Mathematics Question Thread

[fun_jirachi]

Already been answered, but here's my two cents. Pretty sure that given you slice it out of a circle, the bottom is empty since you can't have a circle appear out of nowhere to form the base (that's my interpretation at least). The area of the sector should be the surface area of the cone, so subbing into half x r^2 x theta, you get 25/6 pi. For the volume, you need the height of the cone, which by Pythagoras' theorem (subbing in slant height and radius calculated from arc length) is 5root35/6, the radius of the cone is just 5/6, so sub that into the formula for volume of a cone and you get 125/648 x sqrt 35 x pi. Hope this helps

EDIT: to be fair though, if you have a closed cone it's just one extra step using the radius 5/6 to find the area of the base. Regardless of this, the volume remains the same.

[annabeljxde]

That is a fair question, since they don't seem to tell you whether we should treat it as a 'sealed-off' cone or an empty cone. So why not try out both methods and see which one works?

(However, it is a bit tricky. Recall that the net of a cone looks like what's shown on this website. Note that the sector is wrapped around to form the apex of the cone. Therefore, that radius of 5 becomes the slant height of the cone. Furthermore, the arc length \( \ell = 5\times \frac\pi3\) becomes the circumference of the attached circle.)

Although having said that, this kind of ambiguity would not appear in the HSC. Where did this question originally come from?

It was a question from my textbook, HSC Maths in Focus as one of the harder questions of the 'arc length' exercise.
The answers are:

SA= 175π/36 cm²
V= 125√35/648 cm^3

[annabeljxde]

Already been answered, but here's my two cents. Pretty sure that given you slice it out of a circle, the bottom is empty since you can't have a circle appear out of nowhere to form the base (that's my interpretation at least). The area of the sector should be the surface area of the cone, so subbing into half x r^2 x theta, you get 25/6 pi. For the volume, you need the height of the cone, which by Pythagoras' theorem (subbing in slant height and radius calculated from arc length) is 5root35/6, the radius of the cone is just 5/6, so sub that into the formula for volume of a cone and you get 125/648 x sqrt 35 x pi. Hope this helps

EDIT: to be fair though, if you have a closed cone it's just one extra step using the radius 5/6 to find the area of the base. Regardless of this, the volume remains the same.

You got the Volume answer correct (thank you so much for the working ) but I also got the same answer for the surface area but apparently it says otherwise in the answers in the textbook... why is that??

[annabeljxde]

I figured it out RuiAce and fun_jirachi !

So apparently, it is assumed that the cone is closed off and you have to use the formula for surface area (SA= pi x r(r+l))

I figured out the radius to be 5/6 from the circumference of the cone (which is the arc length of the cut-off sector) and I substituted this value to r in the SA formula.

The radius of the sector becomes the slant height of the cone (which is 5)

My substitution looked like:

SA = pi x 5/6(5/6 + 5)

My answer was correct (according to the textbook) = 175pi/36 cm^2

I really hope these questions don't pop up in future trial or HSC exams. It took me a whole week (and your guys' help!) to finally figure it out!!

Anyways, thanks for the help you guys  

[RuiAce]

I figured it out RuiAce and fun_jirachi !

So apparently, it is assumed that the cone is closed off and you have to use the formula for surface area (SA= pi x r(r+l))

I figured out the radius to be 5/6 from the circumference of the cone (which is the arc length of the cut-off sector) and I substituted this value to r in the SA formula.

The radius of the sector becomes the slant height of the cone (which is 5)

My substitution looked like:

SA = pi x 5/6(5/6 + 5)

My answer was correct (according to the textbook) = 175pi/36 cm^2

I really hope these questions don't pop up in future trial or HSC exams. It took me a whole week (and your guys' help!) to finally figure it out!!

Anyways, thanks for the help you guys  

Typical maths in focus and its lack of clarity.

Glad you solved it but yeah, in the HSC exam this kind of ambiguity is guaranteed to not appear. (Else the examiners get fired.)

[alexnero7]

Can somebody please help me with these. I can't seem to get the correct answer. Thanks.

[RuiAce]

Can somebody please help me with these. I can't seem to get the correct answer. Thanks.

Both of them rely on the formula \( \int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} +C \). Please provide working out showing usage of those formulas for further assistance. Or alternatively you may ask about how that formula provided works if necessary.

[alexnero7]

Both of them rely on the formula \( \int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)} +C \). Please provide working out showing usage of those formulas for further assistance. Or alternatively you may ask about how that formula provided works if necessary.

Thank you Rui!! Really helped, got both answers right this time. 

[alexnero7]

Hey everyone, does anybody know what to do when given questions like the ones attached? When there is one line at x?? Thanks.

[jamonwindeyer]

Hello! You need to find the intercept of the graph with the x-axis as the other limit of your integral! You do this by setting y=0 and then solving for x!

What you really need to do is sketch these curves and see the area they are talking about. These are all standard curves, parabolas and semicircles, so you should be able to do a rough sketch of them and find the relevant intercepts (if you can’t, it is worth revising the Function sketches part of your Prelim textbook!)

[jamonwindeyer]

Help please!
What is the formula for solving this? Please provide working out!
Bill thinks he can afford a mortgage payment of $800 each month. How much can he borrow, to the nearest $100, over 25 years at 11.5% p.a.?

Hey there! So this is actually a tricky question because it is our standard question type, but the unknown is actually the principal! So the setup looks much the same as normal, except \(P\) is now an unknown. I'm considering \(1+r=1.0098...\approx1.01\) to make it easier to type, but you'll probably want to be more precise:



We know \(A_{300}=0\) if the loan is repaid in 25 years, so now just substitute \(n=12\times25=300\) and then solve for \(P\)!

[eeshab7]

Hi could I get some help with this question. I thought I had it but the answers say something completely different.

[david.wang28]

Hi could I get some help with this question. I thought I had it but the answers say something completely different.

You need to integrate this to find the total volume of the container at a point of time. After integrating, let t = 0, so the C becomes 15000. Then let t = 10, and you should be there. Hope this helps

[eeshab7]

okay I got it, thanks for the help

[david.wang28]

Hi there,
I'm not doing 2u right now but I stumbled across this random challenging 2u question, and I have no idea how to do it. Can anyone please help me out? Thanks