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Author Topic: VCE Methods Question Thread!  (Read 5738044 times)  Share 

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alchemy

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Re: VCE Methods Question Thread!
« Reply #3360 on: December 25, 2013, 03:32:19 pm »
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....You can now, apply the formula to find the distance.
Alternatively, try not to think of 'formulas' and take a more practical approach to understanding this problem. Such an approach comes especially in the future, when understanding topics such as locus, coordinate geometry, etc.

How do you find the distance between, say , where you're standing and the nearest door? You could measure it using a ruler, counting feet, palm measures, etc. Now try to imagine yourself and the door positioned on the graph, where positions are relative to the origin . Believe it or not, this makes measuring all that bit easier for you. Your position is . The doors position is and it's exactly 1 unit above you, but some unknown distance to the right or left of you. You can find the vertical distance from yourself and the door since you know yours and it's y-coordinates. To find this vertical distance you simply subtract the doors y-coordinates from your own y- coordinates. That means: . Thus, the door is exactly two units above you. You can apply the same approach to find the horizontal distance between you and the door. That means: is the horizontal distance between you and the door. This might scare you at first since you don't know what a is, but all is well still because the question asks to find an expression for the distance "in terms of the given variables". The given variable in this case is .

Now think about the information that you have gathered. You've got the vertical and horizontal distances between you and the door. You should be able to imagine a right angle triangle, at least in terms of distances between the points. If being an 'unknown' is preventing you from doing this just imagine it to be any close point along the x-axis. The thing that should ring a bell now is the 'Pythagorean Theorem'. This allows you to find the hypotenuse of a triangle. The hypotenuse, in our case, is the distance between you and the door. So if you substitute the values for horizontal & vertical distances, the hypotenuse should equate to:







Phy124

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Re: VCE Methods Question Thread!
« Reply #3361 on: December 25, 2013, 09:10:14 pm »
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Hey thanks, i've done like a whole chapter worth of these questions but because I've done so many its like i've remembered what to rather knowing what I'm doing.  I actually don't really understand the question
Well I feel conceptual understanding is paramount to success in mathematics and as such I advise you to really try and think about the question at hand rather than just jumping into it based on similarity to other questions.

I'm not really sure of any particular methods to aid in understanding the situations more thoroughly so hopefully someone else can you help you there.

So are we trying to find an equation that crosses/intersects with y=-3x+4 and 5x-3y+40=0?  If so i understand we need a gradient and x1 and y1 so we can use the formula am i right?  But i don't understand which gradient we find an which x1 and y 1 we find
We are trying to find the equation of a line that goes through the point of intersection of the lines y=-3x+4 and 5x-3y+40=0.

So you have the two lines y=-3x+4 and 5x-3y+40=0 which intersect at (x,y) = (-2,10).

This is our point (x,y) = (x1,y1) i.e. x1 = -2, y1 = 10

But as you said to find the equation of a line we need to know the gradient of the line.

We were told that the line for which are trying to find the equation is parallel to another line and such it will have the same gradient as that line.

The gradient for that other line which went through the points (-8,5) and (0,4) would be -1/8 and hence the gradient of the line for which we are finding the equation would be m = -1/8.

Now we have the gradient of the line and a point and can find the equation using the formula y-y1= m(x-x1)
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Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3362 on: December 25, 2013, 09:58:05 pm »
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Alternatively, try not to think of 'formulas' and take a more practical approach to understanding this problem. Such an approach comes especially in the future, when understanding topics such as locus, coordinate geometry, etc.

How do you find the distance between, say , where you're standing and the nearest door? You could measure it using a ruler, counting feet, palm measures, etc. Now try to imagine yourself and the door positioned on the graph, where positions are relative to the origin . Believe it or not, this makes measuring all that bit easier for you. Your position is . The doors position is and it's exactly 1 unit above you, but some unknown distance to the right or left of you. You can find the vertical distance from yourself and the door since you know yours and it's y-coordinates. To find this vertical distance you simply subtract the doors y-coordinates from your own y- coordinates. That means: . Thus, the door is exactly two units above you. You can apply the same approach to find the horizontal distance between you and the door. That means: is the horizontal distance between you and the door. This might scare you at first since you don't know what a is, but all is well still because the question asks to find an expression for the distance "in terms of the given variables". The given variable in this case is .

Now think about the information that you have gathered. You've got the vertical and horizontal distances between you and the door. You should be able to imagine a right angle triangle, at least in terms of distances between the points. If being an 'unknown' is preventing you from doing this just imagine it to be any close point along the x-axis. The thing that should ring a bell now is the 'Pythagorean Theorem'. This allows you to find the hypotenuse of a triangle. The hypotenuse, in our case, is the distance between you and the door. So if you substitute the values for horizontal & vertical distances, the hypotenuse should equate to:





Thanks for the explanation but i still don't get it and also someone said the same thing as you did don't look at questions with formulas etc, but why not?  Isn't that the reason for the formula?
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grannysmith

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Re: VCE Methods Question Thread!
« Reply #3363 on: December 25, 2013, 10:05:28 pm »
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Because if you're just relying on formulas, it's going to make it harder to solve more abstract questions

Only Cheating Yourself

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Re: VCE Methods Question Thread!
« Reply #3364 on: December 25, 2013, 10:33:14 pm »
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Because if you're just relying on formulas, it's going to make it harder to solve more abstract questions

so what should i do now?
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Re: VCE Methods Question Thread!
« Reply #3365 on: December 25, 2013, 11:16:43 pm »
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so what should i do now?
I feel your problem has to do with the use of unknowns in the working out.

