VCE Methods Question Thread!

[Zealous]

what about in further? if the question didnt ask we should leave in standard?

They'll always give you rounding information in Further - so don't worry about exact values. Just look at some of the questions and you'll see.

(try to keep your Further questions to the other boards!)

I have two multiple choice question that needs assistance.

I don't have a lot of time, but i'll try explain them quickly. These questions remind me of kilbaha =p

So for Q3, I would simply let a=3 for example, then sketch it on my CAS and find the minimum value, then find the multiple choice that matches it. Then I'd let a=5 or something and then confirm the multiple choice answer. Because it's multiple choice, I'd do something like that quickly which could take less than a minute.

Though if you want a more foolproof method, you can split f(x) into a hybrid function over three different domains and find the minimum value of each section (through calculus) within it's domain.

Here's a graph:
https://www.desmos.com/calculator/6olq8gvomg

For Q13, the graph is a square root function reflected across the y-axis. If you take left rectangles, you are going to overestimate - the left rectangles will actually be above the curve as it's decreasing over it's maximal domain. Do a quick sketch and see.

[IndefatigableLover]

Hey guys,

I have two multiple choice question that needs assistance.

No idea how to do Q3 (finding the minimum value of f)

What I did was taking the negative parts of all of the modulus.
Thus, f(x)= (-2x)-(2x-a)-(2x+a)
= -6x

Which wasn't one of the options provided.

I chose A out of a frenzy but answer is B. Anyone explain please?


The other question Q13 MC,

I chose D. Doesn't the left rectangle method understimate the area because the rectangles are below the curve? Answer says it is nan overestimate, which I find makes no sense because the rectangles are below the curve.. But I checked with the CAS and it is indeed an overestimation? How do you overestimate when rectangles are below curve?

Thanks

Q3:
Well since they're just MCQ so I assume you can use a CAS. Really I'd just sub in values of 'a' and then see what fits. You're only looking for the minimum value of 'f' so no need to differentiate (overkill). Just sub in values for 'a' (say a=3). When you graph it out and find the local min, you'll see that the minimum value is 6 (which is 2a or B).

Q13:
Decreasing function therefore it'll be an overestimation (see the graph when you CAS it out)!

EDIT: Beaten by Zealous

[IndefatigableLover]

Just a quick question! It's attached below
Answer is 24 as well.

[silverpixeli]

what about in further? if the question didnt ask we should leave in standard?

yeah you'll have to check with someone who did further, but methods (and specialist) always require exact values unless otherwise specified

also, a container the shape of a cone, with its circular face on the ground, is filled to a third of its depth. what fraction of the total volume is filled with water?

you're going to need to find the volume occupied by the water and then divide this by the total volume to get the fraction

the total volume is easy, V_total = 1/3 area of base * height = 1/3 * pi * r^2 * h
(hopefully those variables will cancel when we do the division and we'll get a fraction)

the volume filled is a little more complicated, but can be found if you consider the entire cone and then subtract a shorter cone with height 2/3 of the total height (this also means it will have 2/3 the radius, by similar triangles)
so for this shorted cone height = 2/3 h and radius = 2/3 r where h and r are the dimensions of the container, as defined in the volume formula we got before

V_filled = V_total - V_smaller_cone
= V_total - 1/3 area of base * height
= V_total - 1/3 * pi * (2/3 r)^2 * (2/3 h)
= V_total - 8/81 * pi * r^2 * h

= (1/3 * pi * r^2 * h) - (8/81 * pi * r^2 * h)

so the fraction is
V_total / V_filled
= [(1/3 * pi * r^2 * h) - (8/81 * pi * r^2 * h)] / (1/3 * pi * r^2 * h)

and luckily we can cancel r^2, h, and pi leaving

= [(1/3) - (8/81)] / (1/3)
= 19/27
i think

sorry for the lack of latex!

ando. A cylinder of Radius R and Height H has volume V. the volume of a cylinder with radius 3R and heigh H is.... answer is 27V but i get 9V?

I get 9V too

the volume of a cylinder is the area of the base times the height of the cylinder. in this case, the height remains the same but the radius triples which means the area increases by a factor of 9 (area depends on r^2) which gives 9V

could easily be a mistake in the book, or you could have misread the question (though i cant think of a simple error that'd cause this issue)

[silverpixeli]

Just a quick question! It's attached below
Answer is 24 as well.

What's the probability that any loaf has less than 16 slices?

p = Pr(Slices = 14) + Pr(Slices = 15) = 0.59

Let's call this a success, because we're interested in it happening/not happening in a sample space of size n loaves. we can use a binomial distribution to solve this sort of problem.

