VCE Specialist 3/4 Question Thread!

[Willba99]

I just did that exact question reiko and didn't even know where to start - do we have to know how to do differential equations like that? Could someone please talk me through it? I got 90% on the rest of the exam I just had literally no clue how to do any of that question. Part a (the one you asked for help with) stumped me especially

[VanillaRice]

One more thing, I got the correct answer but I tried another way as well but I didnt get the same answer, I'm not sure why https://imgur.com/a/T4wc0

In part ci) they tell us x(t) and when you take its derivative, the first fraction represents dx/dt out, so couldn't you also take the integral of this from 0 to 10 to get the answer for part e? This gives me 580, rather than 51.6.

I don't think the two expressions are actually equivalent (from part b and part c).

I just did that exact question reiko and didn't even know where to start - do we have to know how to do differential equations like that? Could someone please talk me through it? I got 90% on the rest of the exam I just had literally no clue how to do any of that question. Part a (the one you asked for help with) stumped me especially

Part a: Concentration = mass/volume. The mass of the chemical is represented by x (so that's your numerator). The volume initially contains 10L of water, which is your '10' in the denominator. However, each minute, 20L of the chemical solution goes in, and 10L goes out, which results in a net gain of +10L per minute. t is the variable representing time in minutes, so after t minutes, there will be 10t more litres of solution. So, we get 10 + 10t as the denominator.

Hope this helps

[Willba99]

I don't think the two expressions are actually equivalent (from part b and part c).
Part a: Concentration = mass/volume. The mass of the chemical is represented by x (so that's your numerator). The volume initially contains 10L of water, which is your '10' in the denominator. However, each minute, 20L of the chemical solution goes in, and 10L goes out, which results in a net gain of +10L per minute. t is the variable representing time in minutes, so after t minutes, there will be 10t more litres of solution. So, we get 10 + 10t as the denominator.

Hope this helps

legend.

[Rieko Ioane]

Hi,

How would I do this question? https://m.imgur.com/a/DwFiK
Wasn't very sure but I'm getting 4/3 which isn't correct. My working is shown in the image.

Thanks

[VanillaRice]

Hi,

How would I do this question? https://m.imgur.com/a/DwFiK
Wasn't very sure but I'm getting 4/3 which isn't correct. My working is shown in the image.

Thanks

You need to set up equations for the forces with respect to the object on which the forces are acting i.e. they must be with respect to the block. In other words, your angle should be next to the block.
The right-hand angle of the triangle can be given by

This angle is the same as the angle which you have drawn in (in pencil) just to the right of the block.
Similarly, we can find the angle of the left-hand corner of the triangle

I will now work with only the forces acting parallel to the horizontal bar (i.e. directly left and right), assuming the system is in equilibrium.


So then we have

as required

Note you can also work with the force perpendicular to the horizontal bar (i.e. up and down), and you should get the same answer.
Hope this helps

[Rieko Ioane]

Hi,

Could someone help me with Q6eii) on the 2016 Sample Exam 2?

The answer says ..."Ho is not true but is not rejected", so are they comparing this to the results found in parts a,b,c? Since those questions say that Ho is not rejected?

I'm getting confused in regards what the answer in eii is referring to.

Thanks

[VanillaRice]

Hi,

Could someone help me with Q6eii) on the 2016 Sample Exam 2?

The answer says ..."Ho is not true but is not rejected", so are they comparing this to the results found in parts a,b,c? Since those questions say that Ho is not rejected?

I'm getting confused in regards what the answer in eii is referring to.

Thanks

Yes, part e.ii is referring to your previous answers. In part e.i, you found that the p-value is greater 0.05 (operating under the same hypotheses as part a), hence suggesting that the null hypothesis not be rejected. However, the question in part e.i states that the mean is actually 9.5. Recall your null hypothesis: mu = 10. If mu is actually 9.5, our tests should've meant we rejected the null hypothesis, and conclude that mu is less than 10, but the p-value tells you not to reject! So, the type II error comes as the p-value we found in part e.i meant we did not reject the null hypothesis (and conclude that mu = 10), but they tell us that mu is actually = 9.5, and that we actually should have rejected the null.

Hope this helps

[DailyInsanity]

Hey Guys,

Was wondering if someone could help me out with 2017 NHT Exam 2 - Qn 3 part b. I'm having trouble integrating dt/dP and then solving for P in terms of t, which I'm assuming you have to do to see lim t-->infinity to find the limiting (maximum) value of P. Thanks in Advance!

[Eric11267]

Hey Guys,

Was wondering if someone could help me out with 2017 NHT Exam 2 - Qn 3 part b. I'm having trouble integrating dt/dP and then solving for P in terms of t, which I'm assuming you have to do to see lim t-->infinity to find the limiting (maximum) value of P. Thanks in Advance!

The limiting value of P occurs when dP/dt=0
Hope this helps

[exit]

 i was wondering for spesh if they give you dy/dx in terms of y how do you find d2y/dx2?

[MathMethdz99-R]

you'd just take the derivative with respect of x of both sides and mulitply by dy/dx since it's implicit. And you only already know dy/dx so you can simplify it.

[Rieko Ioane]

Hi,

Could I have some help with these Qs https://imgur.com/a/89rsp

Q2) Why doesn't the range change in a way that it's just the implied domain of arcsin(x) * dilation from the x-axis in the given function?

I remember doing a similar question but it's was for k*arctan(x), it's range/asymptote values would just be +-kpi/2, so I thought this would follow for arcsin?

Q13) it's not entirely clear to me how D is the correct option?

I tried tracing the solution of the D.E but might've been inaccurate, B still seems like a possible solution.

Thanks

[Eric11267]

Hi,

Could I have some help with these Qs https://imgur.com/a/89rsp

Q2) Why doesn't the range change in a way that it's just the implied domain of arcsin(x) * dilation from the x-axis in the given function?

I remember doing a similar question but it's was for k*arctan(x), it's range/asymptote values would just be +-kpi/2, so I thought this would follow for arcsin?

Q13) it's not entirely clear to me how D is the correct option?

I tried tracing the solution of the D.E but might've been inaccurate, B still seems like a possible solution.

Thanks

2. 2-x isn't really the same as using k, since its a variable rather than a constant. If you looked at the equation, ignoring the 2-x, you would realise that the endpoints occur at x=-2 and x=2. But when x=2, the entire function equals zero. Honestly If I were to get that question in an exam I would just put the equation into my CAS.

13. Honestly seems like a stupid question which you'll never encounter. Usually they would give you the equation.

[Rieko Ioane]

2. 2-x isn't really the same as using k, since its a variable rather than a constant. If you looked at the equation, ignoring the 2-x, you would realise that the endpoints occur at x=-2 and x=2. But when x=2, the entire function equals zero. Honestly If I were to get that question in an exam I would just put the equation into my CAS.

13. Honestly seems like a stupid question which you'll never encounter. Usually they would give you the equation.

Thanks!

[exit]

you'd just take the derivative with respect of x of both sides and mulitply by dy/dx since it's implicit. And you only already know dy/dx so you can simplify it.

Yay nice. Very appreciated.