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August 24, 2019, 07:58:45 am

Author Topic: VCE Methods Question Thread!  (Read 2409050 times)  Share 

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Phy124

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Re: VCE Methods Question Thread!
« Reply #60 on: January 02, 2012, 05:19:55 pm »
0
Is the answer, no solution ??

cause i find it easiest to just put these into a matrix..

1  4    3
2  1   -1
3 -1  -4

then find the determinant, which for this one = 0. so no solution.. ?
just becareful that if the determinant is 0 it does not necessarily mean no solutions, could mean infinite solutions as well ;)

is there a way to proceed onwards from there to figure out which one it is using matrices ? 

Determinant

= (1 x 1 x -4) + (4 x -1 x 3) + (3 x 2 x -1) - (3 x 1 x 3) - (4 x 2 x -4) - ( 1 x -1 x -1)

= -4 + -12 + -6 - 9 - -32 -1

= -4 - 12 - 6 - 9 + 32 - 1

= 0

Wait, I think you may have been asking something else, my mistake.
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TrueTears

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Re: VCE Methods Question Thread!
« Reply #61 on: January 02, 2012, 05:20:30 pm »
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Is the answer, no solution ??

cause i find it easiest to just put these into a matrix..

1  4    3
2  1   -1
3 -1  -4

then find the determinant, which for this one = 0. so no solution.. ?
just becareful that if the determinant is 0 it does not necessarily mean no solutions, could mean infinite solutions as well ;)

is there a way to proceed onwards from there to figure out which one it is using matrices ? 

yeah there is, but i won't get into the details, it'd involve too many definitions and concepts beyond vce for now lol, but yeah basically you just have to reduce to (reduced) row echelon form and you can usually figured it out from there, dw about this, like i said way beyond vce methods.

Deceitful Wings

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Re: VCE Methods Question Thread!
« Reply #62 on: January 03, 2012, 04:11:18 pm »
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Can you help with this question?

i) Use calculus to find in the form kx + ly = n, where k,l and n are whole numbers, the equation of the normal to the curve at x=-1
ii) Find the point on the curve such that the equation of the normal is 20y+4x=51

Hope you can help :P

TrueTears

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Re: VCE Methods Question Thread!
« Reply #63 on: January 03, 2012, 04:12:28 pm »
+4
Can you help with this question?

i) Use calculus to find in the form kx + ly = n, where k,l and n are whole numbers, the equation of the normal to the curve at x=-1
ii) Find the point on the curve such that the equation of the normal is 20y+4x=51

Hope you can help :P
for i) rearrange to make y the subject, find dy/dx, then rmb the normal gradient is -1/m

Deceitful Wings

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Re: VCE Methods Question Thread!
« Reply #64 on: January 04, 2012, 06:11:34 pm »
0
Can you help with this question?

i) Use calculus to find in the form kx + ly = n, where k,l and n are whole numbers, the equation of the normal to the curve at x=-1
ii) Find the point on the curve such that the equation of the normal is 20y+4x=51

Hope you can help :P
for i) rearrange to make y the subject, find dy/dx, then rmb the normal gradient is -1/m

Thanks :P but two things,
I don't know what the y value for when x=-1 :P
could you help with ii) ?
« Last Edit: January 04, 2012, 06:21:22 pm by Deceitful Wings »

Phy124

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Re: VCE Methods Question Thread!
« Reply #65 on: January 04, 2012, 06:25:19 pm »
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Can you help with this question?

i) Use calculus to find in the form kx + ly = n, where k,l and n are whole numbers, the equation of the normal to the curve at x=-1
ii) Find the point on the curve such that the equation of the normal is 20y+4x=51

Hope you can help :P
for i) rearrange to make y the subject, find dy/dx, then rmb the normal gradient is -1/m

Thanks :P
and what about ii)?
Was there an equation for the curve given with this question, or am I missing something?

From a quick read I think you should do the following:

- Rearrange so you have y in terms of x
- Get the gradient of that equation, we'll call that "m"
- So the derivative of the curve will = -1/m, at the point of intersection (the place the normal occurs)
- Make the derivative of the curve = -1/m, this will give you x coordinate
- put x coordinate back into either equation to give y coordinate
« Last Edit: January 04, 2012, 06:29:34 pm by Phy124 »
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Deceitful Wings

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Re: VCE Methods Question Thread!
« Reply #66 on: January 04, 2012, 06:36:45 pm »
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Can you help with this question?

i) Use calculus to find in the form kx + ly = n, where k,l and n are whole numbers, the equation of the normal to the curve at x=-1
ii) Find the point on the curve such that the equation of the normal is 20y+4x=51

Hope you can help :P
for i) rearrange to make y the subject, find dy/dx, then rmb the normal gradient is -1/m

Thanks :P
and what about ii)?
Was there an equation for the curve given with this question, or am I missing something?

