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October 01, 2025, 09:25:53 pm

Author Topic: VCE Methods Question Thread!  (Read 5707371 times)  Share 

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Planck's constant

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Re: VCE Methods Question Thread!
« Reply #1455 on: January 20, 2013, 02:19:54 am »
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We use parameters if there are going to be infinite solutions. When are there infinite solutions?
* if you've got more variables to solve for, than you have equations to solve with (and thus, you will have to use replace one of the variables with a parameter)



You can actually have 2 equation in 3 unknowns and no solutions (parallel planes).

Or you can have more equations than variables and still have infinite solutions. Think of 3 or more planes all intersecting along a straight line (in a star formation). That system also has infinite solutions.

However, in Methods 3/4 you will only be asked to deal with the 2 most common situations.

1) 3 planes intersecting at a point (unique solution)
2) 2 planes intersecting at a line (infinite solutions, requiring a single parameter)

EDIT: just had a look at the Essentials textbook, and it turns out that Worked Example 12, in section 2.7 is in fact ....

3) 3 planes intersecting at a line (infinite solutions, requiring a single parameter). This is the star formation scenario.
« Last Edit: January 20, 2013, 02:38:10 am by argonaut »

zvezda

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Re: VCE Methods Question Thread!
« Reply #1456 on: January 20, 2013, 10:40:40 am »
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wouldnt it make more sense if there are infinite solutions when the 3 planes are the same? like with equations with 2 variables?
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polar

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Re: VCE Methods Question Thread!
« Reply #1457 on: January 20, 2013, 11:10:56 am »
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both an intersection on a  plane and a line represent an infinite number of solutions

consider


they intersect on the plane , but in this case, you can't just use one parameter. you'll need to use two. let , then, , hence the set of solutions is - there are an infinite number of solutions here.

if they intersect on a line, there is also an infinite number of solutions, but using one parameter.

ashoni

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Re: VCE Methods Question Thread!
« Reply #1458 on: January 21, 2013, 03:01:46 pm »
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Could someone help me with this question and it explain it thoroughly please :)


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Re: VCE Methods Question Thread!
« Reply #1459 on: January 21, 2013, 04:49:34 pm »
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Firstly we write out the equations in matrix form, that is


Now if the determinate of the 3x3 matrix is zero, then we will either have no solutions or infinite solutions, so we find the determinate using the calculator.

Now for i), we want the determinate to be anything other than zero, so we have


For ii), we want the determinate to be zero, but we then need to distringuish between no solutions and infinite solutions by substituting our result back in.

Since the first two equations can never be the same we have no solutions, so

Hope that helps.
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darklight

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Re: VCE Methods Question Thread!
« Reply #1460 on: January 21, 2013, 09:18:24 pm »
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A piece of wire of length 1m is bent into the shape of a sector a circle of radius a cm and sector angle theta. Let the area of the sector be A cm^2.
a) Find A in terms of a and theta
b) Find A in terms of theta.
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Stick

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Re: VCE Methods Question Thread!
« Reply #1461 on: January 21, 2013, 09:39:10 pm »
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A piece of wire of length 1m is bent into the shape of a sector a circle of radius a cm and sector angle theta. Let the area of the sector be A cm^2.
a) Find A in terms of a and theta
b) Find A in terms of theta.

I'm doing this in degrees.

a)

b)















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darklight

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Re: VCE Methods Question Thread!
« Reply #1462 on: January 21, 2013, 10:08:48 pm »
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I'm doing this in degrees.

a)

b)

- Wait, why can A be expressed as 1-2a?

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Re: VCE Methods Question Thread!
« Reply #1463 on: January 21, 2013, 10:14:22 pm »
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Oh, I'll explain that to you. That's not what I meant. What I've done is created an equation that works out the curved length of a sector. Basically, that curved bit is equal to 1 metre minus the two radii and that can be found by multiplying the total circumference by the fraction next to it.
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darklight

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Re: VCE Methods Question Thread!
« Reply #1464 on: January 21, 2013, 10:28:26 pm »
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Oh, I'll explain that to you. That's not what I meant. What I've done is created an equation that works out the curved length of a sector. Basically, that curved bit is equal to 1 metre minus the two radii and that can be found by multiplying the total circumference by the fraction next to it.

I think I get it now... I just never knew that that was how you worked out the curved length. Are we supposed to know it for methods? As in be able to find arc lengths, etc?
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Stick

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Re: VCE Methods Question Thread!
« Reply #1465 on: January 21, 2013, 10:51:33 pm »
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Well I learnt it but forgot it in GMA and just logically worked it out again. Might be a good idea to take note of it. :)
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darklight

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Re: VCE Methods Question Thread!
« Reply #1466 on: January 22, 2013, 01:18:59 pm »
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No idea where to begin :(
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polar

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Re: VCE Methods Question Thread!
« Reply #1467 on: January 22, 2013, 01:35:21 pm »
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hint: draw a diagram, you'll see that they form a right-angled triangle whose side lengths change

Spoiler
assign positive directions - north and east
start with the first ship - moving west at , hence, .
the other ship - moving south at hence, .

now, use pythagoras' theorem , hence ,

solve for on a CAS, otherwise:



substitute t=1.16h back into c to get c=1.2km.
« Last Edit: January 22, 2013, 01:41:57 pm by polar »

shaiga95

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Re: VCE Methods Question Thread!
« Reply #1468 on: January 22, 2013, 01:54:01 pm »
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why does solving give the answer?
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darklight

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Re: VCE Methods Question Thread!
« Reply #1469 on: January 22, 2013, 01:56:16 pm »
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Thanks polar!  ;D

At a minimum, the gradient is 0. Hence, you have to make the gradient equation equal to 0, to solve for t.
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