VCE Specialist 3/4 Question Thread!

[Cort]

Thank you both! I'll be sure to ask more questions. Really helpful, thanks.

[Cort]

While looking at parametric equations for circles & ellipses (by the way, how important is knowing how to transpose from Cartesian -> Parametric?) it states that the domain/range of the Cartesian is determined by the parametric equations.  My question is: How and why does the example actually 'get' the domain and range of the equation? I can see a pattern of using the parametric equation and doing the opposite of addition/subtraction in the parametric and they're just adding/subbing the values without the cos/sin. Is there a reason why this occurs? Sorry if it's confusing.

[b^3]

We know that our value is given by , but we want to know what values our can take given that  .

For , will range from to . To find the domain we need to find the maximum and minimum values that can take. So when is a minimum? When , i.e. .
When is a maximum? When , i.e. .
Hence the domain is .

So really they're adding and subtracting the values without the cosine because in this case that will give you the maximum and minimum values for and . Just note that this will work as long as the values allowed for are for a whole period of the trigonmetric funcitons, otherwise you will need to check what values and will take and adjust accordingly.

A similar situation goes for the range and possible values.

Hope that helps!

[Cort]

We know that our value is given by , but we want to know what values our can take given that  .

For , will range from to . To find the domain we need to find the maximum and minimum values that can take. So when is a minimum? When , i.e. .
When is a maximum? When , i.e. .
Hence the domain is .

So really they're adding and subtracting the values without the cosine because in this case that will give you the maximum and minimum values for and . Just note that this will work as long as the values allowed for are for a whole period of the trigonmetric funcitons, otherwise you will need to check what values and will take and adjust accordingly.

A similar situation goes for the range and possible values.

Hope that helps!

Ah! God damn trigo values. Thanks a lot-- I'll keep my eyes peeled for what x can take in.

[Cort]

I've came across an interesting thingo whilst continuing on with my work. The image states that t between -pi/2 and pi/2 is the RIGHT side of the hyperbola, and the other side is the LEFT side of the hyperbola. Is there a reason why it's the right side (when it's clearly left if you view it normally)? Is it necessary to understand that, or it won't be tested much in spesh? Furthermore, do I also have to write down the left side of the hyperbola, even when it just asks for the domain (which is clearly the right side) ?

How important is this in regards to the spesh exam?

[Stick]

That explanation really isn't worded well at all.

Think about your general hyperbola graph of . You have two branches - one is on the positive side of the x axis and the other is on the negative side of the x axis. Now, when we define a hyperbola using parametric equations, we make x and y subjects of two separate equations using a common third parameter. If you have a look at the graph of the parametric equation describing the x values, we can see that the part between and is on the positive side of the x axis (which is the vertical axis in this case). Hopefully you can now understand why this part of the graph corresponds with the right hand side of the hyperbola - they are both on the side with positive x values.

That explanation sort of implies that the part between and will always be the right branch. That is not the case.

To answer your other question, yes, this is an important concept that regularly pops up in Specialist Maths assessments. Not only do these questions arise as is, but later in the course you will be applying this skill in vector functions.

I hope this helps.

[BubbleWrapMan]

Firstly, the parametrisation of a curve isn't unique. For one, you could easily define , which will reflect the whole hyperbola in the -axis, and of course give you the same curve. This is the main reason you're not asked to give a parametrisation of a curve in an exam; rather, they're given to you, and you answer questions about them.

That said, you could reasonably be asked which values of give you the right/left branches of a hyperbola. But before I explain, could you clarify what you mean by "it's clearly left if you view it normally"?

[Stick]

Firstly, the parametrisation of a curve isn't unique. For one, you could easily define , which will reflect the whole hyperbola in the -axis, and of course give you the same curve. This is the main reason you're not asked to give a parametrisation of a curve in an exam; rather, they're given to you, and you answer questions about them.

That said, you could reasonably be asked which values of give you the right/left branches of a hyperbola. But before I explain, could you clarify what you mean by "it's clearly left if you view it normally"?

I think he means that the bit between and is the left half of the given graph and so is reasoning that it should represent the left branch of the hyperbola.

[Cort]

Firstly, the parametrisation of a curve isn't unique. For one, you could easily define , which will reflect the whole hyperbola in the -axis, and of course give you the same curve. This is the main reason you're not asked to give a parametrisation of a curve in an exam; rather, they're given to you, and you answer questions about them.

That said, you could reasonably be asked which values of give you the right/left branches of a hyperbola. But before I explain, could you clarify what you mean by "it's clearly left if you view it normally"?

Right, sorry about that. Stick clarifies it below. And just to reinforce the new stuff I'm learning here: it isn't unique because the hyperbola is just the same as , correct? So since it's the same curve, it's not, erm, special? That makes sense.

In another light though, I've only noticed that only when you're given a parametric equation does the book ask you to find the domain and range. Is there a reason for this? Can you not find a domain/range in a normal (general) equation? Does that mean most of the questions I've been asked in tests/books been a parametric equation? Or is there some sort of magic I'm not being told that parametric equation = realm of domains and ranges? Sorry about the questions.

[lzxnl]

I think the point of the book's questions is to see whether or not you can find the domain or range given parametric equations which may themselves have a restricted domain for the parameter t. Finding maximal domains and ranges of normal relations is probably not prioritised as much in spesh. That's my theory.

[BubbleWrapMan]

A good way to understand parametrisations is to try and plot one out from scratch. Trig functions can be a bit messy to start off with, so take something simpler like and . Pick different values of and plot the corresponding points on an - graph. For example, for you get and , so plot the point (1,2). Then imagine running through all values of , and you'll see you end up with a continuous curve.

If you think of as time, then this can be thought of as the motion of an object along a path. From experience, you can move along a path fast or slow, and that's why a parametrisation isn't unique. If and , you get the same curve in the - plane as in the example above, but this one is traced out 'faster', if you imagine as time.

For your question, the given curve is drawn on - axes - it's a completely different curve to the hyperbola in question. The purpose of the - graph is to show what values takes for given values of . From the graph, is positive for . Again thinking of as time, this tells you that in that period of time, the particle is travelling on the right side of the - plane, since the -values are positive. The horizontal () coordinate is irrelevant in this regard, since left or right is determined by the value of only.

But the particle could have just as easily travelled on the left side first. That's what you get if you let . Then is negative for , but you still get the same path. The reason you get the same path is because the hyperbola is symmetric about the -axis (as well as the -axis). So by making all the -values negative, you get the same path, but you get different points for the same values of .

[Isithalal]

Find the equation of the circle which passes through, (3,1), (8,2), (2,6)

simultaneous equations? I have no idea, tried and failed

answer is
(x-5)^2+(y-4)^2=13

Thanks in advance

[Eugenet17]

Substitute the coordinates into the form of (x-h)^2+(y-k)^2=r^2 and you'll get 3 equations. From there you can equate the equations together and find the values of k in terms of h which will allow you to find the value of h (and hence k). Substitute h and k into one of the original equations and you'll find the radius

[Eugenet17]

Couple of Questions:

1. Let z = cis(theta), Show that 1/z=cis(-theta)

2. What is the conjugate of -3cis(2pi/3)?

[abcdqd]

Couple of Questions:

1. Let z = cis(theta), Show that 1/z=cis(-theta)

Use De Moivre's thereom:

2. What is the conjugate of -3cis(2pi/3)?

We can re-write as , which, using polar form multiplication properties, is equal to .
Now, since we are reflecting the complex number in the x-axis when taking its conjugate, we simply place a negative in front of the principle argument.
So the final answer is