VCE Specialist 3/4 Question Thread!

[Stevensmay]

Can you please help me with this question? I've been stuck on it for ages.

"Prove that the diagonals of a parallelogram bisect each other." (It's a vector proof question.)

Thanks in advance!

EDIT: Skipped over the bit about it being a vector proof. My apologies.

https://www.khanacademy.org/math/geometry/quadrilaterals-and-polygons/v/proof---diagonals-of-a-parallelogram-bisect-each-other

Pretty good explanation. Video form is generally better for these style questions.

[devilsadvocate]

https://www.khanacademy.org/math/geometry/quadrilaterals-and-polygons/v/proof---diagonals-of-a-parallelogram-bisect-each-other

Pretty good explanation. Video form is generally better for these style questions.

Thanks for the video
But, as the question is a vector proof question, I'd like to know how the concept can be proved using vectors; the video only explains how the result can be achieved through congruency.

[Lasercookie]

Can you please help me with this question? I've been stuck on it for ages.

"Prove that the diagonals of a parallelogram bisect each other." (It's a vector proof question.)

Thanks in advance!

So say we have a parallelogram ABCD. What are we trying to prove? That the diagonals bisect each other. What does bisect mean? To divide into equal parts. So this means that the diagonals intersect with each other at their midpoint. So rephrasing, what are we trying to prove? We want to show that the diagonals, that is the vectors AC and BD have the same mid-point. Let's label the midpoint of AC to be and the midpoint of BD to be .

A picture helps here, so here's one from google


It's clear that the vector that stretches from A to M_1 is half the length of AC. That's just pulling things out from the definition. Similarly, take advantage of the other piece of information we have, that our shape is a parallelogram.

I don't want to give the game away, so I'll leave it here for now. If you have more trouble just type up your thoughts and we can help you out.

[devilsadvocate]

So say we have a parallelogram ABCD. What are we trying to prove? That the diagonals bisect each other. What does bisect mean? To divide into equal parts. So this means that the diagonals intersect with each other at their midpoint. So rephrasing, what are we trying to prove? We want to show that the diagonals, that is the vectors AC and BD have the same mid-point. Let's label the midpoint of AC to be and the midpoint of BD to be .

A picture helps here, so here's one from google
(Image removed from quote.)

It's clear that the vector that stretches from A to M_1 is half the length of AC. That's just pulling things out from the definition. Similarly, take advantage of the other piece of information we have, that our shape is a parallelogram.

I don't want to give the game away, so I'll leave it here for now. If you have more trouble just type up your thoughts and we can help you out.

Haha, got it thanks

[M_BONG]

Can anyone please me out with this vector proof question?

"Using a vector method only, prove that any two major diagonals of a cube bisect each other".

I had a go and checked the solutions manual for my textbook; somehow they defined a midpoint "O" in the cube to prove that they bisect. I thought the point of a bisect proof is to show a midpoint exists, not to define it at the start.

Help please? Thanks!

[lzxnl]

Can anyone please me out with this vector proof question?

"Using a vector method only, prove that any two major diagonals of a cube bisect each other".

I had a go and checked the solutions manual for my textbook; somehow they defined a midpoint "O" in the cube to prove that they bisect. I thought the point of a bisect proof is to show a midpoint exists, not to define it at the start.

Help please? Thanks!

What I would say is that firstly, you prove the two major diagonals intersect, and then you show they bisect. The book seems slightly dodgy because strictly speaking, you DO need to prove that the two lines intersect. Two non-parallel lines don't have to intersect in 3D space.

[brightsky]

What I would say is that firstly, you prove the two major diagonals intersect, and then you show they bisect. The book seems slightly dodgy because strictly speaking, you DO need to prove that the two lines intersect. Two non-parallel lines don't have to intersect in 3D space.

The proof is actually slightly easier than that. All you need to do is define X to be the midpoint of one diagonal, and Y to be the midpoint of the other diagonal, and then show that X and Y coincide.

[lzxnl]

That works too

[drake]

The proof is actually slightly easier than that. All you need to do is define X to be the midpoint of one diagonal, and Y to be the midpoint of the other diagonal, and then show that X and Y coincide.

brightsky, you are the smartest of them all !

