VCE Methods Question Thread!

[jessss0407]

Personally I wouldn't spend time typing up the questions onto my CAS because like achre said, if it's different parts on different days so no real benefit... at our school we have the same thing except it's 3 parts and the 'parts' relate to a story where all the questions are totally different from each other so yeah...
However what I would do beforehand would be to have some sort of guide on how to use your CAS properly (so any nifty tricks you could use that you might forget in the real exam because for me it has come in handy before)

Ahh I see. . Yeah my school lets us get part 2 back on the day they give is part 3.. its basically like one part done over two different days.. Thanks for ur advice! Btw, what did you mean by "nifty tricks"?

[jessss0407]

Hey guys! Can someone help me with this question please:

An inverted cone has a depth of 80cm and 40cm.  Water is poured into the container at the rate of 20cm³/s. Find the rate at which the depth (h) is increasing when h=5cm

I was also wondering how I would be able to efficiently check my answer?
Thanks!

[nhmn0301]

Hey guys! Can someone help me with this question please:

An inverted cone has a depth of 80cm and 40cm.  Water is poured into the container at the rate of 20cm³/s. Find the rate at which the depth (h) is increasing when h=5cm

I was also wondering how I would be able to efficiently check my answer?
Thanks!

Write down what you need to find first:
dh/dt= dh/dV x dV/dt
imagine having an inverted cone with water filling up and halve the cone vertically. you can see that the height and radius at any time will create a triangle which is similar to half of the cone. Hence, we say that 2 triangles are similar and obtain the property: h/r=80/40=2 => r=h/2
Since now we have an expression of r in terms of h, we can substitute in the formula to find dh/dV.
V (cone) = 1/3 pi r^2 h
             = 1/3 pi (h/2)^2 h
             = 1/3 pi (h^3/4)
dV/dh= pi x h^2/4
Hence, come back to our original equation: dh/dt = dh/dV x dV/dt = 4/(pi x h^2) x 20 = 80/ (pi x h^2)
when  h=5, dh/dt = 80/ ( pi x 25) = 16/(pi/5) (cm/s)

[M_BONG]

Hey in the printscreen below:

Did my teacher make a mistake with the average rate of change question?

I think it's supposed to be -In(6) but she wrote -2In(6)?

Can anyone clarify this?

Also avg. ROC does not require the gradient/first derivative right?

[Stevensmay]

Yes it should be -Ln(6), from a quick look it appears in the first line they have written 4ln instead of 2ln?

Avg ROC does not require the first derivative, as that would be giving us the instantaneous rates of change.

[Rishi97]

How do I differentiate tan(2x +1 )  ?
Thanks

[Conic]

You can use the chain rule:









Edit: As stated later in the thread, this is also equal to . Both forms are on the formula sheet so either should be fine.

[Einstein]

Ive got 3 questions, if anyone could answer them that would be awesome!

1) if i have the equation 15 = 20t - 5t^2, how to i rearrange to find t?
2) if you look at question number 5 in the attachment, why is it D and not C?
3) If you look at Q7, why is it E, you cant have [-infinity... Wouldnt the range be (-infinity, 4]

[Stevensmay]

Ive got 3 questions, if anyone could answer them that would be awesome!

1) if i have the equation 15 = 20t - 5t^2, how to i rearrange to find t?
2) if you look at question number 5 in the attachment, why is it D and not C?
3) If you look at Q7, why is it E, you cant have [-infinity... Wouldnt the range be (-infinity, 4]

1)



Simply factorize the polynomial to find t.

2) We will have two real solutions if the determinant is > 0.
So






https://s3.amazonaws.com/grapher/exports/fvri5yzyin.png
Where the red graph is if c = 26/8 and the blue one if c = 24/8.
If c>25/8 then we can see that it never intercepts the x-axis at all, but if c<25/8 we can see it happens twice.

3) I think you are correct here, it should be inclusive of 4 and exclusive of -infinity, giving us

[Zealous]

Ive got 3 questions, if anyone could answer them that would be awesome!

1) if i have the equation 15 = 20t - 5t^2, how to i rearrange to find t?
2) if you look at question number 5 in the attachment, why is it D and not C?
3) If you look at Q7, why is it E, you cant have [-infinity... Wouldnt the range be (-infinity, 4]

2)


3)
There might be an error in that question...

[edit] beaten by steve!

[paper-back]

Two cars move away from the intersection of 2 perpendicular straight roads. Car A travels at 60km/hr and Car B at 80km/h. If both cars are at the intersection initialy , at what rate are they moving apart after 6 min?

[Zealous]

Two cars move away from the intersection of 2 perpendicular straight roads. Car A travels at 60km/hr and Car B at 80km/h. If both cars are at the intersection initialy , at what rate are they moving apart after 6 min?

I believe you can just use pythagoras to complete this question.



If we let X_c be the distance between the two cars at point time t (rates are in km/h):



If you need more explanation:



So the rate in which they are moving apart, regardless of time is a constant 100km/h. If both cars are moving away from the intersection at a constant rate, the velocity between them should also be constant.

[Einstein]

2)


3)
There might be an error in that question...

[edit] beaten by steve!

With c < 25\8 could you also have c > -25/8?

[Einstein]

3) I think you are correct here, it should be inclusive of 4 and exclusive of -infinity, giving us

so is it defiantly that and not (- infinity, 4), because <--- was what my friend say but i said (- infinity, 4]

[Conic]

With c < 25\8 could you also have c > -25/8?

Those aren't necessarily the same thing. Try sketching them on a number line. You can have values of c that fit in with both inequalities, but the sets of values for both inequalities are different.