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November 28, 2020, 07:05:16 am

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#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18900 on: October 24, 2020, 10:27:08 pm »
+8
Hi

How do I differentiate this (attached below)
Would I have to split the log up first?

No need - just apply the chain rule.

$y=-\ln\left(\frac{x}{2}\right)\\ u=\frac{x}{2}\\ y=-\ln(u) \\ \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}\\ \frac{dy}{dx}=-\frac{1}{u}\times\frac{1}{2}\\ \frac{dy}{dx}=-\frac{2}{x}\times\frac{1}{2}\\ \frac{dy}{dx}=\frac{-1}{x}$

Let me know if you're confused. If you're curious about future problems, I highly recommend derivative-calculator.net - it will include steps for any derivative you want the answer to! See https://www.derivative-calculator.net/#expr=-ln%28x%2F2%29
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#### ArtyDreams

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##### Re: VCE Methods Question Thread!
« Reply #18901 on: October 24, 2020, 10:29:37 pm »
+6
Hi

How do I differentiate this (attached below)
Would I have to split the log up first?

You can use the chain rule!

Let u = x/2
du/dx = 1/2

y = -ln(u)
dy/du = -1/u

dy/dx = dy/du * du/dx
=1/2 * -1/u
= -1/2u

Then sub 'u' back in
dy/dx = -1/(2(x/2))
= -1/x

Hope this helps!

EDIT: beaten by keltingmeith with a much easier to follow solution - will just leave this here anyway.
« Last Edit: October 24, 2020, 10:31:41 pm by ArtyDreams »
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My Short Guide to Mathematical Methods
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#### svnflower

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##### Re: VCE Methods Question Thread!
« Reply #18902 on: October 24, 2020, 10:40:10 pm »
0
Both working outs were super clear, thanks so much to you both

#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18903 on: October 25, 2020, 01:56:50 pm »
0
Hey guys
I've tried this problem twice and come to the same incorrect solution on both occasions.
The working is real messy but hopefully you can see the process
Any help is appreciated,
Corey

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18904 on: October 25, 2020, 02:14:18 pm »
+4
Hey guys
I've tried this problem twice and come to the same incorrect solution on both occasions.
The working is real messy but hopefully you can see the process
Any help is appreciated,
Corey

I mean, to me it just looks like you've just mis-written the quadratic formula. It should be:

$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$
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#### Corey King

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##### Re: VCE Methods Question Thread!
« Reply #18905 on: October 25, 2020, 02:47:43 pm »
0
I mean, to me it just looks like you've just mis-written the quadratic formula. It should be:

$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$

Well whoops :p
At least there weren't any holes in my process.
Most of my incorrect answers seem to come from simple mistakes like this still unfortunately. Is this a common experience for you guys?

#### S_R_K

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##### Re: VCE Methods Question Thread!
« Reply #18906 on: October 25, 2020, 03:43:48 pm »
+6
Hi

How do I differentiate this (attached below)
Would I have to split the log up first?

As others have mentioned, you can differentiate by using chain rule.

However, splitting up the logarithm first is a nice method because you get $-\log(\frac{x}{2})=-\log(x) + \log(2)$ and since $\log(2)$ is a constant, you immediately get the derivative $\frac{-1}{x}$

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18907 on: October 25, 2020, 05:45:59 pm »
+8
Well whoops :p
At least there weren't any holes in my process.
Most of my incorrect answers seem to come from simple mistakes like this still unfortunately. Is this a common experience for you guys?

So, I don't know your personal circumstances. I don't know how you approach answering questions, checking your answers, etc. So sorry if I make an assumption that's wrong

But have you ever tried to proof-read one of your own essays? Ever notice how you proof-read it, and you maybe catch 1 or 2 errors, but most of it looks fine - then you get it back covered in red pen? It's because we suck at finding our own errors. This is common practice, extremely normal for everyone.

I also notice you posting a lot asking questions - which is fine, that's why the thread is here - but quite a few times, they're not "I don't understand this concept" questions, they're "I'm not getting the answer right and I don't know why" questions. And when it's the second type, they're usually examples of you making a silly little mistake that I'm sure you could've caught yourself, because you seem like a very bright kid.

So, how do you stop making these silly little mistakes? Get better at spotting your own mistakes. How do you get better at something nobody is good at to begin with? Practice. Before coming here to ask why you got a question wrong, scour the question. I don't mean just look at it, I don't mean give it a couple of looks over. I mean annotate each line. Make sure each number you put there is correct. Quadruple check all your formula, one pronumeral at a time. It will suck, and it will be long at first, but you will slowly get better at finding your mistakes. And once you've found all your mistakes, by yourself, you'll slowly start to stop making them, because you will have hard-wired those mistakes out of your brain. Just like riding a bicycle. But, if you don't learn to spot your own mistakes, and just keep asking us for help, you won't get any better at it.

Like, it's still fine to ask us questions, go ahead - we're here to help. But it might be worth your time to lean on us a little less, so that you get better at spotting your own mistakes, because that will lead you to stop making them.
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#### james.lhr

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##### Re: VCE Methods Question Thread!
« Reply #18908 on: October 25, 2020, 06:23:57 pm »
+7
Hey Corey!

