VCE Specialist 3/4 Question Thread!

[zvezda]

You have shown that for there to be a line that is parallel to the y axis (this will include ) that . Now we also have a relationship between and , which is . Substituting in will give a second equation with just and . You know have two equations with and only, and those two unknowns, so you should be able to solve for .

Hope that helps.

EDIT:

Spoiler

 

Is there a reason why you have 216^4 instead of 216^6???

[b^3]

Yes there is... it's an error on my part. I'll fix it up now, but you'll probably get those factors multiplied by another factor that doesn't have a solution, hence why the answers came out correct.

Fixed it now, if should be noted that if this was a tech active assessment then it would be best to calc it, rather than going through all of that.

[Stick]

If someone could provide a methodical, foolproof way of tackling questions like the one above, I'd really appreciate it. I worked it out but it was probably more luck than Maths. The solutions supplement is not helpful, because it just tells you which transformations to apply. -.- You can tell that this style of extended response question is a lot more reasonable (and more likely to come up on an exam) than some of the stuff Essentials throws at you - seriously, a lot of this stuff requires a lot more than just a thorough knowledge of the content. XD

[brightsky]

sketch the graph of f:[-pi/3,pi/3] --> R, f(x) = sec(x)
you will see that you need to perform a reflection in the x-axis. this creates the following graph:
f:[-pi/3,pi/3] --> R, f(x)=-sec(x)
now you want the archway to be 4 m in height. currently it is 1 m in height. so dilate from the x-axis by a factor of 4. the result is:
f:[-pi/3,pi/3] --> R, f(x) = -4 sec(x)
now you want the archway to be 6 m in width. currently it is 2pi/3 m in width. so dilate from the y-axis by a factor of 9/pi. the result is:
f:[-3,3] --> R, f(x) = -4 sec(pi/9*x)
now perform appropriate transformations. currently the bottom left point on the archway is (-3,-8). so we need to translate the graph 3 units in the pos dir of the x-axis and 8 units in the pos dir of the y-axis. the result is:
f:[0,6] --> R, f(x) = -4 sec(pi/9*(x-3))+8= -4sec(pi/9*x - pi/3) + 8
so a = -4, b= pi/9, c = -pi/3 and d = 8

always approach these questions systematically; apply reflections and dilations before translations.

[Stick]

Thanks.

[saba.ay]

Could someone please help with this question?
I have to sketch the complex relation given in e) on a complex plane.

I'm having trouble trying to deduce the Cartesian equation. :/ The equation I get doesn't match the graph provided in the answers.

Please and thank you.

[abcdqd]

sqrt(x^2+(y+1)^2) / sqrt((x-1)^2+y^2) = sqrt 2
(x^2+(y+1)^2)/((x-1)^2+y^2)=2
(x^2+(y+1)^2)= 2*((x-1)^2+y^2)
x^2+y^2+2y+1=2(x^2-2x+1+y^2)
x^2+y^2+2y+1= 2x^2 - 4x +2+2y^2
0= x^2-4x+y^2-2y+1
complete the square for x
0= X^2-4x+4-4+y^2-2y+1
4= (x-2)^2+(y-1)^2

[saba.ay]

sqrt(x^2+(y+1)^2) / sqrt((x-1)^2+y^2) = sqrt 2
(x^2+(y+1)^2)/((x-1)^2+y^2)=2
(x^2+(y+1)^2)= 2*((x-1)^2+y^2)
x^2+y^2+2y+1=2(x^2-2x+1+y^2)
x^2+y^2+2y+1= 2x^2 - 4x +2+2y^2
0= x^2-4x+y^2-2y+1
complete the square for x
0= X^2-4x+4-4+y^2-2y+1
4= (x-2)^2+(y-1)^2

Thank you so much !

[Limista]

What exactly is meant by a "unit" vector? How is it different from a normal vector?

I know that the formula for unit vector = (vector)/(size of vector)..........what is actually happening in this formula?

[pi]

A unit vector has a length of 1 (ie. magnitude = 1).

[silverpixeli]

A unit vector is defined as a vector in any direction that has a magnitude (size, length) of one unit
The formula is vector/size as you have said, here's what is actually happening.
If you take say a vector (imagine an actual arrow, that helps me) that is 3 units right. it has a size (3) and a direction (right)
we want to find it's unit vector. so we get the vector (3 units right) and divide it by its size (3)
if you break a vector that is 3 units long into 3 parts, (dividing it by 3) you get a vector that is one unit long in the same direction. This is the unit vector.
It also works if the vector's length starts off less than one unit. say you have a vector that is defined as 1/2 up. it has a size (half a unit) and a direction (up)
if you take the vector (1/2 up) and divide it by the size (1/2) you end up with (1/2)/(1/2) up which is the same as 1 up
This new vector (1 up) is a vector in the same direction as the original, but with a length of 1 unit.
Most of the time we're dealing with vectors in more than 1 dimension, though (the above examples are obviously only in one dimension) but the same holds for vectors in 2 and 3 dimensions.
say you have a vector 20 units long at 30 degrees from true north, if you divide it by 20 you end up with a vector 1 unit long 30 degrees from true north.

[polar]

What exactly is meant by a "unit" vector? How is it different from a normal vector?

I know that the formula for unit vector = (vector)/(size of vector)..........what is actually happening in this formula?

also, lets say that then, ,

take the length of the vectors on either side,

since just scales the vector, we can take it out of the modulus signs to get

now, since (thanks 2/cos(c) great name lol), this simplifies to

therefore, dividing a vector by its length gives the unit vector which is a vector in the same direction as the original vector but with a magnitude of 1 unit.

[Limista]

^ Explanation was great, thanks Polar, but can I deduce the following from what you said: the size of a unit vector is always going to be 1? That's why it's a UNIT (as in 1 unit) vector?

[Phy124]

now, since

I think you meant

^ Explanation was great, thanks Polar, but can I deduce the following from what you said: the size of a unit vector is always going to be 1? That's why it's a UNIT (as in 1 unit) vector?

The magnitude of a unit vector will always be 1, yes, for the reason you gave

[Limista]

Thanks 2/cos(c)! It was a bit silly of me not to realize before -- it's kind of like "unit circle" in circular functions/trigonometry anyway!