[dpagan]

brightsky, you are the smartest of them all !

Couldn't agree more!!!!!! Rosh, you're not too bad yourself though, mate!

[Sanguinne]

Exercise 3c in essential textbook
q8

Not sure how to approach these kind of quesstions

Given that the domain of sinx and cosx are restricted to [-pi/2, pi/2] and [0,pi] respectively, define the implied domain and range of each of the following where y is equal to:
a) sin^-1(cosx)
g) cos(tan^-1x)

[lzxnl]

Exercise 3c in essential textbook
q8

Not sure how to approach these kind of quesstions

Given that the domain of sinx and cosx are restricted to [-pi/2, pi/2] and [0,pi] respectively, define the implied domain and range of each of the following where y is equal to:
a) sin^-1(cosx)
g) cos(tan^-1x)

So, for these questions, the output of the inside function must fit into the input of the outside function. For a, the range of cos x must be a subset of the domain of inverse sine x. But the range of cos x over its maximal domain fits perfectly into the domain of inverse sine x, so the domain of sin^-1(cos x) is the maximal domain of cos x, [0, pi] The implied range of the first function is thus [-pi/2,pi/2]

For the second one, the output of inverse tan x must be a subset of the domain of cos x. If cos x is only defined for [0,pi], you can't sub in x=-2, for instance. But, the domain of inverse tan x over its maximal domain is (-pi/2,pi/2) which is outside the domain of cos x. Therefore you must limit this as little as possible to fit into [0,pi]. Evidently, this is done by restricting the range of inverse tan x to be [0,pi/2), which yields a domain of [0, inf). That is the implied domain of the second one. As the cosine function now has a domain of [0,pi/2), its range is (0,1]. Be wary of the bracket shapes.

[Cort]

Sorry, but try your best not to laugh here --

I'm currently struggling with understanding on how to transform f(x) --> 1/f(x) reprical graphs. I'm using the Maths Quest 12, and I'm doing Chapter 1B. Hah.

The problem is, I'm barely getting the grasp of it. I really can't explain what I'm not understanding (sorry), but...
 -- I can find the vertical asymptotes and the intercepts (no problemo)
 -- I cannot understand why in some cases that when x approaches whatever, the 'y' could swing all the way in the positive or negative region. I mean -- besides substituting values into the rules and slowly plotting it, I'm not getting the conceptual grasp that I'm aiming for. In the first picture you see that when x --> -3-, it 'swings' up. When x--> -3+, it swings down. I'm just wondering, how do you people figre this out? The textbook example is making me run around in circles. What's your thinking process like?

Sorry if I'm not explaining right. This is just a huge "woah" to me.

EDIT: If vertical asymptote for the reciprocal is the x-intercepts of the f(x), does this logical also follow  for the horizontal asymptote as well? That is, y-intercept = horizontal asymptote of the recip. Or is horizontal recip. & y-intercept be different in this case?

[brightsky]

Don't think of it as a transformation. You are not really performing any transformations. The theory behind the topic of reciprocal graphs is actually pretty straightforward. Suppose you had a function y1 = f(x). Suppose that this function looks nothing like anything you've previously seen. How would you sketch it on a Cartesian graph? Well, you would select an x-value within the domain of the function, plug it into the function to see what the corresponding y-value is, and then plot the point on the Cartesian graph. Then you would select another x-value within the domain of the function, determine the corresponding y-value, and then plot a second point on the Cartesian graph. You would repeat this procedure a billion times until you get a sufficiently large set of points to enable you to make out what the shape of the graph is.

Suppose now you were asked to sketch the graph of y2 = 1/f(x). If you knew nothing about what the graph of y1 = f(x) looks like, then you would use the technique outline above and start plotting individual points. But the thing is, you now know what the graph of y1 = f(x) is. So we need not start from scratch. Observe that y2 = 1/y1. This means that all we have to do now is go to every x-value in the domain of y1 = f(x), look at the original y-value, and take the reciprocal of it, and then plot the new y-value corresponding to the x-value we selected. We do this for every single x-value in the domain of y1 = f(x). The graph that results will be y2 = 1/f(x).

