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March 29, 2024, 09:06:39 am

Author Topic: VCE Methods Question Thread!  (Read 4802806 times)  Share 

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Corey King

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Re: VCE Methods Question Thread!
« Reply #18885 on: October 24, 2020, 01:01:17 pm »
0
I'm gonna be honest - I have no clue what you're asking. What I do know is, looking at your attachment, that's wrong.

Everything is fine up until line 4. When moving from line 3 to line 4, you've removed the square from the brackets. Where has it gone? Remember - things can't just disappear in maths, they have to move somewhere or you must change both sides of the equation. On top of that, when you moved that 17/4 to the other side, you forgot to change its sign to +17/4. To remove the square from the brackets (and this is what the book is doing), you should take the square root of both sides. This will cancel out the squaring of the (x+5/2)^2, and leave a square root on the other side. However, since you don't know if +sqrt(17)/2 or -sqrt(17)/2 led to (x-5/2)^2, you will need to leave the second square root as +/-, just as the book does. The plus/minus thing might sound complicated, so I will come back to it. First, here's how the working should look, for full clarity:




On the plus/minus thing, imagine your equation is x^2=4. This is real simple, right - you square a number, it equals 4. What is that number? Well, there are two numbers that become 4 when you square them - 2^2=4, because 2*2=4. However, (-2)^2 ALSO equals 4, because a negative times a negative makes a positive. Hence, (-2)^2=-2*-2=4. So, the solution to x^2=4 ISN'T x=2, it's actually x=-2 OR 2. We commonly write this as x=-2,2, or as x=plus/minus 2.

Some of that you may have already known, but again, had no clue what you were actually asking, so had to make some guesses. Let me know if I haven't solved your confusion, and maybe try rewording what you're having trouble understanding?

Hey Kelting :)
In your solution here, you find only one solution for x. In parabolas like these however, there are two x intercepts.
How do you find the second following your method?


dogwhip

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Re: VCE Methods Question Thread!
« Reply #18886 on: October 24, 2020, 01:36:18 pm »
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Hi! This is a question from a past VCE examination, Exam 2. The question is as follows:



I attempted to solve this by solving the simultaneous equation:



However, my CAS is unable to give me an exact answer. The output was k = 0.367879 which corresponds to the answer 1/e but I remember seeing this as an extended response question too somewhere else and I would have lost marks in that case because my CAS wasn't able to determine the exact answer.

Does anyone here have any suggestions on what to input into the CAS to get an exact answer?
A screenshot of my CAS input is attached.

The Cat In The Hat

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Re: VCE Methods Question Thread!
« Reply #18887 on: October 24, 2020, 01:40:48 pm »
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Hi! This is a question from a past VCE examination, Exam 2. The question is as follows:



I attempted to solve this by solving the simultaneous equation:



However, my CAS is unable to give me an exact answer. The output was k = 0.367879 which corresponds to the answer 1/e but I remember seeing this as an extended response question too somewhere else and I would have lost marks in that case because my CAS wasn't able to determine the exact answer.

Does anyone here have any suggestions on what to input into the CAS to get an exact answer?
A screenshot of my CAS input is attached.
Ooh! I got stuck on that exact question too! (Or a close facsimile.) I also want the answer. :)
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Corey King

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Re: VCE Methods Question Thread!
« Reply #18888 on: October 24, 2020, 01:50:16 pm »
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Hey guys. Im looking at how the quadratic formula is made through the process of conpleting the square on the general quadratic equation.

I dont understand the algebra at the start though. It looks to me like they are forgetting to put a square sign on the a?
Am I the one making the mistake?
Many thanks,
Corey

james.358

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Re: VCE Methods Question Thread!
« Reply #18889 on: October 24, 2020, 01:59:02 pm »
+7
Hey Kelting :)
In your solution here, you find only one solution for x. In parabolas like these however, there are two x intercepts.
How do you find the second following your method?