From a quick read I think you should do the following:

- Rearrange so you have y in terms of x
- Get the gradient of that equation, we'll call that "m"
- So the derivative of the curve will = -1/m, at the point of intersection (the place the normal occurs)
- Make the derivative of the curve = -1/m, this will give you x coordinate
- put x coordinate back into either equation to give y coordinate
i think they might have left out something..
thanks for your help :P

TrueTears

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Re: VCE Methods Question Thread!
« Reply #67 on: January 04, 2012, 07:54:26 pm »
+2
Can you help with this question?

i) Use calculus to find in the form kx + ly = n, where k,l and n are whole numbers, the equation of the normal to the curve at x=-1
ii) Find the point on the curve such that the equation of the normal is 20y+4x=51

Hope you can help :P
for i) rearrange to make y the subject, find dy/dx, then rmb the normal gradient is -1/m

Thanks :P
and what about ii)?
Was there an equation for the curve given with this question, or am I missing something?

From a quick read I think you should do the following:

- Rearrange so you have y in terms of x
- Get the gradient of that equation, we'll call that "m"
- So the derivative of the curve will = -1/m, at the point of intersection (the place the normal occurs)
- Make the derivative of the curve = -1/m, this will give you x coordinate
- put x coordinate back into either equation to give y coordinate
i think they might have left out something..
thanks for your help :P
they didnt ;D here's the working if you're still stuck:





gradient of normal is

the normal also passes through x=-1 so

So you have the point (-1, \frac{k+n}{l}) on the normal and you know the gradient of the normal is , now just plug into y = mx+c and you're done ;)

Deceitful Wings

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Re: VCE Methods Question Thread!
« Reply #68 on: January 04, 2012, 08:14:00 pm »
+2
Can you help with this question?

i) Use calculus to find in the form kx + ly = n, where k,l and n are whole numbers, the equation of the normal to the curve at x=-1
ii) Find the point on the curve such that the equation of the normal is 20y+4x=51

Hope you can help :P
for i) rearrange to make y the subject, find dy/dx, then rmb the normal gradient is -1/m

Thanks :P
and what about ii)?
Was there an equation for the curve given with this question, or am I missing something?

From a quick read I think you should do the following:

- Rearrange so you have y in terms of x
- Get the gradient of that equation, we'll call that "m"
- So the derivative of the curve will = -1/m, at the point of intersection (the place the normal occurs)
- Make the derivative of the curve = -1/m, this will give you x coordinate
- put x coordinate back into either equation to give y coordinate
i think they might have left out something..
thanks for your help :P
they didnt ;D here's the working if you're still stuck:





gradient of normal is

the normal also passes through x=-1 so

So you have the point (-1, \frac{k+n}{l}) on the normal and you know the gradient of the normal is , now just plug into y = mx+c and you're done ;)

Thanks heaps!
sorry for the constant spam of questions but i have another one :P
I am trying to plough through my holiday analysis task so i can actually begin my revision :D

the number of stationary points the cubic function y=dx3+ax2+bx+c has will depend on the values of a,b and d. there are three possible cases
Case 1: 2 stationary points
Case 2: 1 stationary point
Case 2: no stationary points

Explain why a2-3bd > 0                  <-- i have no idea where they got this from and what to do this

dc302

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Re: VCE Methods Question Thread!
« Reply #69 on: January 04, 2012, 08:17:02 pm »
+2
That expression is the determinant of the derivative. If the determinant > 0, then you get 2 real solutions and so 2 stationary points, so this only applies to case 1.
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Insa

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Re: VCE Methods Question Thread!
« Reply #70 on: January 08, 2012, 02:36:05 am »
0
Hey guys,

Can anyone tell me how the turning point of this equation is (3/2,-8)? I got (3,-8).

(2x-3)^2 - 8

 Thanks c:
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dc302

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Re: VCE Methods Question Thread!
« Reply #71 on: January 08, 2012, 03:49:44 am »
0
If y = (2x-3)^2 - 8, then dy/dx = 4(2x-3)

To find the turning point you let dy/dx = 0 and solve for x. This gives x = 3/2. Also, (3,-8) isn't even on the graph so how did you get this?
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Insa

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Re: VCE Methods Question Thread!
« Reply #72 on: January 08, 2012, 04:11:01 am »
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Ah I see. The textbook I use never taught me this method (dy/dx) to find turning points. The book teaches me to use "a(x-b)^2+c" where (b,c) is the turning point.

May I ask another question? How do you know when to use dy/dx in any question?

Thanks for the reply c:
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dc302

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Re: VCE Methods Question Thread!
« Reply #73 on: January 08, 2012, 11:50:10 am »
0
You can still use a(x-b)^2+c to find the turning point, the only problem is your quadratic was not in this form. To get it into this form:

y=(2x-3)^2 - 8
y=(2[x-1.5])^2 - 8
y=4(x-1.5)^2 - 8

b=1.5 = 3/2, c=8

And you should use dy/dx whenever you need to find the gradient. The turning point is when the gradient = 0, so this is why we used it.
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Hutchoo

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Re: VCE Methods Question Thread!
« Reply #74 on: January 09, 2012, 09:42:08 pm »
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Sup.
Stuck on these two questions. I'm usually okay with the 'show this = that' sorta questions, but this one is weird (for me).
I looked at the worked solutions but they make no sense to me.

Anyway, see the attached images.
Thanks.
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