Consider the function y1 = x^2 - 9. The function is easy enough to sketch. It is simply a parabola with x-intercepts (-3,0) and (3,0) and y-intercept (0,-9). Now, suppose we were asked to sketch the graph of y2 = 1/y1 = 1/(x^2-9). How would we proceed? We would follow the procedure outlined above. First, we would select an x-value in the domain of y1 = x^2-9. Suppose we selected x = -5. Then we go to the point x = -5, and we look at the y-value of the original graph. The y-value is y = 16. Now since we know that y2 = 1/y1, all we need to do now is take the reciprocal of this y-value, so y = 1/16, and then plot the new point for the same x-coordinate. So we know that (-5, 1/16) is on y2 = 1/(x^2-9). Now we select a new x-value. Say x = -3. We go to the point x = -3, and we look at the y-value of the original graph. The y-value is y = 0. Now since we know that y2 = 1/y1, all we need to do now is take the reciprocal of this y-value, and the plot the new point. Now the reciprocal of 0 is undefined. So we have a vertical asymptote x = -3. We repeat this procedure until we've sketched the whole graph of y2 = 1/(x^2-9).

Another way to think of it is this. You need to sketch the graph y = 1/f(x). Now you can see that if the original f(x) is small, then 1/f(x) is large, and if the original f(x) is big, then 1/f(x) is small. The extreme occurs when f(x) is positive or negative infinity, in which case 1/f(x) is 0. If f(x) is 0, then 1/f(x) is undefined so we have a vertical asymptote. You can also appreciate that 1/f(x) = f(x) when f(x) = 1 or -1. So the original graph and the new graph intersect at y = +-1.

Hope this is at least of some use to you!

[lzxnl]

Sorry, but try your best not to laugh here --

I'm currently struggling with understanding on how to transform f(x) --> 1/f(x) reprical graphs. I'm using the Maths Quest 12, and I'm doing Chapter 1B. Hah.

The problem is, I'm barely getting the grasp of it. I really can't explain what I'm not understanding (sorry), but...
 -- I can find the vertical asymptotes and the intercepts (no problemo)
 -- I cannot understand why in some cases that when x approaches whatever, the 'y' could swing all the way in the positive or negative region. I mean -- besides substituting values into the rules and slowly plotting it, I'm not getting the conceptual grasp that I'm aiming for. In the first picture you see that when x --> -3-, it 'swings' up. When x--> -3+, it swings down. I'm just wondering, how do you people figre this out? The textbook example is making me run around in circles. What's your thinking process like?

Sorry if I'm not explaining right. This is just a huge "woah" to me.

EDIT: If vertical asymptote for the reciprocal is the x-intercepts of the f(x), does this logical also follow  for the horizontal asymptote as well? That is, y-intercept = horizontal asymptote of the recip. Or is horizontal recip. & y-intercept be different in this case?

Ah these questions. Fun right?
I know what you mean. So our function is y=(x^2-9)^-1
If x approaches +3 from the right, x^2-9 is larger than zero. Therefore, x^2-9 approaches zero from the positive side, so taking the reciprocal of this, the function approaches positive infinity.
However, if x approaches +3 from the left, x^2-9 is lower than zero. x^2-9 then approaches from the negative side and its reciprocal diverges to negative infinity. That explains why the graph changes sign across the asymptote (if that makes sense). I trust you can reason the cases for x approaching -3 now.

The y intercept is when x=0. Let us define y=f(x) and 1/y=g(x). Then, the y intercept of g(x) is g(0) which is 1/f(0). The y intercept of the reciprocal function is just the reciprocal of the original function.
As for a horizontal asymptote, that is when x=>infinity. I'm going to use crude notation here but hopefully the point gets across. f(infinity)=1/(g(infinity)) if there is an asymptote, so the horizontal asymptotes are also reciprocals of each other.

Now if only I didn't have to go downstairs, then my reply would have been up earlier than brightsky's. Oh well.