VCE Methods Question Thread!

[zhen]

Thank you! But do u know why exactly cos (-x) = cos(x) and elepaborate on what you mean by identity

If you take cos(π/3) or cos(-π/3). They are both the same. This is because the "x-values", which are the values of cos in a unit circle are the same. Another way to think about it is that if you reflect across the x-axis of a unit circle, the x-values remain the same. The attached image should help you better visualise it. By trig identity I mean that it's like a rule that we just have to remember.

[LPadlan]

Find the equation of the straight line which passes through the point (1,6) and is: perpendicular to the line with equation y=2x+3

[zhen]

Find the equation of the straight line which passes through the point (1,6) and is: perpendicular to the line with equation y=2x+3

The line perpendicular to y=2x+3 will have a gradient of -1/2
So, y=-1/2x+c
Sub in (1,6)
6=-1/2+c
c=13/2
So y=-(1/2)x+13/2

[LPadlan]

The length of the line segment joining A(2,-1) and B(5,y) is 5 units. Find y

[MightyBeh]

The length of the line segment joining A(2,-1) and B(5,y) is 5 units. Find y

Two ways to do this one; both are pretty much the same.

1. Draw a triangle and use pythagoras theorem (this question is very easy if you notice it's a pythagorean triad: 3-4-5)

2. Use the distance formula, sub in your values and solve algebraically. \(D = \sqrt{(x_2 - x_1)^2 + (y_2 + y_1)^2}\). This will give you \(5 = \sqrt{(5-2)^2 + (y+1)^2}\). Doing this you might end up with only \(y=3\), but it's important to note that \(y=-5\) is an equally valid answer. If you only get one answer, you need to go back and consider that \(y\) is to the power of two, so there will be two solutions.

Let me know if I wasn't clear enough, or you would like a line by line solution.

[TheCommando]

If you take cos(π/3) or cos(-π/3). They are both the same. This is because the "x-values", which are the values of cos in a unit circle are the same. Another way to think about it is that if you reflect across the x-axis of a unit circle, the x-values remain the same. The attached image should help you better visualise it. By trig identity I mean that it's like a rule that we just have to remember.

Ahhhci get it, thank you

[LPadlan]

Hi can someone tell me a way to break down Application problems? I really don't know how to tackle the problems.
Example:
A car journey of 300 km lasts 4 hours. Part of this journey is on a freeway at an average speed of 90 km/h. The rest is on country roads at an average of 70 km/h. Let T be the time(in hours) spent on the freeway.
b)i) state the distance travelled on the freeway in terms of T
  ii) state the distance travelled on country road in terms of T
c)i)Find T
 ii)Find the distance travelled on each type of road

[QueenSmarty]

Hi can someone tell me a way to break down Application problems? I really don't know how to tackle the problems.
Example:
A car journey of 300 km lasts 4 hours. Part of this journey is on a freeway at an average speed of 90 km/h. The rest is on country roads at an average of 70 km/h. Let T be the time(in hours) spent on the freeway.
b)i) state the distance travelled on the freeway in terms of T
  ii) state the distance travelled on country road in terms of T
c)i)Find T
 ii)Find the distance travelled on each type of road

First of all, since the journey lasts for 4 hours, 4-T hours are spent on country roads

b)i) The distance travelled on the freeway in terms of T would be 90T km (since 90km/h for T hours gives a distance of 90 x T km)
b)ii) 70km/h for 4-T hours gives a distance of 70(4-T) km
c)i) 90T + 70(4−T) = 300 km.
20T=20
therefore, T=1
c)ii) 90(1)= 90km on the freeway
70(4-1)= 210km on country roads

Hope this helped!

[Wota]

Hi, how do you do question 5bii? And why is the answer

Spoiler

0<x<12

?

Thanks!!

[zhen]

Hi, how do you do question 5bii? And why is the answer

Spoiler

0<x<12

?

Thanks!!

