VCE Methods Question Thread!

[brightsky]

When using matrices for transformations when you have more than one transformation applied to a point does it matter when you combine the matrices together which order you put them next to each other like say you have to apply more than one transformation to a point and you want to combine those transformations together does it matter what order you put them next to each other as matrices.

Like for example in this question.
Combine a dilation of factor 2 from the x-axis and factor 3 from the y-axis to find the new
coordinates of the point (x,y)

The order does matter. In general, in order to perform a linear transformation, you must pre-multiply rather than post-multiply. Consider the following sequence of transformations:

1. Dilate by a factor of 2 from the x-axis.
2. Translate 1 unit in the negative direction of the y-axis.
3. Dilate by a factor of 3 from the y-axis.

The matrix equation associated with this sequence is as follows (where semicolon denotes the beginning of a new row):

[x';y'] = [3,0;0,1] ([1,0;0,2][x;y] + [0;-1])

Do you see how, for the last transformation, I pre-multiplied the entire expression ([1,0;0,2][x;y] + [0;-1]) by the transformation matrix? I'd be wrong to do anything else, e.g. post-multiply by the transformation matrix, only pre-multiply [1,0;0,2][x;y] by the transformation matrix, etc..

[knightrider]

The order does matter. In general, in order to perform a linear transformation, you must pre-multiply rather than post-multiply. Consider the following sequence of transformations:

1. Dilate by a factor of 2 from the x-axis.
2. Translate 1 unit in the negative direction of the y-axis.
3. Dilate by a factor of 3 from the y-axis.

The matrix equation associated with this sequence is as follows (where semicolon denotes the beginning of a new row):

[x';y'] = [3,0;0,1] ([1,0;0,2][x;y] + [0;-1])

Do you see how, for the last transformation, I pre-multiplied the entire expression ([1,0;0,2][x;y] + [0;-1]) by the transformation matrix? I'd be wrong to do anything else, e.g. post-multiply by the transformation matrix, only pre-multiply [1,0;0,2][x;y] by the transformation matrix, etc..

Thanks brightsky
so you have to apply the transformations in the order they give you right?

[knightrider]

for this question i dont get how they simplified the matrix to nI

[brightsky]

Thanks brightsky
so you have to apply the transformations in the order they give you right?

Indeed.

for this question i dont get how they simplified the matrix to nI

[n,0;0,n] = n [1,0;0,1] = nI

They simply factorised the n out.

[AirLandBus]

The area given in the worked solutions is correct. Perimeter of a circle is , perimeter for quarter of a circle is and the radius is hence both AB and CD have length .

Then .

You know area of quarter circle is , so you have the area of the two quarter circles at the ends. You also know that as found earlier so you can find the area of the rectangle in terms of x (in terms of x and y it is A = xy).

The domain Euler has suggested is completely different?

[brightsky]

Solve



If a and b are real, positive numbers.

Nigh-impossible to solve by hand.

If you're desperate for an answer, you can try plugging coefficients into the cubic formula:



But tbh...

[knightrider]

Indeed.

[n,0;0,n] = n [1,0;0,1] = nI

They simply factorised the n out.

Thanks Brightsky 
so are you allowed to factor out of matrices?

[knightrider]

How would you do this question?

The point P (3, 4) is transformed to become the point P′ .

Prove that this transformation represents an anticlockwise rotation of θ ° about the origin.

 Calculate the value of θ in degrees.

[keltingmeith]

The domain Euler has suggested is completely different?

My apologies, there - I interpreted "length" wrong, thinking they meant the horizontal distance between the two furthest points. However, they meant the perimeter of the fence.

Thanks Brightsky 
so are you allowed to factor out of matrices?

Yep, nothing wrong with it. As long as you take it out of everything:

How would you do this question?

The point P (3, 4) is transformed to become the point P′ .

Prove that this transformation represents an anticlockwise rotation of θ ° about the origin.

 Calculate the value of θ in degrees.

So we want to show that the transformation (3,4)--->(-2sqrt(3), sqrt(13)) is a simple rotation, AND find the amount of degrees we have rotated. If we have a rotation, then the distance from the origin should be the same - this should be sufficient to show a rotation has occurred. (someone, feel free to argue, though~)

The distance from the origin of the first is 5 (I know by 3-4-5 triangles, you can use the distance between two points to prove it, though). The second point:



And so we have a rotation! Now, the tricky part - finding the angle. So, we draw the points on a cartesian plane, and use some geometry to find our angle. I can draw up a picture if you want, but for now I'll let you try it out yourself. You should get:



Where theta 1 is the angle made by the point (3, 4) with the positive x-axis, and theta 2 is the angle made by the second point with the negative x-axis.

[AirLandBus]

My apologies, there - I interpreted "length" wrong, thinking they meant the horizontal distance between the two furthest points. However, they meant the perimeter of the fence.

Cool. With that in mind. Would you be able to work through the rest of it after the bit where you found your error. Thanks for the help dude.

[dankfrank420]

Hey Euler my Essentials textbook tells me that the angle between two lines is simply theta 2 - theta 1... where did the pi come from?

[knightrider]

The distance from the origin of the first is 5 (I know by 3-4-5 triangles, you can use the distance between two points to prove it, though). The second point:



And so we have a rotation! Now, the tricky part - finding the angle. So, we draw the points on a cartesian plane, and use some geometry to find our angle. I can draw up a picture if you want, but for now I'll let you try it out yourself. You should get:



Where theta 1 is the angle made by the point (3, 4) with the positive x-axis, and theta 2 is the angle made by the second point with the negative x-axis.

