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March 29, 2024, 11:31:30 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164464 times)  Share 

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VanillaRice

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9240 on: April 04, 2018, 03:48:32 pm »
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From Cambridge spec.

f(x) = xe^x. Find the area enclosed by the tangent to the point P(-1,y), the curve and the y-axis.

The tangent is y=-1/e
So I've subtracted the absolute value of the integral from 0 to -1 of xe^x from the rectangle formed under the curve, which is of dimensions 1*1/e.
So the area is 1/e - int(-1,0) xe^x dx.
Hence my question! :)
I would say that this is probably a calculator question. There's no direct way for you to calculate that integral in VCE. Hope that helps :)
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Ssuper_19

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9241 on: April 12, 2018, 11:05:12 pm »
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This question is slightly different. I have the answer and I just need an explanation as to why it works.
The original question is:
If Arg z = Pi /4 and Arg (z-3) = Pi/2, find Arg(z-6i) .

Down below I attached a picture with the “worked solution”. There are two Rays but I don’t understand why the intersection of the two rays is z and why an arg translation eg. Arg(z+3) means a translation three units in negative x-axis (I guess how the translation is the translation. Is z is midpoint or.... I’m not exactly sure what it means conceptually I guess)
Thanks heaps in advanced.
Source: Cambridge specialist

jazzycab

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9242 on: April 13, 2018, 06:39:59 am »
+1
This question is slightly different. I have the answer and I just need an explanation as to why it works.
The original question is:
If Arg z = Pi /4 and Arg (z-3) = Pi/2, find Arg(z-6i) .

Down below I attached a picture with the “worked solution”. There are two Rays but I don’t understand why the intersection of the two rays is z and why an arg translation eg. Arg(z+3) means a translation three units in negative x-axis (I guess how the translation is the translation. Is z is midpoint or.... I’m not exactly sure what it means conceptually I guess)
Thanks heaps in advanced.
Source: Cambridge specialist


If \(\text{Arg}\left(z\right)=\frac{\pi}{4}\) then its real and imaginary components are the same (i.e. \(z=a+ai,a\in\mathbb{R}\).
Therefore \(z-3=a-3+ai\) and \(\text{Arg}\left(a-3+ai\right)=\frac{\pi}{2}\), which means that \(z-3\) is a pure imaginary number (i.e. \(\text{Re}\left(z\right)=a-3=0\) giving \(a=3\).
This then gives \(z-6i=3+3i-6i=3-3i\), which is in quadrant 4, so \(\text{Arg}\left(z\right)\in\left(-\frac{\pi}{4},0\right)\).
Finally, this gives \(\text{Arg}\left(z-6i\right)=-\tan^{-1}{\left(\frac{3}{3}\right)}=-\tan^{-1}{\left(1\right)}=-\frac{\pi}{4}\)

Homer

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9243 on: April 15, 2018, 02:11:21 pm »
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Need help with part C  :o
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Buoyancy

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9244 on: April 19, 2018, 10:41:32 pm »
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Hey guys,

I'm having trouble finding the normal gradient of the tangent gradient (2x-y)/(y+x). I've been trying to do -1/tangent gradient to get the normal gradient but I don't get the correct answer.

The correct answer is: (x+y)/(y-2x)
but I end up getting -(y+x)/(2x-y)

appreciate the help :)
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VanillaRice

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9245 on: April 20, 2018, 06:32:37 am »
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Hey guys,

I'm having trouble finding the normal gradient of the tangent gradient (2x-y)/(y+x). I've been trying to do -1/tangent gradient to get the normal gradient but I don't get the correct answer.

The correct answer is: (x+y)/(y-2x)
but I end up getting -(y+x)/(2x-y)

appreciate the help :)
Those answers are the same, you simply factor out a negative from the denominator to get one to look like the other.

Hope that helps :)
« Last Edit: April 20, 2018, 08:47:50 pm by VanillaRice »
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Yertle the Turtle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9246 on: April 24, 2018, 12:27:18 pm »
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I've been really struggling with this question on vectors, didn't even really know how to start. Could someone please help me?
Thanks :)
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jazzycab

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9247 on: April 24, 2018, 02:20:25 pm »
+1
I've been really struggling with this question on vectors, didn't even really know how to start. Could someone please help me?
Thanks :)


Let \(\mathbf{c}=\alpha\mathbf{i}+\beta\mathbf{j}+\gamma\mathbf{k}\).
From the information and supporting diagram, we can see that \(\left|\mathbf{a}\right|=\left|\mathbf{b}\right|=\left|\mathbf{c}\right|\), therefore \(\sqrt{\alpha^2+\beta^2+\gamma^2}=\sqrt{1^2+1^2+1^2}=\sqrt{3}\), which gives \(\alpha^2+\beta^2+\gamma^2=3\ \left(1\right)\).
Because \(\mathbf{b}\) is the vector with magnitude \(\sqrt{3}\) that bisects \(\mathbf{a}\) and \(\mathbf{c}\), we get:

There is an angle of \(2\theta\) between \(\mathbf{a}\) and \(\mathbf{c}\), giving:

We now have 3 simultaneous equations that can be solved to find \(\mathbf{c}\):
« Last Edit: April 24, 2018, 02:22:29 pm by jazzycab »

Mattjbr2

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9248 on: April 25, 2018, 06:11:07 pm »
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Is it valid to say that if \[\frac{x^{2}+x}{x} = \frac{1}{y}\]
Then \[x\neq0 \: and \: y\neq0?\]

(0,0) exists though?
« Last Edit: April 25, 2018, 06:13:11 pm by Mattjbr2 »
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jazzycab

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9249 on: April 25, 2018, 10:11:06 pm »
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Is it valid to say that if \[\frac{x^{2}+x}{x} = \frac{1}{y}\]
Then \[x\neq0 \: and \: y\neq0?\]

(0,0) exists though?

Do you mean that \(\left(0,1\right)\) exists, rather than \(\left(0,0\right)\)?
\(y\ne0\) regardless of how you look at this problem. If what you have written is defined as given in the question, then the left-hand-side gives us the 'indeterminate form' \(\frac{0}{0}\) if \(x=0\).
In VCE Maths, we would certainly say that this is not defined for \(x=0\). We can, however, investigate the limit as \(x\) approaches 0:

So the limit as \(x\) approaches 0 gives \(y\) approaches 1, but this doesn't mean that we don't exclude \(x=0\).

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9250 on: April 25, 2018, 11:46:30 pm »
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Do you mean that \(\left(0,1\right)\) exists, rather than \(\left(0,0\right)\)?
\(y\ne0\) regardless of how you look at this problem. If what you have written is defined as given in the question, then the left-hand-side gives us the 'indeterminate form' \(\frac{0}{0}\) if \(x=0\).
In VCE Maths, we would certainly say that this is not defined for \(x=0\). We can, however, investigate the limit as \(x\) approaches 0:

So the limit as \(x\) approaches 0 gives \(y\) approaches 1, but this doesn't mean that we don't exclude \(x=0\).

At uni, suitable subjects will disregard the 'issue' at x = 0 and just let you sub in x = 0 (I'm looking at you, complex analysis). It's really a matter of pedantry to not identity x/x as 1.
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jazzycab

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9251 on: April 26, 2018, 06:40:02 am »
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At uni, suitable subjects will disregard the 'issue' at x = 0 and just let you sub in x = 0 (I'm looking at you, complex analysis). It's really a matter of pedantry to not identity x/x as 1.
This is true, and why I specified "in VCE Maths..."

Yertle the Turtle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9252 on: May 01, 2018, 09:13:08 am »
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Can I get a hand with this question about trig? I've really got no idea how to start.
Thanks guys. :)
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9253 on: May 01, 2018, 04:13:54 pm »
+3
Can I get a hand with this question about trig? I've really got no idea how to start.
Thanks guys. :)

Hey Yertle the Turtle,

Any time you have the sum of two inverse trig functions, it's a good sign that angle addition formulae may be involved. For part a:
\begin{align}
\tan^{-1}(\frac{1}{m}) + \tan^{-1}(\frac{1}{n}) &= \frac{\pi}{4}\\
\implies \tan(\tan^{-1}(\frac{1}{m}) + \tan^{-1}(\frac{1}{n})) &= \tan(\frac{\pi}{4})\\
\implies \frac{\tan(\tan^{-1}(\frac{1}{m})) + \tan(\tan^{-1}(\frac{1}{n}))}{1-\tan(\tan^{-1}(\frac{1}{m})) \tan(\tan^{-1}(\frac{1}{n}))} &= 1 \text{ (using angle addition on LHS)}\\
\implies \frac{\frac{1}{m} + \frac{1}{n}}{1-\frac{1}{mn}} &= 1\\
\implies \frac{\frac{n+m}{mn}}{\frac{mn-1}{mn}} &= 1\\
\implies n+m &= mn-1\\
\implies mn-m-n &= 1\\
\implies mn-m-n+1 &= 2\\
\therefore (m-1)(n-1) &= 2 \text{, as required}\\
\end{align}

Part b can be solved by finding n in terms of k, and then substituting into the equation given in the question.
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Yertle the Turtle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9254 on: May 08, 2018, 05:24:54 pm »
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snip
Thanks heaps man! :D

Can anyone give me a hand with this question on complex numbers?
Thanks :)
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