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September 15, 2026, 01:58:50 pm

Author Topic: VCE Methods Question Thread!  (Read 6239351 times)  Share 

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JackSonSmith

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Re: VCE Methods Question Thread!
« Reply #9285 on: March 15, 2015, 02:36:26 pm »
0
For the line y = x to be tangent to the curve with equation  y= k / x-1   , What must k equal?

I tried letting k/ x-1 = x but have not found success.

Answer is supposed to be k= -1/4

Can anyone help me?
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soNasty

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Re: VCE Methods Question Thread!
« Reply #9286 on: March 15, 2015, 02:43:13 pm »
+1
find the derivative of y=k/x-1
then make it equal to 1 as y=x (the tangent) has an obvious gradient of 1
solve that for x, once u have x, y is easy to find because at a certain point the tangent line and curve share a point
in this case once you find x, the y value will be the same

after understanding that and getting both x and y values sub both into y=k/x-1 and youll find k

knightrider

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Re: VCE Methods Question Thread!
« Reply #9287 on: March 15, 2015, 03:40:53 pm »
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How come    but    isnt equal to 4x?

Gentoo

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Re: VCE Methods Question Thread!
« Reply #9288 on: March 15, 2015, 03:44:25 pm »
+1
It is equal to 4x.

A few dumb questions:

How come 0/0 isn't equal to 1? Isn't the whole meaning of a fraction (and division for that matter) how many times the denominator goes into the numerator? 0 goes into 0 once, doesn't it? Obviously you can't divide by 0 in situations where the numerator is not equal to 0 because there is no combination of 0s that will add up to any other real number, but isn't this different? EDIT: Wait I'm dumb, 0 could go into 0 any number of times, not just once. >.<

Do we even need to use the matrix method to show the number of solutions that two linear equations (with constants in them) will have? Because I just realised that even referencing and calculating the determinent and letting it equal to zero (let alone if you wanted to show the whole xy matrix times the constants equals the solution matrix bit) is a bit superfluous. The reason why ad-bc = 0 gives infinite/no sols is because it represents when the ratio of the x and y co-efficients are the same which indicates the same gradient; you don't need matrices to do it. e.g.

ax + by = whatever
cx + dy = whatever

You know the lines will have the same gradient (and thus have no/infinite sols) when a/b = c/d ---> ad = bc ---> ad - bc = 0

The whole matrix/letting the determinent = 0 thing seems to purport some other reason for this being the case (like the inverse matrix not existing due to 1/det being on the RHS which would lead to undefined when notionally solving for the xy matrix; which, while also true, is a more roundabout way of doing it) but do we even have to acknowledge that this is the process we're using or does VCAA allow just going straight to the ratios without even mentioning the determinent?

Also, about square roots: According to google, the definition of a square root of a number is such that (where y represents the square roots). This means that you have to consider the negative as well. So why then is f(x)=root(x) even a function? According to that definition we'd have to consider the negative which would make it just the inverse relation of a parabola (more accurately, x^2). And for that matter, why is the square root of x^2 equal to the modulus of x and not +-x?

Does VCAA just define it differently? Because obviously if you're solving x^2 = 25 for example then x=+-5 which VCAA acknowledges. But they wouldn't say that the square roots of 25 are 5 and -5, it'd just be 5. So yeah, what gives?
« Last Edit: March 15, 2015, 06:43:44 pm by Gentoo »

warya

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Re: VCE Methods Question Thread!
« Reply #9289 on: March 15, 2015, 03:46:06 pm »
0
http://i.imgur.com/VK9S9ET.gif

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Gentoo

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Re: VCE Methods Question Thread!
« Reply #9290 on: March 15, 2015, 04:04:52 pm »
+1
It's 2 to the power of (the power, which, when 2 is raised by, equals 4x), which equals 4x. Try it on your CAS calculator if you don't believe me. :p

knightrider

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Re: VCE Methods Question Thread!
« Reply #9291 on: March 15, 2015, 04:12:16 pm »
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How would you solve ?

IntelxD

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Re: VCE Methods Question Thread!
« Reply #9292 on: March 15, 2015, 04:20:50 pm »
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Prove it then

Eulerfan proved that it was an illogical statement. I don't think it is necessary for me to reiterate his point (especially since my reasoning was very similar to his).
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soNasty

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Re: VCE Methods Question Thread!
« Reply #9293 on: March 15, 2015, 04:23:26 pm »
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Cogglesnatch Cuttlefish

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Re: VCE Methods Question Thread!
« Reply #9294 on: March 15, 2015, 04:38:48 pm »
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Just what my opinion is, I have always wondered why the answer cannot equal zero, I mean i proved it above right?
Also literally speaking division can be read as how many times does 0 go into 'n', assuming it is any number, zero would go into it 0 times, right? Some deep stuff here..
Using your logic:
not k=0
So you havent really proved anything  ???
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Cogglesnatch Cuttlefish

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Re: VCE Methods Question Thread!
« Reply #9295 on: March 15, 2015, 04:49:50 pm »
-1
You need to work on your transposing skills there, buddy.



Is equivalent to



HENCE K=0





So you haven't really proved me wrong o.O

hmm... I remember you asking something about trig before.
So is 1/cos(x)=0 equivalent to cos(x)=0 now? So mathematicians have defined sec(x) because they got bored of referring to cos(x)
« Last Edit: March 15, 2015, 04:52:08 pm by Cogglesnatch Cuttlefish »
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Gentoo

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Re: VCE Methods Question Thread!
« Reply #9296 on: March 15, 2015, 05:06:22 pm »
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240*t1=x (1)
320*t2=x (2)
t1+t2=7/12 (3)

Re-arranging equation (3) gives us t1=7/12 - t2

Subbing into (1) gives us 240(7/12 - t2)= x

Then you can solve ^that^ and (2) simultaneously and go from there.

knightrider

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Re: VCE Methods Question Thread!
« Reply #9297 on: March 15, 2015, 05:13:51 pm »
0




and







Thanks soNasty  :)

How did you get from to ?

wobblywobbly

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Re: VCE Methods Question Thread!
« Reply #9298 on: March 15, 2015, 05:17:02 pm »
+1
Thanks soNasty  :)

How did you get from to ?

Negative powers reciprocate (i.e. flip the fraction) of what's inside
:)

cosine

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Re: VCE Methods Question Thread!
« Reply #9299 on: March 15, 2015, 05:18:06 pm »
+1
Thanks soNasty  :)

How did you get from to ?



Multiply the whole expression by -1, leaving the power to 3/5 only:



Now we know if we raise to the power of negative 1, you must reciprocate it, which will get 32/243

and then simplify further :)
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