Thanks eulerfan101 

How do you know which angle is the one made with the positive x-axis and the one with the negative x-axis.
and also how do you work out theta 1 and theta 2.
Also could you please draw up a picture 

[keltingmeith]

Cool. With that in mind. Would you be able to work through the rest of it after the bit where you found your error. Thanks for the help dude.

This will take a while - I'll edit it into this post after helping the other two so that they're not waiting too long. Check back in about half an hour or so.

EDIT: Already, here's the fix:

a) To find the area, it is easiest to split the shape up into a bunch of little shapes we can work with. For this shape, I'm going to split it up into the two curved bits, and then the long rectangle in the middle.

Now, these two curved bits look like a quarter circle each - which the question confirms. So, their individual area will be . Next, the long rectangle - which has area .

Finally, to get the area of the whole thing, we just add up each section:



b) The length (or perimeter) of the fence is 100 m - to find this length, we need to add up each length around the outside, remembering the blue wall. So, we start with the two curved segments. These are quarters of a circle, so we can use the perimeter of a circle divided by four: . Next is the straight length, much easier - just add y. So, we get:



i) Using the above, we can see that .
ii) Using (i), we have that
iii) Now, y and x MUST be positive, so x,y>0. Since y>0, then 100-pi*x>0 ==> 100>pi*x ==> 100/pi>x, so 0<x<100/pi. Finally, we need to ensure that the AREA isn't negative or zero, and a quick sketch shows that for 0<x<100/pi, the area is positive. So, we take our domain to be 0<x<100/pi

c) We want the area=1000, so we do A(x)=1000 and solve for x. Just do this via your CAS. If you want to know how to do it by hand, it's a matter of taking the 1000 to the other side and using the quadratic formula - which I can promise you will be very tedious.

d) Now we take the problem up one dimension! Here's a handy little guide:

length*new length = area
area*new length = volume
volume*new length = fourth dimensional measurement, theorised by many a physicist to be time. (don't ask me, I do maths. :3 )

So, with this in mind:

i) We want the volume, so we do area*new length=A(x)*x/50. So:



ii) Just use the x value that gives you an area of 1000m^2, and chuck this into the volume equation.

iii) Do V(x)=500m^3, and solve for x. Do this by CAS, do not even ATTEMPT to solve the cubic otherwise.

Hey Euler my Essentials textbook tells me that the angle between two lines is simply theta 2 - theta 1... where did the pi come from?

Depends on how you define the angles - see my picture for how I've defined my angles differently, leading to the formula I've got.

Thanks eulerfan101 

How do you know which angle is the one made with the positive x-axis and the one with the negative x-axis.
and also how do you work out theta 1 and theta 2.
Also could you please draw up a picture 

See attachment.

Now, by drawing the two points, we can turn each of them into a triangle, as shown. Then, finding the angle is just a matter of doing , using all values as a positive value (see the coloured numbers next to each side). Then, the angle of rotation (or angle between the two lines) is going to be 180 degrees (or pi radians) minus the angles in the triangles.

[knightrider]

Now, by drawing the two points, we can turn each of them into a triangle, as shown. Then, finding the angle is just a matter of doing , using all values as a positive value (see the coloured numbers next to each side). Then, the angle of rotation (or angle between the two lines) is going to be 180 degrees (or pi radians) minus the angles in the triangles.

Thanks eulerfan101 
i was just wondering how did you get for the use of finding angles.How is this derived for finding angles.

[AirLandBus]

This will take a while - I'll edit it into this post after helping the other two so that they're not waiting too long. Check back in about half an hour or so.

EDIT: Already, here's the fix:

a) To find the area, it is easiest to split the shape up into a bunch of little shapes we can work with. For this shape, I'm going to split it up into the two curved bits, and then the long rectangle in the middle.

Now, these two curved bits look like a quarter circle each - which the question confirms. So, their individual area will be . Next, the long rectangle - which has area .

Finally, to get the area of the whole thing, we just add up each section:



b) The length (or perimeter) of the fence is 100 m - to find this length, we need to add up each length around the outside, remembering the blue wall. So, we start with the two curved segments. These are quarters of a circle, so we can use the perimeter of a circle divided by four: . Next is the straight length, much easier - just add y. So, we get:



i) Using the above, we can see that .
ii) Using (i), we have that
iii) Now, y and x MUST be positive, so x,y>0. Since y>0, then 100-pi*x>0 ==> 100>pi*x ==> 100/pi>x, so 0<x<100/pi. Finally, we need to ensure that the AREA isn't negative or zero, and a quick sketch shows that for 0<x<100/pi, the area is positive. So, we take our domain to be 0<x<100/pi

c) We want the area=1000, so we do A(x)=1000 and solve for x. Just do this via your CAS. If you want to know how to do it by hand, it's a matter of taking the 1000 to the other side and using the quadratic formula - which I can promise you will be very tedious.

d) Now we take the problem up one dimension! Here's a handy little guide:

length*new length = area
area*new length = volume
volume*new length = fourth dimensional measurement, theorised by many a physicist to be time. (don't ask me, I do maths. :3 )

So, with this in mind:

i) We want the volume, so we do area*new length=A(x)*x/50. So:



ii) Just use the x value that gives you an area of 1000m^2, and chuck this into the volume equation.

iii) Do V(x)=500m^3, and solve for x. Do this by CAS, do not even ATTEMPT to solve the cubic otherwise.

Thanks mate so much. Once i got the domain correct, the rest of it was a breeze. So when you say x and y has to be positive, that is >0, this is because the area of the box can only be positive? And as a result can only be when x and y are both positive? The reason why i was getting the wrong domain was cause i was using the Area formula and solving x for 0, which got me 200/pi as the other root when graphed. Why is this wrong?