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April 27, 2024, 03:37:44 pm

Author Topic: VCE Methods Question Thread!  (Read 4832551 times)  Share 

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Shadowxo

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Re: VCE Methods Question Thread!
« Reply #14610 on: March 20, 2017, 03:44:01 pm »
0
Why finding the derivative of trig functions, why must the angle be in radians?why not degrees?

Angle in degrees is radians*180/pi, so this means when you multiply by the derivative of the inside of the bracket, it'll result in a completely different answer. The derivative of trig functions needs it to be in radians for this reason, so you must use radians (or convert degrees to radians) in order to use this rule. Hard to explain but, basically, it's only accurate for radians.
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Gogo14

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Re: VCE Methods Question Thread!
« Reply #14611 on: March 20, 2017, 03:52:50 pm »
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Angle in degrees is radians*180/pi, so this means when you multiply by the derivative of the inside of the bracket, it'll result in a completely different answer. The derivative of trig functions needs it to be in radians for this reason, so you must use radians (or convert degrees to radians) in order to use this rule. Hard to explain but, basically, it's only accurate for radians.
But why must you use radians?like fundamentally, why cannot you use degreess?
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #14612 on: March 20, 2017, 11:32:00 pm »
+3
Why finding the derivative of trig functions, why must the angle be in radians?why not degrees?

It comes from the derivation of the derivative of sin x.

Let

Then
(I won't prove this identity; uses the fact that sin(a+b) - sin(a-b) = 2 sin(b) cos (a), and then let x + h = a + b, x = a - b, solve for a and b and plug into above formula)

The derivative limit definition then is


The cosine term can be moved out by continuity of the cosine function. This gives


It can be proved, using a geometric argument comparing the area of an inscribed triangle with the corresponding sector and a tangential triangle, that the final limit is 1 if and only if h is measured in radians. If it's measured in degrees, you'll need a conversion factor here.
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Perryman

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Re: VCE Methods Question Thread!
« Reply #14613 on: March 21, 2017, 10:13:55 am »
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Can someone please help with this question:

Find the coordinates of the points of intersection of the line with equation y=x+k and the parabola with equation y=x^2+2x, where k>0

I have no idea on how to go with this and have asked teachers and they aren't that sure either!

Thanks!

evandowsett

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Re: VCE Methods Question Thread!
« Reply #14614 on: March 21, 2017, 10:29:37 am »
+2
Can someone please help with this question:

Find the coordinates of the points of intersection of the line with equation y=x+k and the parabola with equation y=x^2+2x, where k>0

I have no idea on how to go with this and have asked teachers and they aren't that sure either!

Thanks!

Hey Perryman,

Here's a worked solution for it. All you need to recognise is that when you do points of intersection between two graphs, you simply need to make the functions equal each other. Then solve for x and sub those values back into one of the equations to get the y coordinate. I know it looks strange with the k value, but just treat it like any other number and you'll be fine.

« Last Edit: March 21, 2017, 10:31:21 am by evandowsett »
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Perryman

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Re: VCE Methods Question Thread!
« Reply #14615 on: March 21, 2017, 01:52:20 pm »
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Ok thanks!

I jst got confused on solving the equation because the cas calculator obviously wouldnt factorise it to find x with the k in the equation....
and i totally forgot about that x= equation!

thanks again!

evandowsett

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Re: VCE Methods Question Thread!
« Reply #14616 on: March 21, 2017, 02:31:15 pm »
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Ok thanks!

I jst got confused on solving the equation because the cas calculator obviously wouldnt factorise it to find x with the k in the equation....
and i totally forgot about that x= equation!

thanks again!

Yeah, it can be confusing especially when you're not used to these questions, but you can actually use the solve function (MENU > 3 > 1) to find x, then simply put x back into the function to find y.

No worries! :)
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Guideme

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Re: VCE Methods Question Thread!
« Reply #14617 on: March 21, 2017, 07:06:48 pm »
0

Can anyone help me with these questions pls thank you! Even when I plug them into my calculator I can't figure it out! Thank you.
:0 :)

TooLazy

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Re: VCE Methods Question Thread!
« Reply #14618 on: March 21, 2017, 07:12:38 pm »
+2
I was doing the same question and my CAS was giving me trippy answers.
My teacher showed me but i forgot :/
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evandowsett

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Re: VCE Methods Question Thread!
« Reply #14619 on: March 22, 2017, 09:10:58 am »
+1
(Image removed from quote.)
Can anyone help me with these questions pls thank you! Even when I plug them into my calculator I can't figure it out! Thank you.

See below :)




Let me know if you have any questions!
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Sine

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Re: VCE Methods Question Thread!
« Reply #14620 on: March 22, 2017, 11:13:31 am »
0
(Image removed from quote.)
Can anyone help me with these questions pls thank you! Even when I plug them into my calculator I can't figure it out! Thank you.
21.
Questions like this should never be done mathematically but logically to conserve time.

First define the function g(x) on your calculator.
Simply input all the different options until you get a result that says "true".

22.
Also best done logically and not mathematically. I would be subbing in different values of a into the equation and solving until you have restricted yourself to a particular answer.

imo it may seem nicer to do it properly but it's a MCQ so you can get away with these sort of methods. It saves a lot of time since you aren't really thinking about how to tackle the question. (also a really good tactic if you are stuck)

evandowsett

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Re: VCE Methods Question Thread!
« Reply #14621 on: March 22, 2017, 01:53:12 pm »
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Sine, clearly it's good practice to get used to your calculator when doing these sorts of questions, but I think it's extremely important to first have a deep understanding of what the question is asking rather than simply subbing in numbers until you get the correct answer.

Guideme, the fact that you needed help with this tells me that you weren't sure how to do this mathematically (which is obviously completely fine and I'm more than happy to help!). However, before diving into using your calculator for shortcuts, for this question or any others in future, it's important that you know how to solve it mathematically. If you only have a 'rote-learned' surface level of understanding, you'll falter when questions in future are slighly different. That's why I wrote out the proper solutions for you.

Once you know how to do these questions properly with confidence and accuracy, that's when you should start perfecting your calculator skills to save time in exams and SACs :)
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deStudent

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Re: VCE Methods Question Thread!
« Reply #14622 on: March 22, 2017, 06:28:43 pm »
0
http://m.imgur.com/a/hy6Nt

C)I)Am I doing something wrong or is this question not possible? The book's worked solution is the same with what I have but they used log base 10 for some reason.

Regardless, they both have this extra 1/log(2) term.

Thanks

evandowsett

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Re: VCE Methods Question Thread!
« Reply #14623 on: March 22, 2017, 06:43:02 pm »
+2
http://m.imgur.com/a/hy6Nt

C)I)Am I doing something wrong or is this question not possible? The book's worked solution is the same with what I have but they used log base 10 for some reason.

Regardless, they both have this extra 1/log(2) term.

Thanks

Hey deStudent, I'm not actually too sure where you pulled that log(2) from in your third last step. It looks like it's just been inserted there for no particular reason :) check out this and let me know if you have any questions!

« Last Edit: March 22, 2017, 06:45:12 pm by evandowsett »
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Perryman

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Re: VCE Methods Question Thread!
« Reply #14624 on: March 22, 2017, 07:20:41 pm »
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any help on this will be appreciated!

Solve the following equation for x, where c is a positive constant:
x^2 - 3cx = 10c^2

i have got it to x = (10c^2)/(x-3c)

but cant extract the the x to the left hand side.........is there a way of doing this??or is this the closest its going to be??