Hey Corey,

If you look carefully, there are two solutions which is denoted by the "±" symbol.

Hey guys. Im looking at how the quadratic formula is made through the process of conpleting the square on the general quadratic equation.

I dont understand the algebra at the start though. It looks to me like they are forgetting to put a square sign on the a?
Am I the one making the mistake?
Many thanks,
Corey

Yep this is a mistake on the textbook. Great job exploring the formulas and not taking them for granted  ;D. This "inquiry based" mindset will set you up well for your future studies.

Hi! This is a question from a past VCE examination, Exam 2. The question is as follows:



I attempted to solve this by solving the simultaneous equation:



However, my CAS is unable to give me an exact answer. The output was k = 0.367879 which corresponds to the answer 1/e but I remember seeing this as an extended response question too somewhere else and I would have lost marks in that case because my CAS wasn't able to determine the exact answer.

Does anyone here have any suggestions on what to input into the CAS to get an exact answer?
A screenshot of my CAS input is attached.

Hey Dogwhip,

Sometimes the CAS cannot solve for exact solutions unfortunately. However, if an exact solution is required, it is always possible to solve it by hand.

Let Eq1 be ekx = x, Eq2 be k×ekx = 1.

By subbing Eq1 into Eq2, we have k×x = 1, x = 1/k

Now we can sub this back into Eq1: ek×1/k = x, hence x = e, k = 1/e

Sometimes when trigonometric or exponential equations are involved, the CAS cannot give u an exact answer, so it is worth learning how to do these by hand.

Hope this helps!
James
« Last Edit: October 24, 2020, 02:08:56 pm by james.lhr »
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dogwhip

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Re: VCE Methods Question Thread!
« Reply #18890 on: October 24, 2020, 04:03:55 pm »
+1
Hey Corey,

If you look carefully, there are two solutions which is denoted by the "±" symbol.

Yep this is a mistake on the textbook. Great job exploring the formulas and not taking them for granted  ;D. This "inquiry based" mindset will set you up well for your future studies.

Hey Dogwhip,

Sometimes the CAS cannot solve for exact solutions unfortunately. However, if an exact solution is required, it is always possible to solve it by hand.

Let Eq1 be ekx = x, Eq2 be k×ekx = 1.

By subbing Eq1 into Eq2, we have k×x = 1, x = 1/k

Now we can sub this back into Eq1: ek×1/k = x, hence x = e, k = 1/e

Sometimes when trigonometric or exponential equations are involved, the CAS cannot give u an exact answer, so it is worth learning how to do these by hand.

Hope this helps!
James

Wow! I didn't realise the calculation was so easy by hand ::) Thanks!

Corey King

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Re: VCE Methods Question Thread!
« Reply #18891 on: October 24, 2020, 06:15:45 pm »
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Thanks james, both helpful answers :)



svnflower

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Re: VCE Methods Question Thread!
« Reply #18892 on: October 24, 2020, 10:11:02 pm »
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Hi :)

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

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18893 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

ArtyDreams

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Re: VCE Methods Question Thread!
« Reply #18894 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 »

svnflower

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Re: VCE Methods Question Thread!
« Reply #18895 on: October 24, 2020, 10:40:10 pm »
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Both working outs were super clear, thanks so much to you both :)

Corey King

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Re: VCE Methods Question Thread!
« Reply #18896 on: October 25, 2020, 01:56:50 pm »
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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 :P
Any help is appreciated,
Corey

keltingmeith

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Re: VCE Methods Question Thread!
« Reply #18897 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 :P
Any help is appreciated,
Corey

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


Corey King

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Re: VCE Methods Question Thread!
« Reply #18898 on: October 25, 2020, 02:47:43 pm »
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I mean, to me it just looks like you've just mis-written the quadratic formula. It should be:



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?

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Re: VCE Methods Question Thread!
« Reply #18899 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 \(\frac{-1}{x}\)