2y+16+2x=64
y>x (from the diagram)
2y+2x=48
Assuming that y is a number that is only a tiny bit bigger than x.
You get 4x+an infinitely small value=48
So, 4x<48
x<12
The inequality appears, as if x cannot equal 12, as if x=12, then y=12, which defies the propery y>x
To find the other limit, you just think that if x=0, then the shape cannot exist. But if x approaches 0 (becomes a really small positive number) than the shape does exist.
So x>0
Combining these 0<x<12
So the general idea is to find the maximum and minimum allowable values for x and y by looking at the largest and smallest possible values for x and y

[TheCommando]

Like my teacher always says, no maths question is dumb. I'll just start you off with the first few.
a) There's an identity that says cos(-x)=cos(x) and sin(-x)=-sin(x). So, cos(-a)=cos(a)=0.6
b) sin(π/2+a)=cos(a), which is another trig identity. Basically sin(π/2-a)=sin(π/2+a)=cos(a) because they are the same triangle on the unit circle. For these questions just be careful of the quadrants and signs. So, the answer is sin(π/2+a)=cos(a)=0.6
c) tan(-θ)=-tan(θ)=-0.7
d) cos(π/2-x)=sin(x)=0.3
I may have made a mistake so someone else should check, but from this, you should be able to do the rest.

I know how to do these questions but i dont particularly understand these concepts as the book explanation is very brief and doesnt explain it.
So why is -sin (pie/2 - theta) =a= cos theta
- cos(pie/2 -theta)= b =sin theta
Why is sin (pie/2 + theta)=a =cos theta
Cos (pie/2 +theta) =-b =-sin theta

[LPadlan]

For each of the following, state a transformation which maps the graph of y=f(x) to the graph of y=f1(x):

a)f(x)=x2, f1(x)=(x+5)^2

Can someone please break this down and explain?

[Syndicate]

For each of the following, state a transformation which maps the graph of y=f(x) to the graph of y=f1(x):

a)f(x)=x2, f1(x)=(x+5)^2

Can someone please break this down and explain?

Assuming:
y = x^2 and y' = (x'+5)^2,

y = y'
Therefore, no transformations has been applied to y.

x = x'+5
x' = x-5 (I made x' the argument in order to figure out what transformations have been applied to x)
Therefore, there is a translations of 5 units in the negative direction of x-axis (which is true. Check by graphing the function if you want to).

[LPadlan]

Assuming:
y = x^2 and y' = (x'+5)^2,

y = y'
Therefore, no transformations has been applied to y.

x = x'+5
x' = x-5 (I made x' the argument in order to figure out what transformations have been applied to x)
Therefore, there is a translations of 5 units in the negative direction of x-axis (which is true. Check by graphing the function if you want to).

Hey, thanks for the reply. But how you calculated those things are a bit unclear for me. Can you clarify by doing another example? f(x)=(1/x^2) -3, f1(x)=1/x^2. I tried your formula for this question but it did not work. Thanks

[Syndicate]

Hey, thanks for the reply. But how you calculated those things are a bit unclear for me. Can you clarify by doing another example? f(x)=(1/x^2) -3, f1(x)=1/x^2. I tried your formula for this question but it did not work. Thanks

No problem 

In this case:
y =\( \frac{1}{x^2} - 3 \) and y' = \(\frac{1}{x'^2} \)

Take 3 to the other side, as it is not part of the original function (\(\frac{1}{x^2} \)).
Therefore y' =  y+3 (Notice that it is already in the form we require it in).

and x' = x (no transformation has been applied to x).
There was only one transformation, which was a translation of 3 units in the positive direction of the y-axis.

For these questions, it is important to figure out, what parts can be taken to y, and what parts must stay with x (like I did with y =\( \frac{1}{x^2} - 3 \) by taken 3 to the other side).
If f(x) =\( \frac{2}{x^2} -3 \) , then I would add 3 to y, and divide it by 2. So \(\frac{y+3}{2} = \frac{1}{x^2}\), and equate this to y' = \( \frac{1}{x'^2} \).

EDIT: 600th post!