$ and $2xy = b$. So far this is the same as the usual method, and the next step would be to solve simultaneously by substituting the second equation into the first.

However, we can generate another useful equation by noting that $\lvert z^2 \rvert = \lvert z \rvert ^2$ for any complex number $z$ and hence we have: \begin{aligned}\lvert (x+yi)^2 \rvert &= \lvert x+yi \rvert ^2 \\ &= x^2 + y^2\end{aligned}.
And now, since $\lvert (x+yi)^2\rvert = \lvert a+bi \rvert$, we have $x^2 + y^2 = \sqrt{a^2 + b^2}$.

So we now have two equations in $x^2$ and $y^2$ that are easily solved by elimination: $x^2 - y^2 = a$ and $x^2 + y^2 = \sqrt{a^2 + b^2}$.

Once you have solved for $x^2$ and $y^2$, then you find $x$ and $y$ by taking square roots. To know which signs of the roots should be taken, use the equation $2xy = b$: if $b > 0$, then $x$ and $y$ have the same sign; if $b < 0$, then $x$ and $y$ have opposite signs.

#### TheEagle

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9661 on: July 09, 2020, 01:11:07 am »
0
A particle moves in a straight line with a constant velocity of 20 m/s for 10 seconds. It is then subjected to a constant acceleration of 5 m/s2 in the opposite direction for T seconds, at which time the particle is back to its original position.

How do you determine the original position??

#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9662 on: July 09, 2020, 01:20:04 am »
+1
A particle moves in a straight line with a constant velocity of 20 m/s for 10 seconds. It is then subjected to a constant acceleration of 5 m/s2 in the opposite direction for T seconds, at which time the particle is back to its original position.

How do you determine the original position??

How do you mean? The original position is where it started at, there's not really anything to determine
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#### TheEagle

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9663 on: July 09, 2020, 01:24:07 am »
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How do you mean? The original position is where it started at, there's not really anything to determine

Woah that was a quick reply!

As in, how did they get-5(T-4) as the velocity for the "original position"  (I attached the graph)

#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9664 on: July 09, 2020, 02:12:26 am »
+4
Woah that was a quick reply!

As in, how did they get-5(T-4) as the velocity for the "original position"  (I attached the graph)

Right. Well, using constant acceleration formula, this is quite easy:

v = u + at

a=-5 (as we're moving backwards)
t=T
u=20

v = 20 - 5T = -5(T - 4)

You can also do this by integration:

a=-5

Integrate with respect to time to get:

v=-5t+c

At time t=10, v=20:

20=-50+c
c=70

At time t=T+10 (this is T seconds after turning around):

v=-5(T+10)+70=-5(T+10)-5*(-14)=-5(T+10-14)=-5(T-4)

Note that in the first case, I decided to use my constant acceleration formula as if I was starting from t=10, not t=0. As you can see, it still works out fine, and makes the maths a little easier. This kind of maths is what we like to call "time homogenous" - basically, it doesn't matter if things happen 50 seconds from starting the graph, or 5 hours. As long as the interval is the same length, everything will work out fine
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#### TheEagle

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9665 on: July 09, 2020, 02:39:14 am »
+1
Right. Well, using constant acceleration formula, this is quite easy:

v = u + at

a=-5 (as we're moving backwards)
t=T
u=20

v = 20 - 5T = -5(T - 4)

You can also do this by integration:

a=-5

Integrate with respect to time to get:

v=-5t+c

At time t=10, v=20:

20=-50+c
c=70

At time t=T+10 (this is T seconds after turning around):

v=-5(T+10)+70=-5(T+10)-5*(-14)=-5(T+10-14)=-5(T-4)

Note that in the first case, I decided to use my constant acceleration formula as if I was starting from t=10, not t=0. As you can see, it still works out fine, and makes the maths a little easier. This kind of maths is what we like to call "time homogenous" - basically, it doesn't matter if things happen 50 seconds from starting the graph, or 5 hours. As long as the interval is the same length, everything will work out fine

Omg thanks so much! That helped me understand it much better, awesome

#### a weaponized ikea chair

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9666 on: July 09, 2020, 07:54:56 pm »
0
Hey  ,

I am stuck on this question:

Prove that -((x)^2 - xy^2) = yx^2 and y = |x| + 1 do not intersect.

Any help is appreciated.
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#### 1729

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9667 on: July 09, 2020, 08:21:44 pm »
+2
Hey  ,

I am stuck on this question:

Prove that -((x)^2 - xy^2) = yx^2 and y = |x| + 1 do not intersect.

Any help is appreciated.
I don't actually know if there's an easier way to do it, but after looking at the graph of the two equations, I'd try to prove that there aren't any intersections with y = x+1, as that's easier to work with, and then you can work with the half where x < 0. With the former, you can just substitute $y = x+1$ into $-x^2 + xy^2 = yx^2$ and see if you get a contradiction. With the latter, if you simplify $-x^2 + xy^2 = yx^2$, you'd end up with $x = \frac{y^2}{y+1}$

Now see if you can deduce anything about y if x < 0, and try to relate that to y = -x+1 for x < 0.

Hope this helps!
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#### TheEagle

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9668 on: July 09, 2020, 08:23:30 pm »
+1
Hey  ,

I am stuck on this question:

Prove that -((x)^2 - xy^2) = yx^2 and y = |x| + 1 do not intersect.

Any help is appreciated.

Have you written the equation of the first one correctly?

