VCE Specialist 3/4 Question Thread!

[Will T]

Essentials 12B Question 9c ii.
I cannot get the answer the book gets and I continually get 277.86 degrees.
Help pls.

[Ancora_Imparo]

It may help to draw a quick sketch of the points on a set of axes on some paper.
You know that the initial position is at .
You know that P is at .
Join the two points with a line.
The bearing of P from the initial position is evidently 270 degrees plus the angle made between the line you just drew and the negative x-axis.
To find this angle, you use trig. Let the angle be degrees. You know that , so:

Solve for using tan inverse on your calculator. Make sure you're in degrees. You should get 15.44 degrees (2 decimal places).
Add this to 270 degrees and you get 285.44 degrees, which is the answer!
Hope this helps!

[Jeggz]

Can someone please explain to me how you would find the domain and range of a parametric equation?
For instance, x= sec(t) and y=tan(t) where t is an element of (pi/2 , 3pi/2)
I found the equation to be x^2 - y^2 = 1. But I don't know how to get the domain and range of this fuction?

[Jaswinder]

i also have trouble understanding that!

btw just a total guess but is the range R and domain (-infinity to -1]

[Jeggz]

i also have trouble understanding that!

btw just a total guess but is the range R and domain (-infinity to -1]

CORRECT and CORRECT 
Now please explain to me how you got that!!

[Ancora_Imparo]

For the domain, you look at the expression involving .
, where t is an element of (pi/2 , 3pi/2).
Now, find the range of this expression within the given domain of . You can do this by sketching sec(t) from pi/2 to 3pi/2 (you need to what the graphs of inverse trig functions look like too!) and simply looking at the highest and lowest points on the graph.
Lowest points: At t=pi/2 and t=3pi/2, x=-infinity
Highest point: At t=pi, x=-1
Therefore the domain is (-infinity,-1].

Similarly, for the range, you look at the expression involving .
, where t is an element of (pi/2 , 3pi/2).
Again, find the range of this expression within the given domain of .
Lowest point: At t=pi/2, y=-infinity
Highest point: At t=3pi/2, y=infinity
Therefore the range is (-infinity, infinity) or R.

[Jeggz]

For the domain, you look at the expression involving .
, where t is an element of (pi/2 , 3pi/2).
Now, find the range of this expression within the given domain of . You can do this by sketching sec(t) from pi/2 to 3pi/2 (you need to what the graphs of inverse trig functions look like too!) and simply looking at the highest and lowest points on the graph.
Lowest points: At t=pi/2 and t=3pi/2, x=-infinity
Highest point: At t=pi, x=-1
Therefore the domain is (-infinity,-1].

Similarly, for the range, you look at the expression involving .
, where t is an element of (pi/2 , 3pi/2).
Again, find the range of this expression within the given domain of .
Lowest point: At t=pi/2, y=-infinity
Highest point: At t=3pi/2, y=infinity
Therefore the range is (-infinity, infinity) or R.

LEGEND! THANKS 

[Bad Student]

Let a = 2i -3j + k and b = 3i + j + 5k.

Find a vector which is perpendicular to both a and b.

[BubbleWrapMan]

Where did you get that question? It's most efficiently done by the cross product but that's not part of the course.

[Planck's constant]

Let a = 2i -3j + k and b = 3i + j + 5k.

Find a vector which is perpendicular to both a and b.

Using methods which are available to you in VCE maths, the best you can do is:

Let the vector in question be: v = xi + yj + zk

The dot product of v with each of the two vectors a, b must be zero, giving rise to the following two equations:

2x - 3y + z = 0
3x + y + 5z = 0

Two equations in 3 unknowns give rise to a parametric solution. Let z = lamda etc. (as per Maths Methods 3/4, which may or may not be assumed knowledge for Spesh. I can't remember). All the solutions will be parallel vectors, all perpendicular to a, b

However if the problem had been a little more specific in asking for a UNIT vector which is perpendicular to a, b then you could use a third equation, namely,

X^2 + y^2 + z^2  = 1

This third equation should cut the number of solutions to 2. Two unit vectors, facing in opposite directions, both perpendicular to a, b.

[lzxnl]

@Anonymiza

8x^3-4x-1=0

You're given that 1/4*(sqrt 5 +1) is a root, so you know that 1/4 (-sqrt 5 + 1) is a root as well due to the integer coefficients.
Therefore you know that 8x^3-4x-1=(x-1/4*(sqrt 5+1))(x-(1/4(-sqrt 5 +1))*(8x-8a) where a is the third root of the equation.
Expanding the first two brackets yields ((x-1/4)^2 -5/16)(8x-8a)=(x^2-1/2 x-1/4)*(8x-8a)=8x^3-4x-1
Letting x=0, we find that 2a=-1, a=-1/2 and the fact that x=-1/2 is actually a root shows that our initial reasoning that 1/4 (-sqrt 5 +1) is a root is correct.
Now from your knowledge of cosine values, you know that cos pi/5 is certainly not negative. Two of the roots that you do have are negative, so 1/4*(sqrt 5 + 1) is the only one left. As sqrt 5 is roughly 2.2, 1/4 (sqrt 5 +1) is roughly 0.8, which makes sense given that cos pi/5 is slightly smaller than cos pi/6=sqrt 3/2 or roughly 0.866.

I hope this helps!

[saba.ay]

Can someone please show how I would simplify following. I keep getting the wrong answer. :/

tan(A+B) – tan(A-B)

The answer is :

(2 tan(A) x (1 + (tan(b))^2)) / (1 - ((tan(A)^2) x ((tan(b))^2))

[b^3]

Are you sure its minus between them? The answer you gives results from a plus between them.





Otherwise for the actual question you gave you would get

[zvezda]

Hey, question 16 in exercise 12B in essentials, would it be possible to find a cartesian equation for this?
r(t)=ti+3tj+tk

[Ancora_Imparo]

Hey, question 16 in exercise 12B in essentials, would it be possible to find a cartesian equation for this?
r(t)=ti+3tj+tk

Well, the cartesian plane is 2D, and since this vector involves three dimensions, finding a cartesian equation would not be possible.
The question asks to 'describe the motion'. That does not necessarily mean that you need only one equation to do this.
In this case, you can make three parametric equations. Let the planes be x, y and z.
So: , and

As you can see, in all three planes, the particle is moving linearly, as all equations involve with a degree of 1.
Thus, the particle's overall motion is also simply linearly. You can see this by substituting values for .
When , the particle is at .
When , the particle is at .
When , the particle is at ... etc.

Hope that helps.