VCE Methods Question Thread!

[brightsky]

assuming that r, b and w are all greater or equal to 3, then:

Pr(RRR) + Pr(BBB) + Pr(WWW) = rC3 * (b+w)C(0)/(r+b+w)C(3) + bC3 * (r+w)C(0)/(r+b+w)C(3) + wC3 * (b+r)C(0)/(r+b+w)C(3)

you might find this page quite helpful: http://en.wikipedia.org/wiki/Hypergeometric_distribution.

[soNasty]

thanks brightsky

[Blondie21]

Hey guys, I'm still pretty confused about finding the anti diff when r = -1
How would you solve the following equations for the antiderrivative?

a)

b)

[brightsky]

For the first one, note that x/(x+1) = 1 - 1/(x+1) so the antiderivative is x - ln|x+1| + c.

For the second one, convert fraction to (x+2)^(-2). The antiderivative is (x+2)^(-1)/(-1) = -1/(x+2) + c.

Hope this clears things up for you!

[Blondie21]

For the first one, note that x/(x+1) = 1 - 1/(x+1) so the antiderivative is x - ln|x+1| + c.

For the second one, convert fraction to (x+2)^(-2). The antiderivative is (x+2)^(-1)/(-1) = -1/(x+2) + c.

Hope this clears things up for you!

Thanks!!

[achre]

Just to clarify brightsky's point about splitting up the fraction for the first anti-diff, the intermediate step just requires you to play around with the numerator to make it into something that cancels down. For future ref:

[Anchy]

A manufacturer decided to build a open right-circular cone container with height 3m and radius 2m.
Grain is released from this container into trailers below through a small hole at the vertex of the cone.
Grain falls from the container at a rate of H^1/2 cubic metres/min, where H is the height of the grain above the vertex at any time t

Find the volume of grain in the container in terms of H at any time t

Thanks

[Brunette15]

Hi everyone,
I was just wondering if we use specialist knowledge in a methods exam , such as proving things with second derivatives will we lose marks?

[LiquidPaperz]

When's everyone starting their methods exams?

[Rishi97]

Using linear approximation :
1) If a spherical balloon has a radius of 5 cm, find the increase in volume of the balloon when the radius expands by 0.02cm
I can't figure out the x value

Thanks

[keltingmeith]

Hi everyone,
I was just wondering if we use specialist knowledge in a methods exam , such as proving things with second derivatives will we lose marks?

Sorry for the late response - I have been told that if you do something that's mathematically correct, you won't be penalised for it. However, the general consensus is that you should try to use methods techniques in your methods exam wherever possible.

Using linear approximation :
1) If a spherical balloon has a radius of 5 cm, find the increase in volume of the balloon when the radius expands by 0.02cm
I can't figure out the x value

Thanks

For a sphere, it's volume can be represented by the function . From here, we can take two methods.

First method:
Find the volume for both r=5 and r=5.02 and subtract the final from the initial:


Second method:
Use linear approximation - which believe it or not, is much faster:


Obviously, the values will be slightly different based on your method. If this is multiple choice and you use linear approximation, go for the closest value. If this is short answer/extended response, I don't think the examiner will penalise you if they see you've used a method like linear approximation which is going to sacrifice exactness - as long as you're relatively close to the true value.

[Rishi97]

Thanks EulerFan but why have we used 3 for the radius?

[keltingmeith]

Because it's early and I mis-read - let me substitute it for a 5.

[LiquidPaperz]

Alright, so i need help with I, J, K ,L. For I, J i dont know what they have done (this is circular functions symmetry) and i thought you would have to use the points P(there), P(180- theta) etc.. but i dont think they have?

With I and J. I i know tan(180) can be tan(180-0), but would the answer be zero? why have they done 0 degrees.

And with J, i know sin 270 is at point (0,-1) but why have they made it -1 degrees? isnt it just -1 if anything?

Thanks

[LiquidPaperz]

and why have they done 2pi +... , in the symmetry circle theirs only 2 pi - ... , pi +..., pi - ... and pi