If we were asked to find the distance between the points: (a,1) (2,3)

If we were told that "a" was equal 1, then we'd have the points (1,1) and (2,3) which would make the answer easier to find right?

Let's say we have 2+3+3, that's 8.
What's 2+3+a? 5+a
Well we don't know because "a" is unknown, so we find the simplest way we can express it.
If "a" changes, then 5+a will change.
If we were told "a" then we could easily calculate but for now all we can do is simplify.

So back to the question, we don't know one of the points so we decide to label it "a".

So we can put "a" into the distance formula and simplify to find the distance in terms of "a". If we change "a", the distance is going to change.

Hope this helps in some way!


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Re: VCE Methods Question Thread!
« Reply #3366 on: December 25, 2013, 11:20:36 pm »
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I feel your problem has to do with the use of unknowns in the working out.

If we were asked to find the distance between the points: (a,1) (2,3)

If we were told that "a" was equal 1, then we'd have the points (1,1) and (2,3) which would make the answer easier to find right?

Let's say we have 2+3+3, that's 8.
What's 2+3+a? 5+a
Well we don't know because "a" is unknown, so we find the simplest way we can express it.
If "a" changes, then 5+a will change.
If we were told "a" then we could easily calculate but for now all we can do is simplify.

So back to the question, we don't know one of the points so we decide to label it "a".

So we can put "a" into the distance formula and simplify to find the distance in terms of "a". If we change "a", the distance is going to change.

Hope this helps in some way!

Yea it does thanks BUT

this question lost me!


Find the distance between each of the following pairs of points in terms of the given variables

(5,b) (0,6)

So as you said we can't have a direct answer as we don't know b so we put it in the simplest form

5-0=5square =25+ 6-b=6b square= 36bsquare
square root 25+36bsquare

But the answer is wrong its 12b?
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alchemy

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Re: VCE Methods Question Thread!
« Reply #3367 on: December 25, 2013, 11:38:25 pm »
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Thanks for the explanation but i still don't get it and also someone said the same thing as you did don't look at questions with formulas etc, but why not?  Isn't that the reason for the formula?

That's how formulas are formed. That's how algorithms are created. Have you heard of Informatics? Here are some stuff that will blow your mind: http://orac.amt.edu.au/cgi-bin/train/problem.pl?set=simple1&problemid=372

Because if you're just relying on formulas, it's going to make it harder to solve more abstract questions

This is what I mean. The way to solve that problem is the same way we've discussed. Of course, with just knowledge of formulas you're going to have a very tough time. Just shows how practical thinking really comes in handy.
« Last Edit: December 25, 2013, 11:41:21 pm by Sheldon Cooper »

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Re: VCE Methods Question Thread!
« Reply #3368 on: December 25, 2013, 11:53:13 pm »
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Yea it does thanks BUT

this question lost me!


Find the distance between each of the following pairs of points in terms of the given variables

(5,b) (0,6)

So as you said we can't have a direct answer as we don't know b so we put it in the simplest form

5-0=5square =25+ 6-b=6b square= 36bsquare
square root 25+36bsquare

But the answer is wrong its 12b?
Make sure you keep your pairs of coordinates labelled as coordinate set 1 or 2.
You used x1-x2 then used y2-y1. You see how you've suddenly changed the 2 and 1.
Use (b-6) now.
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Re: VCE Methods Question Thread!
« Reply #3369 on: December 25, 2013, 11:54:31 pm »
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Make sure you keep your pairs of coordinates labelled as coordinate set 1 or 2.
You used x1-x2 then used y2-y1. You see how you've suddenly changed the 2 and 1.
Use (b-6) now.

it becomes the same thing in the ends doesn't it?
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grannysmith

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Re: VCE Methods Question Thread!
« Reply #3370 on: December 26, 2013, 05:40:53 am »
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No it doesn't; if you do x1-x2 then it also has to be y1-y2

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Re: VCE Methods Question Thread!
« Reply #3371 on: December 26, 2013, 10:01:38 am »
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No it doesn't; if you do x1-x2 then it also has to be y1-y2

can you work it out please?
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grannysmith

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Re: VCE Methods Question Thread!
« Reply #3372 on: December 26, 2013, 10:52:46 am »
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d=sqrt{(0-5)^2+(6-b)^2}
=sqrt{25+(6-b)^2}
=5+6-b
=11-b


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Re: VCE Methods Question Thread!
« Reply #3373 on: December 26, 2013, 11:20:21 am »
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d=sqrt{(0-5)^2+(6-b)^2}
=sqrt{25+(6-b)^2}
=5+6-b
=11-b

Umm, I think you may have made a mistake there.

cannot be simplified to

Think of it as putting the whole equation to the power of like this
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Re: VCE Methods Question Thread!
« Reply #3374 on: December 26, 2013, 12:16:31 pm »
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guys theres a 12b for the answer of that question?
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