VCE Methods Question Thread!

[JR_StudyEd]

Why do sin and cos graphs look like squiggly lines? And why does the graph of the tan function look how it does? Is there an explanation?

[Matthew_Whelan]

Hey with probability questions, is there a preference of answering with decimals or as a percentage?

Why do sin and cos graphs look like squiggly lines? And why does the graph of the tan function look how it does? Is there an explanation?

It's to do with circles; sin is vertical, cos horizontal, tan is gradient - when you move 90 degrees (anticlockwise) from the starting position (right side of circle), its vertical so a sin(x) graph will show y=1 for x=pi/2. cos(x) graph on the other hand will have y=1 at x=0 and x=pi since thats when the position on the circle is horizontal. Tan(x) has asymptotes when the position on the circle is vertical (undefined), so when x=pi/2 for instance.
At least thats how I think of it, hope that helps

[^^^111^^^]

For Methods will we need to learn about osculating circles? I'm confused, because some of the students at my school are learning them, (idk if it's for vce methods or something else). If not, can someone pls tell me where I can find VCE Maths Methods Study design (current)?

Thank you

[AlphaZero]

Hey with probability questions, is there a preference of answering with decimals or as a percentage?

You must give an exact value unless otherwise stated.

The question will state whether to give your answer as a percentage or not.

Almost all of the time, you will be giving an answer of a numerical value between 0 and 1.

[AlphaZero]

For Methods will we need to learn about osculating circles? I'm confused, because some of the students at my school are learning them, (idk if it's for vce methods or something else). If not, can someone pls tell me where I can find VCE Maths Methods Study design (current)?

Thank you

(Sorry for the double post, but this is a different question).

Could you elaborate by perhaps providing an example of a question?

The specific concept of osculating circles isn't mentioned in the study design, but it wouldn't be too hard for VCAA to integrate it into a question.

The current VCE Mathematics study design can be found here.

[JR_StudyEd]

Someone please help me with the questions in the attached pdf. I know you need to use the distance formula for question 1a, but after that, my composure and confidence in myself to do the rest of the questions falls apart.

Even just some explanation on how to get started on Questions 1a-f would be much appreciated. If someone can go through some possible methods for approaching the remaining questions, that would be even more awesome!

[BAH0003]

Hi,

I was working through this question and have completed a few parts from it, but I am stuck on the following parts. If anyone could go through these parts that would be great.

Thanks

[BAH0003]

Hi,

I was working through this question and have completed a few parts from it, but I am stuck on the following parts. If anyone could go through these parts that would be great.

Thanks

[BAH0003]

Hi,

I was working through this question and have completed a few parts from it, but I am stuck on the following parts. If anyone could go through these parts that would be great.

Thanks

Sorry, there was a bit of a glitch and this posted twice please ignore one

[Matthew_Whelan]

Someone please help me with the questions in the attached pdf. I know you need to use the distance formula for question 1a, but after that, my composure and confidence in myself to do the rest of the questions falls apart.

Even just some explanation on how to get started on Questions 1a-f would be much appreciated. If someone can go through some possible methods for approaching the remaining questions, that would be even more awesome!

So distance can be found using trig for 150/cos(angle), but the question wants it in terms of x so I used pythag to find d = (x^2 -400x + 62500)^1/2, although I could be wrong.
on the grass the velocity is 1m/s so t = d/v = d, so time and distance are equivalent on the grass.
Part c is time on grass + time from X to B, so d/v (AX) + d/v (BX) = d + x/2 =(x^2 -400x + 62500)^1/2 + x/2.
I don't have time to do the rest but I hope that these ones are correct. 

Sorry, there was a bit of a glitch and this posted twice please ignore one

You can remove and/or edit your posts to tidy it up.

[AlphaZero]

Someone please help me with the questions in the attached pdf. I know you need to use the distance formula for question 1a, but after that, my composure and confidence in myself to do the rest of the questions falls apart.

Even just some explanation on how to get started on Questions 1a-f would be much appreciated. If someone can go through some possible methods for approaching the remaining questions, that would be even more awesome!

Part a
The length \(AX\) is the hypotenuse of a right-angled triangle \((\triangle ACX)\), so we have \[d=\sqrt{150^2+(200-x)^2}=\sqrt{x^2-400x+62500}\]
Part b
The time taken is given by the distance divided by the speed. Thus, \[T_1=\frac{d}{1}=\sqrt{x^2-400x+62500}\]
Part c
The total time is given by the time taken to travel \(AX\) plus the time taken to travel \(BX\).  Using the same idea as in part b, can you write an expression for the time taken to travel \(BX\) in terms of \(x\)?

Spoiler

\[T(x)=\sqrt{x^2-400x+62500}+\frac{x}{2}\]

Part d
What is a useful domain of \(x\)? Hint: is there much point travelling to the right when at point \(A\)?


Part e.i
Now that your have \(T(x)\), use you CAS.


Part e.ii
This is a standard optimisation problem. What is the value of \(\dfrac{dT}{dx}\) at the point where \(T\) is minimum?


Part f
Now that you have the value of \(x\) for which \(T\) is minimum, what is the minimum value of \(T\)?


Part g
This question tells you what to do. Pick values slightly to the left and slightly to the right of the critical value found in part e.ii. Determine the values of \(\dfrac{dT}{dx}\) at these values. What do these tell you about the nature of the stationary point found?


Part h
\(x=200\)


Part i
\(x=0\)


Part j
\(x=100\)


Part k.i
Use your CAS to sketch it. Remember that \(T\) has a domain.


Part k.ii
Use the graph you just sketched.
Okay, I'm going to leave it here. Part L onwards requires a bit of thought. Make a start with what I've given.

[AlphaZero]

Hi,

I was working through this question and have completed a few parts from it, but I am stuck on the following parts. If anyone could go through these parts that would be great.

Thanks

Part b.ii
Use the remaining point to find \(d\). You know that \(x=0\) and \(y=31.5\).


Part b.iii
The maximum height of the larger hill occurs when \(\dfrac{dy}{dx}=0\). This gives you the \(x\)-coordinate of this point. How can you use this to find the height?


Part c
Let  \(y=10\).

[JR_StudyEd]

Why is the inverse function of f(x)= e^2x for x ∈ R⁺ equal to

[BAH0003]

Hi,

I am having trouble with part d) of this question. It follows on with other parts of a question which I have attached. I have written that 'p' can indeed have stationary points but am not sure what the possible values of 'p' could be. If anyone can list these possible values that would be great, as this is apart of a big assignment that is due soon. Please note this is from question 4 from the first picture.

Thank You and Kind Regards.

[Matthew_Whelan]

Why is the inverse function of f(x)= e^2x for x ∈ R⁺ equal to

Let f(x)=y.
y= e^2x (switch x and y to find inverse)
x= e^2y
2y = log_e(x)
y = 1/2(log_e(x))

Hi,

I am having trouble with part d) of this question. It follows on with other parts of a question which I have attached. I have written that 'p' can indeed have stationary points but am not sure what the possible values of 'p' could be. If anyone can list these possible values that would be great, as this is apart of a big assignment that is due soon. Please note this is from question 4 from the first picture.

Thank You and Kind Regards.

So if you have a quartic function, then the graph can have either one, two or three stationary points.
Another way to think of it is if you derive ax⁴+bx³+cx²+dx+e you get 4ax³+3bx²+2cx+d which makes it a cubic.
To find possible stationary points you'd make it =0, and cubics can only have 1,2, or 3 values of x where y=0.