Just adding to the fantastic advice by Keltingmeith, you should try to follow "The 15 Minute Rule". Basically when you're stuck on something, you should work on it for about 15 minutes before seeking help. During that time you should try to get unstuck on your own, and note down everything you try. This will hopefully train your mind to become more independent and get better at spotting your own mistakes.

And like Keltingmeith said, there really is no shortcut to avoiding careless errors. Just keep practising, and ensure that you don't make the same mistake twice. For example when doing a particular type of question, you should almost subconsciously think about what you have done wrong on the question type before, and doing your best to avoid it this time.

Hope this helps!
James
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#### svnflower

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##### Re: VCE Methods Question Thread!
« Reply #18909 on: October 25, 2020, 06:52:08 pm »
0
Hello again ,

I'm stuck on this question. Do I need to use the Sn formula?

#### keltingmeith

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##### Re: VCE Methods Question Thread!
« Reply #18910 on: October 25, 2020, 07:09:28 pm »
+2
Hello again ,

I'm stuck on this question. Do I need to use the Sn formula?

Hey, so you might have the wrong thread - this thread is for VCE maths methods help, but this isn't actually in the VCE methods curriculum, so unfortunately nobody here is likely to be able to help you. Perchance you meant to go to the HSC boards?

EDIT: Okay, this question got to me, I couldn't just leave it alone. VCE kids, don't worry, you don't need to know how to do this. Anyway, first thing's first - let's figure out what kind of series it is. It doesn't look like it's an arithmetic series - if it were, then we'd see a lot of pluses and not a lot of square roots. Okay, so it's probably geometric - so we need to figure out the common ratio. First, we'll need to remove the sqrt(5), so let's do that:

$\sqrt{5}\left(1+\sqrt{9}+\sqrt{25}+...+\frac{z}{\sqrt{5}}\right)=225\sqrt{5}$

Hang on a second - this is an arithmetic sequence! How sneaky. Watch what happens when divide by that sqrt(5):

$
1+3+5+...+\frac{z}{\sqrt{5}}=225
$

It was in disguise the whole time, this is the sum of odd numbers! Okay, so now we have an equation for the sum of an arithmetic series, and that's:

$
S_n=\frac{n}{2}(a_1+a_n)
$

Except, we don't know how long the sequence is, and a_n is what we're trying to find... Okay, but we have another formula we can use:

$
S_n=\frac{n}{2}\left[2a_1+(n-1)d\right]
$

Now this, we can work with! So, sub in everything we know, and solve for n:

$
225=\frac{n}{2}\left[2+(n-1)2\right]\\
225=n+n(n-1)\\
225=n+n^2-n\
15=n
$

Rejecting the negative answer because, of course, a series must have a positive number of terms. So, we can now sub this into the equation from before to get the answer:

$
S_n=\frac{n}{2}(a_1+a_n)\\
225=\frac{15}{2}\left(1+\frac{z}{\sqrt{5}}\right)\\
\frac{450}{15}=1+\frac{z}{\sqrt{5}}\\
30-1=\frac{z}{\sqrt{5}}\\
29\sqrt{5}=z
\sqrt{4205}=z
$

In which the last step is turning the number into the same form presented, but you could stop at the step before that (and I would encourage you to do this).

Also, a different formula you could've finished with is:

$
T_n=a+(n-1)d\\
\frac{z}{\sqrt{5}}=1+(15-1)2\\
z=\sqrt{5}\left(1+14\times2\right)\\
z=\sqrt{5}\left(1+28\right)\\
z=29\sqrt{5}
$

Pick your poison with this one
« Last Edit: October 26, 2020, 12:14:39 am by keltingmeith »
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#### a weaponized ikea chair

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##### Re: VCE Methods Question Thread!
« Reply #18911 on: October 27, 2020, 07:38:02 pm »
0
i understand how to find the total number of combinations.

Can someone please explain how to find the number of combinations that contain ham? I know that it is 2^7 but I do not understand why or how.
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#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #18912 on: October 27, 2020, 07:46:50 pm »
+5
Hey there!

They mention you can have any combination of a list of ingredients ie. they can either be in the sandwich or not in the sandwich - notice how these are two options. If you're given that the sandwich has ham, then that 'blocks' it off as a choice (ie. you no longer have to make that choice any more) and thus have to choose from the seven other ingredients. You pick one of two choices for seven ingredients, ie 27

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#### a weaponized ikea chair

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##### Re: VCE Methods Question Thread!
« Reply #18913 on: October 27, 2020, 09:13:40 pm »
0
Thanks for the reply, it helped.

Also, I found every other topic to be easy but when my class started combinations and their applications i just sought of hit a roadblock. Is it more difficult than the other topics or is my internal hatred for probability taking its toll?
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#### fun_jirachi

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##### Re: VCE Methods Question Thread!
« Reply #18914 on: October 27, 2020, 09:24:27 pm »
+5
I'd say a bit of both. The internal hatred part makes you less inclined to study it or make sense of it, even if you don't notice this happening (in theory, anyway). Also, it is a hard topic; both because it's notoriously easy to get questions wrong while actually understanding the theory properly. Setting up an answer properly and understanding the wording of the question properly is mostly what screws people over in questions like these.

Hope that makes sense
Failing everything, but I'm still Flareon up.
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