VCE Methods Question Thread!

[alchemy]

for c why do you need to times whole thing by 4? and not just the b value? also, how did they change the signs? if they divided by -1 wouldnt it be -4a + ...

a) Apply reflection in x-axis:
(x',y') --> (x,-y)
y'=-y
-y= a + b ln(x-2)
y = -a - b ln(x-2)

c) Apply dilation of 4 in x-axis:
(x',y') --> (x,4y)
y'=4y
y=y'/4
y'/4 = -a-b ln(x-2)
y' = -4a -4b ln(x-2)

Apply reflection in y-axis:
(x',y') --> (-x,y)
x'=-x
y' = -4a -4b ln(-x-2)
   = -4a +4b ln (x+2)

Btw, the answer given 4a+4b ln(x+2) would correspond to a reflection in the x-axis....so maybe there's a mistake in the suggested solutions you provided, but someone might want to correct me if I've made a mistake in my working.

[Zues]

i always wonder why

c) Apply dilation of 4 in x-axis:
(x',y') --> (x,4y)

it is 4y? i remember being explained this before but didnt quite grasp the concept. basically , all i know is if i have 4x, the input value is always the same but the output value changes according to the "4" or another value there. e.g. in this case (1,1) in y = x to y=4x is (1,4) ? why do you denote it as 4y, doesnt this suggest the y val of 4 multiplied 4 times?

[knightrider]

How would you do this question?

[keltingmeith]

i always wonder why

c) Apply dilation of 4 in x-axis:
(x',y') --> (x,4y)

it is 4y? i remember being explained this before but didnt quite grasp the concept. basically , all i know is if i have 4x, the input value is always the same but the output value changes according to the "4" or another value there. e.g. in this case (1,1) in y = x to y=4x is (1,4) ? why do you denote it as 4y, doesnt this suggest the y val of 4 multiplied 4 times?

The way I like to think of it is pulling a rope tied to the curve, and we're pulling it AWAY from the axis in question. So, a dilation of 4 in the x-axis means we tie the rope to the curve, and pull up until we're 4x the distance as we were before. Logically, since we pulling the curve up, the y-value only should change, not the x-value. And, so that's why we do the transformations as you've indicated above.

How would you do this question?

First, find the determinant:



Now swap the leading diagonal, make the other two elements negative, and divide by the determinant:



Two fun facts:

1. The method used to find the inverse is a simplified form of what we call the "adjoint method", and can be generalised for higher dimensional matrices. You do not need to know the adjoint method, as long as you know that:



2. The matrix given is actually a special type of transformation matrix called a "rotation matrix", and we can use it like any other form of transformation matrix to rotate any graph counterclockwise by the angle theta. Not needed to know for VCE, and you probably won't see it outside linear graphs (even then, VCAA will give you the matrix), but just some fun tidbits~

[knightrider]

First, find the determinant:



Now swap the leading diagonal, make the other two elements negative, and divide by the determinant:



Two fun facts:

1. The method used to find the inverse is a simplified form of what we call the "adjoint method", and can be generalised for higher dimensional matrices. You do not need to know the adjoint method, as long as you know that:



2. The matrix given is actually a special type of transformation matrix called a "rotation matrix", and we can use it like any other form of transformation matrix to rotate any graph counterclockwise by the angle theta. Not needed to know for VCE, and you probably won't see it outside linear graphs (even then, VCAA will give you the matrix), but just some fun tidbits~

Thanks Eulerfan101 

you know for this part you know that this equals 1 from the Pythagorean identity right?

[Zues]

Find the values of A and k, given that the graph of y = A log10 (kx) passes through the points (1, 1) and (3, 2). Give your answers correct to 4 decimal places where appropriate.

edit: what i did

sub the points
1=Alog10(k)  [1]
 2 = A log10(3k) [2]

[2] - [1]

1 = A log10(3k) - Alog10(k)
1 = Alog10 (3k/k)
1 = Alog10(3)

A = 2.0959

sub into [1]
where [1] is 1=Alog10(k)

1 = 2.0959log10(k)
0.477 = log10(k)
k = 3


Is there another way of doing this? i dont get what the solution did in the 3rd/4th/5th line

[IndefatigableLover]

Find the values of A and k, given that the graph of y = A log10 (kx) passes through the points (1, 1) and (3, 2). Give your answers correct to 4 decimal places where appropriate.

If they're asking for decimal places then you should be using your CAS so..

Using your CAS:
[Menu][3][7][1] and then input your equations in and you'll get a=2.0959 and k=3

EDIT: I didn't see your edit
What the solutions did in the 3rd-5th line was apply basic logarithmic law by splitting the log and subtracting it from the other side.

[Zues]

i always do the working until last step before using cas  more practice

[Zues]

how do you graph these type of questions in q3? if someone can do a for example

and why does the horizontal asymptote change to 1?

thanks

[brightsky]

Consider the basic function y = e^x - 1. This is an exponential function with base e and horizontal asymptote y = -1. Sketch this function. Now, to sketch y = |e^x - 1|, all you need to do is reflect any part of the graph of the basic function y = e^x - 1 that lies beneath the x-axis in the x-axis. Since the horizontal asymptote y = -1 lies beneath the x-axis, you must reflect that in the x-axis as well in order to obtain the horizontal asymptote of the new function  y = |e^x - 1|. Do you see how after you reflect y = -1 in the x-axis, you get y = 1? This is why the horizontal asymptote of  y = |e^x - 1| is y = 1.

Q3 a) is a little harder than the question in the worked example. We are required to sketch y = e^(|x-1|) + 4. Due to the complexity of this function, we need to have a plan of attack:

1. Sketch y = e^(x) + 4. This is an exponential function with base e and horizontal asymptote y = 4.
2. Sketch y = e^(|x|) + 4. All you need to do to sketch this function is delete any part of y = e^(x) + 4 from Step 1 that lies to the left of the y-axis, and then reflect the part which lies to the right of the y-axis in the y-axis.
3. Sketch y = e^(|x-1|) + 4. All you need to do to sketch this function is translate y = e^(|x|) + 4 from Step 2 one unit in the positive direction of the x-axis (i.e. right).

Follow the steps above and you should have no issues sketching the required function.

[AirLandBus]

Can someone walk through this question with me please?

[keltingmeith]

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 of the fence is 100 m - to find the length, we need to measure it from top to bottom. We can see that the blue segment does this quite nicely, so we take the sum of all those lengths (x+y+x=2x+y), and equate that to 100. So, 2x+y=100.

i) Using the above, we can see that y=100-2x.
ii) Using (i), we have that
iii) Now, y and x MUST be positive, so x,y>0. Since y>0, then 100-2x>0 ==> 100>2x ==> 50>x, so 0<x<50. Finally, we need to ensure that the AREA isn't negative or zero, and a quick sketch shows that for 0<x<50, the area is positive. So, we take our domain to be 0<x<50

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.

[AirLandBus]

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 of the fence is 100 m - to find the length, we need to measure it from top to bottom. We can see that the blue segment does this quite nicely, so we take the sum of all those lengths (x+y+x=2x+y), and equate that to 100. So, 2x+y=100.

i) Using the above, we can see that y=100-2x.
ii) Using (i), we have that
iii) Now, y and x MUST be positive, so x,y>0. Since y>0, then 100-2x>0 ==> 100>2x ==> 50>x, so 0<x<50. Finally, we need to ensure that the AREA isn't negative or zero, and a quick sketch shows that for 0<x<50, the area is positive. So, we take our domain to be 0<x<50

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.

Now i am even more confused.
The worked solutions.

[Phy124]

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).

[knightrider]

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)