#### a weaponized ikea chair

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9669 on: July 09, 2020, 09:07:29 pm »
+1

Have you written the equation of the first one correctly?
yes

I don't actually know if there's an easier way to do it, but after looking at the graph of the two equations, I'd try to prove that there aren't any intersections with y = x+1, as that's easier to work with, and then you can work with the half where x < 0. With the former, you can just substitute $y = x+1$ into $-x^2 + xy^2 = yx^2$ and see if you get a contradiction. With the latter, if you simplify $-x^2 + xy^2 = yx^2$, you'd end up with $x = \frac{y^2}{y+1}$

Now see if you can deduce anything about y if x < 0, and try to relate that to y = -x+1 for x < 0.

Hope this helps!
hmm...

when I solved it I got x =0 and y =1, and also tried it on CAS and got the same answer.

But on the desmos graph of the two equations, they do not intersect at (0,1), so I am not sure what is going on.

Hang on.

I did it again, this time differentially and got 4 = 0. This does not explain why they don't intersect though?

------

Rearranging for y in

$-x^2 +xy^2 = yx^2$

gives

$y = (sqrt(x(x+4)) + x)/2$

then:

$x + 1= (sqrt(x(x+4)) + x)/2$
$x + 2 = (sqrt(x(x+4)))$
$(x+2)^2 = x^2 + 4x$
$x^2 + 4x + 4 = x^2 + 4x$
$4 = 0$

Mod edit: posts merged
« Last Edit: July 09, 2020, 10:57:05 pm by insanipi »
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#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9670 on: July 09, 2020, 10:05:05 pm »
+4
Hang on.

I did it again, this time differentially and got 4 = 0. This does not explain why they don't intersect though?

It actually does - this is a classic case of proof by contradiction. The way it works is:

1. Assume some fact that you want to disprove (so in this case, we assume that the two graphs DO intersect)
2. Do some maths until you arrive at something that makes absolutely no sense
3. If the maths if you done in step 2 is all correct, then the reason for that must be that your assumption in step 1 is wrong
4. You're done. You proved it. Walk away happy.

By inserting y=x+1 into your equation, and finishing with 4=0, you've arrived at a contradiction - meaning that the graphs y=x+1 and -x^2+xy^2=x^2y CANNOT be equal.

hmm...

when I solved it I got x =0 and y =1, and also tried it on CAS and got the same answer.

But on the desmos graph of the two equations, they do not intersect at (0,1), so I am not sure what is going on.

The reason for this is that technically -x^2+xy^2=x^2y has an infinite amount of points that satisfy the equation at x=0. Let's see if the point (0,2) is on the graph:

-(0)^2+(0)(2)^2=(0)^2(2)
0+0=0

Since the result is true, that point exists. Let's try (0, 1):

-(0)^2+(0)(1)^2=(0)^2(1)
0+0=0

Result is true. Now, let's try the point (0, c), where c is any real number:

-(0)^2+(0)(c)^2=(0)^2(c)
0 + 0 = 0

Result is true. Technically, the question is wrong in that sense
« Last Edit: July 09, 2020, 10:11:27 pm by keltingmeith »
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#### a weaponized ikea chair

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9671 on: July 09, 2020, 10:41:18 pm »
+1
It actually does - this is a classic case of proof by contradiction. The way it works is:

1. Assume some fact that you want to disprove (so in this case, we assume that the two graphs DO intersect)
2. Do some maths until you arrive at something that makes absolutely no sense
3. If the maths if you done in step 2 is all correct, then the reason for that must be that your assumption in step 1 is wrong
4. You're done. You proved it. Walk away happy.

By inserting y=x+1 into your equation, and finishing with 4=0, you've arrived at a contradiction - meaning that the graphs y=x+1 and -x^2+xy^2=x^2y CANNOT be equal.

The reason for this is that technically -x^2+xy^2=x^2y has an infinite amount of points that satisfy the equation at x=0. Let's see if the point (0,2) is on the graph:

-(0)^2+(0)(2)^2=(0)^2(2)
0+0=0

Since the result is true, that point exists. Let's try (0, 1):

-(0)^2+(0)(1)^2=(0)^2(1)
0+0=0

Result is true. Now, let's try the point (0, c), where c is any real number:

-(0)^2+(0)(c)^2=(0)^2(c)
0 + 0 = 0

Result is true. Technically, the question is wrong in that sense
thanks for the clarification.
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#### withez

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9672 on: July 12, 2020, 05:21:36 pm »
+1
Hello
This is my first time posting on this thread
Wondering if anyone could walk me through this question:
2019
Biol[47]|Methods[44]
2020
Spesh|Physics|Chem|EngLang

#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9673 on: July 12, 2020, 06:25:57 pm »
0
What is known about the velocity of the plane?

#### 1729

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9674 on: July 12, 2020, 06:29:05 pm »
+1
Hello
This is my first time posting on this thread
Wondering if anyone could walk me through this question:
Doesn't seem like they give the initial horizontal velocity. Question is unsolvable with the information given.

EDIT: If hes dropped from the plane that means no initial vert velocity the only thing acting on the person is gravity so use vf^2 = 2ad to find vf after 600 and since the person travels at a constant speed for the rest thats the speed he lands with.

ANOTHER EDIT: That's only half the problem though, the other part you can't do without horizontal velocity but the speed it lands at is possible to identify.
« Last Edit: July 13, 2020, 10:17:41 am by 1729 »
What my avatar is
If you are wondering about my avatar, It was inspired by a problem I did which asked me to prove that the graphs of xy = 1 and y^2 = x^2 + 2 intersected at a 90 degree angle. The resulting figure in the middle kinda like a 8-sided square in hyperbolic space. It also resembles the conformal map of the complex square root.
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