VCE Methods Question Thread!

[Samueliscool223]

i found question 1f of the 2014 exam 2 really confusing, it says find dh/dt in terms of h. i thought it was only possible to find dh/dt in terms of t, i.e. the independent variable? this is really confusing and isnt in my book :/

[fun_jirachi]

Is there an error on the answer sheet here? Shouldn't the last translation be pi/3 units in the negative direction of the x-axis?

There is no mistake; recall that transforming \(f(x)\) to \(f(x-\alpha)\) is a translation of \(f(x) \) by \( \alpha\) units in the positive x-direction.

i found question 1f of the 2014 exam 2 really confusing, it says find dh/dt in terms of h. i thought it was only possible to find dh/dt in terms of t, i.e. the independent variable? this is really confusing and isnt in my book :/

Can't find the question you're referring to so I can't help you here. It is always better to attach the relevant question on the forum (even forcing me to download it is better than a complete lack of context). No clue what you're talking about

[jnlfs2010]

i found question 1f of the 2014 exam 2 really confusing, it says find dh/dt in terms of h. i thought it was only possible to find dh/dt in terms of t, i.e. the independent variable? this is really confusing and isnt in my book :/

Out of current study design. That was when related rates was still in maths methods but now only in specialist maths.

[Samueliscool223]

There is no mistake; recall that transforming \(f(x)\) to \(f(x-\alpha)\) is a translation of \(f(x) \) by \( \alpha\) units in the positive x-direction.

Can't find the question you're referring to so I can't help you here. It is always better to attach the relevant question on the forum (even forcing me to download it is better than a complete lack of context). No clue what you're talking about

i meant 2f in exam 2, my bad.
image of the question and its answer are attached below:

[amyzzwq]

There is no mistake; recall that transforming \(f(x)\) to \(f(x-\alpha)\) is a translation of \(f(x) \) by \( \alpha\) units in the positive x-direction.

Oh ok thank you, must have gotten it the other way round then, transformation is the most confusing topic ever. If something like this comes out in mcq, is it the same as translated pi/3 units to the left? (pls tell me to the left is positive direction)

[Samueliscool223]

Out of current study design. That was when related rates was still in maths methods but now only in specialist maths.

Out of current study design. That was when related rates was still in maths methods but now only in specialist maths.

oh, i see. still, im kind of curious as to how that even makes sense (finding derivatives with respect to a variable other than the independent one). is it related to implicit differentiation or something?

[fun_jirachi]

Oh ok thank you, must have gotten it the other way round then, transformation is the most confusing topic ever. If something like this comes out in mcq, is it the same as translated pi/3 units to the left? (pls tell me to the left is positive direction)

On the cartesian plane the right is the positive x direction. Moving something to the left is moving it in the negative x direction.

oh, i see. still, im kind of curious as to how that even makes sense (finding derivatives with respect to a variable other than the independent one). is it related to implicit differentiation or something?

Read up on the chain rule, it will not only help you understand the question (whether it's still relevant or not; and it apparently is not) but is very useful for a lot of maths.

[Rose34]

Hey guys,

The question is asking to find the derivative of this graph. I just want to confirm my answer. I got a negative parabola its vertex is at (0,0) is that correct?
I am asking this because the answer of the book draws it slightly bellow (0,0) but in one of the examples(d)they explained(they also said the inflection point is point of zero gradient) and they drew the parabola with vertex exactly at (0,0), so now I am confused about why there are different.

Thanks in advance

[fun_jirachi]

You should be correct. It's just a lot harder to work off graphs devoid of detail.

The only reason you should have a derivative that is a 'negative' parabola is if the curve you have is a polynomial of degree 3 and is monotonically decreasing, which does not seem to be the case.

[Rose34]

Hey everyone,

Does anyone know how to do this question? I did solve it but my answer is different from the book. I got 0.95 but the book's answer is 0.933.

Thanks in advance.

[SmartWorker]

Hey everyone,

Does anyone know how to do this question? I did solve it but my answer is different from the book. I got 0.95 but the book's answer is 0.933.

Thanks in advance.

Hi Rose

So you used NormCDF, which you can't use because you don't know that the random variable is normally distributed.
Your values for mean and variance are correct --> Pr(1.17<x<6.16), therefore Pr(2≤x≤5), the only values between 1.17 and 6.16 is 2 to 5.

Hence, 1-1/15 = 14/15 approximates to 0.933

Normal distributions are used for continuous random variables as opposed to discrete random variables

[Rose34]

Thank you so much for clarifying!

Hi Rose

So you used NormCDF, which you can't use because you don't know that the random variable is normally distributed.
Your values for mean and variance are correct --> Pr(1.17<x<6.16), therefore Pr(2≤x≤5), the only values between 1.17 and 6.16 is 2 to 5.

Hence, 1-1/15 = 14/15 approximates to 0.933

Normal distributions are used for continuous random variables as opposed to discrete random variables

[CorkedBoard]

How would one figure out the variables in a situation like this? I don’t understand how to do it if the missing variables are not symmetrical around the mean.

[fun_jirachi]

There's a lot of missing context here:
- Are there any tools that you are allowed to use to solve this question? (CAS, etc.)
- Is there any extra information that we should know? Moreover, is there an overarching problem that is an intermediate problem for? (read: XY problem)

The best I can do here is remind you that \(P(a < X < b) = P(X < b) - P(X < a)\). There are an infinite number of possibilities for a, b in this case given they are not symmetric around the mean (the only condition is that \(P(X < a) < 0.36\)), which leads me to think that there is a lack of context. Fixing one of a or b or expressing a in terms of b or vice versa would make this solvable.

Hope this helps

[LimDimSim]

Hi, I'm having some issues with this problem from my textbook (Methods 1 and 2, chapter 7H).
I would really appreciate some help.

A transformation is defined by the matrix
[0 2]
[-3 0]
Find the equation of the image of
the straight line with equation y = 2x + 3 under this transformation.