We want Pr(Number of 'successful' (less than 16 slices) loaves >= n/2) in a binomial distribution which isn't an easy formula to write out. It involves a different number of binomial formulas added together for each value of n. Luckily, we can use the CAS for this, as in use its binomial cumulative function.
Not sure what calculator you have, but it'd look something like this:

binomCdf(trials = n, p(success)=0.59, lower bound = n/2, upper bound = n)
and then I think you'd have to try different values of n until you found the lowest n value that gives over 0.86 as the answer

depending on your calculator's ability and how it handles the binomial function, you might even be able to solve for n in something like:

solve(binomCdf(n, 0.59, n/2, n)>0.86, n)

but i dont think the tinspire can do that, so trial and error might be your best bet unless your calculator can

[LiquidPaperz]

I get 9V too

the volume of a cylinder is the area of the base times the height of the cylinder. in this case, the height remains the same but the radius triples which means the area increases by a factor of 9 (area depends on r^2) which gives 9V

could easily be a mistake in the book, or you could have misread the question (though i cant think of a simple error that'd cause this issue)

didnt misread the question, it was a question at a lecture i went to and was just looking over notes now. Apparently the ratio of the lengths are (1:3) for volume its 1:27 hence 27 times bigger, but im not sure why it doesnt give 27 times bigger when using the formula for volume??

[IndefatigableLover]

What's the probability that any loaf has less than 16 slices?

p = Pr(Slices = 14) + Pr(Slices = 15) = 0.59

Let's call this a success, because we're interested in it happening/not happening in a sample space of size n loaves. we can use a binomial distribution to solve this sort of problem.

We want Pr(Number of 'successful' (less than 16 slices) loaves >= n/2) in a binomial distribution which isn't an easy formula to write out. It involves a different number of binomial formulas added together for each value of n. Luckily, we can use the CAS for this, as in use its binomial cumulative function.
Not sure what calculator you have, but it'd look something like this:

binomCdf(trials = n, p(success)=0.59, lower bound = n/2, upper bound = n)
and then I think you'd have to try different values of n until you found the lowest n value that gives over 0.86 as the answer

depending on your calculator's ability and how it handles the binomial function, you might even be able to solve for n in something like:

solve(binomCdf(n, 0.59, n/2, n)>0.86, n)

but i dont think the tinspire can do that, so trial and error might be your best bet unless your calculator can

Sweet that's great silverpixeli! I got the same values as what you got except I was wondering whether there was a way to solve it (that is to get 'n=24') since via trial and error I do get to that number eventually but I can't solve for it using the 'Solve Command' on the CAS (I use the TI-nspire) so yeah.. >.<

[psyxwar]

Sketch binomcdf(x,0.59,x/2,x) where pr(1 loaf with less than 16)=0.59=p, n=x, X~Bi(x,0.59), X=no. of loaves he must buy, lower bound is x/2, upper bound is x

For TI-Nspire: Menu 7 (table) -> 1 (split screen table)

Scroll down until you hit a probability that is greater than 0.86. This (first) occurs when x=24.

Hence, 24 loaves.

[IndefatigableLover]

Sketch binomcdf(x,0.59,x/2,x) where pr(1 loaf with less than 16)=0.59=p, n=x, X~Bi(x,0.59), X=no. of loaves he must buy, lower bound is x/2, upper bound is x

For TI-Nspire: Menu 7 (table) -> 1 (split screen table)

Scroll down until you hit a probability that is greater than 0.86. This (first) occurs when x=24.

Hence, 24 loaves.

My gosh that's brilliant! Though when I graph it nothing shows up? That's normal right (I have values from the table of values though which is good )?

[psyxwar]

Yeah happens with me too. I think it's because you get a bunch of discrete data points (ie. dots) that are only defined at integer values, and since there're all so small it's hard to see them. But yeah it's not the graph that's important lol, it's the table you generate

[keltingmeith]

Sweet that's great silverpixeli! I got the same values as what you got except I was wondering whether there was a way to solve it (that is to get 'n=24') since via trial and error I do get to that number eventually but I can't solve for it using the 'Solve Command' on the CAS (I use the TI-nspire) so yeah.. >.<

Graphing that is a lot of work, even though it's pretty cool.

Use nsolve - it's a function based on a numerical method of finding solutions, so it will give you a numerical answer. Just gotta know which way to round.

[psyxwar]

Graphing that is a lot of work, even though it's pretty cool.

Use nsolve - it's a function based on a numerical method of finding solutions, so it will give you a numerical answer. Just gotta know which way to round.

can you explain how I'd use this on CAS? thanks

[keltingmeith]

can you explain how I'd use this on CAS? thanks

It's been a while, but to memory it functions just like your normal solve function. Just write nsolve instead of solve.

[silverpixeli]

be careful with nsolve, i'm not saying you should avoid using it but depending on the complexity of the equation you ask it to compute it can take a lot longer than solving a regular, symbolic equation
for example asking it to solve normal distribution things for varying parameters can take quite a while, or i remember it taking quite a while when I tried to do the last question off some insight exam 2 last year using it.

awesome method for things like this question though!

and yes, it's just nsolve(equation(s), variable) instead of solve(equation(s), variable)

[keltingmeith]

be careful with nsolve, i'm not saying you should avoid using it but depending on the complexity of the equation you ask it to compute it can take a lot longer than solving a regular, symbolic equation
for example asking it to solve normal distribution things for varying parameters can take quite a while, or i remember it taking quite a while when I tried to do the last question off some insight exam 2 last year using it.

awesome method for things like this question though!

and yes, it's just nsolve(equation(s), variable) instead of solve(equation(s), variable)

This is because nsolve is based on (last I checked) Newton's method - which involves differentiating, taking a ratio with the original function, and then checking against the actual answer. For an equation as big as the normal distribution, this is going to require some decent RAM.