Will the real TeeEm please stand up!

Original post by Mutleybm1996

Is there a document showing how many candidates sat each exam?

Posted from TSR Mobile

Graham sent out a vast document with exam statistics, but it is mostly by percentages. The only figures involving numbers of candidates are for each qualification, not each unit.

Reply 1242

OP

Original post by fatart123

What's it on? There's no guarantee I'd even be able to do it if it's one of your horrible sequence product questions...

Original post by physicsmaths

I will try it if i can do it bro.

Posted from TSR Mobile

Here comes the beast.
The finite region bounded by the curve with equation
y = (6 - x)(x -2), the tangent to the curve at the point where x = 3 and the x axis, is rotated by a full turn about the tangent.
Determine the volume generated.
(Calculator is allowed)

I will make another similar question later (easier) because when I made this little did I know how tough it was going to be

(edited 8 years ago)

Reply 1243

Original post by TeeEm

Here comes the beast.
The finite region bounded by the curve with equation
y = (6 - x)(x -2), the tangent to the curve at the point where x = 3 and the x axis, is rotated by a full turn about the tangent.
Determine the volume generated.
(Calculator is allowed)

I will make another similar question later (easier) because when I made this little did I know how tough it was going to be

Well, I've moved the region so that it's 'connected' to the origin. But rotating the axis/curve + tangent about the origin and keeping the equation of the curve correct is pretty horrible

Reply 1244

OP

Original post by fatart123

Well, I've moved the region so that it's 'connected' to the origin. But rotating the axis/curve + tangent about the origin and keeping the equation of the curve correct is pretty horrible

I tried rotation matrices but the new equation was appalling ...
I derived a volume element but I do not know if there is any silly or serious error in my workings... over 3 pages still

Reply 1245

Original post by TeeEm

I tried rotation matrices but the new equation was appalling ...
I derived a volume element but I do not know if there is any silly or serious error in my workings... over 3 pages still

lol, I found this after thinking about matrices, but I'm not going to use/adapt it because it's kinda cheating now:
http://www.emathhelp.net/notes/calculus-2/applications-of-integrals/volume-of-solid-of-revolution-about-slant-line/

Trying to translate to polar coordinates and rotate once I've got it in that system... This quadratic is a pain, but it should be doable

Reply 1246

OP

Original post by fatart123

lol, I found this after thinking about matrices, but I'm not going to use/adapt it because it's kinda cheating now:
http://www.emathhelp.net/notes/calculus-2/applications-of-integrals/volume-of-solid-of-revolution-about-slant-line/

Trying to translate to polar coordinates and rotate once I've got it in that system... This quadratic is a pain, but it should be doable

excellent research...
I can now check my answer

Reply 1247

Craig1998

12

Original post by TeeEm

here it is

Thanks so much, been looking for this for ages now.

Is there any chance you have the OCR markschemes and examiner reports? I do psychology and chemistry and haven't been able to find a single thing on TSR.

Reply 1248

OP

Original post by Craig1998

Thanks so much, been looking for this for ages now.

Is there any chance you have the OCR markschemes and examiner reports? I do psychology and chemistry and haven't been able to find a single thing on TSR.

I am sorry but I am unable to.

I have a friend in a school which teaches EDEXCEL maths and kindly lets me have access to Maths EDEXCEL material only

Reply 1249

Original post by TeeEm

excellent research...
I can now check my answer

What was your answer? Would like to check mine tbh since I'm 50/50 as to whether I made a logic error in my work somewhere

Reply 1250

A Slice of Pi

13

Original post by TeeEm

Here comes the beast.
The finite region bounded by the curve with equation
y = (6 - x)(x -2), the tangent to the curve at the point where x = 3 and the x axis, is rotated by a full turn about the tangent.
Determine the volume generated.
(Calculator is allowed)

I will make another similar question later (easier) because when I made this little did I know how tough it was going to be

It truly is a beastly question! I had an attempt although I haven't checked for mistakes yet (they are no doubt in plentiful supply with the integration, the transformations etc.). I ended up with
7root 5 /6 pi

(edited 8 years ago)

Reply 1251

Craig1998

12

Original post by TeeEm

I am sorry but I am unable to.

I have a friend in a school which teaches EDEXCEL maths and kindly lets me have access to Maths EDEXCEL material only

No problem, I waited the entire year to find out all of my GCSE exam markschemes and examiner reports.

Reply 1252

OP

Original post by fatart123

What was your answer? Would like to check mine tbh since I'm 50/50 as to whether I made a logic error in my work somewhere

my answer is wrong

I am going to go through the calculus method you supplied in a while (still doing other material)

Reply 1253

A Slice of Pi

13

Original post by TeeEm

my answer is wrong

I am going to go through the calculus method you supplied in a while (still doing other material)

Did you not use calculus yourself?

Reply 1254

Original post by A Slice of Pi

It truly is a beastly question! I had an attempt although I haven't checked for mistakes yet (they are no doubt in plentiful supply with the integration, the transformations etc.). I ended up with
7root 5 /6 pi

I'm pretty close to that and got lazy at the end by approximating my integral using Wolfram, so we could be correct. Did you translate and reflect the quadratic so that it makes applying the calculus from that link easier or not?

Reply 1255

OP

Original post by A Slice of Pi

Did you not use calculus yourself?

I did but I got the thickness of the element wrong

Reply 1256

A Slice of Pi

13

Original post by fatart123

I'm pretty close to that and got lazy at the end by approximating my integral using Wolfram, so we could be correct. Did you translate and reflect the quadratic so that it makes applying the calculus from that link easier or not?

My method was a little more complicated I'd say (making it more likely incorrect). I drew a diagram showing the curve, the tangent and the required area. What I then did was introduce a second pair of axes (w and z, as shown) at (1.5, 0) and started trying to express w and z in terms of x and y. I came up with

w = \frac{2x-3}{\sqrt{5}}

and

z = \frac{\sqrt{5}y}{2}

. All I did then was make substitutions in the formula

V = \pi\int_{0}^{\frac{3\sqrt{5}}{2}}w^2 dz

to get it in terms of x and dy. The limits etc. were explained in my full method. Excuse the poor diagram...

(edited 8 years ago)

I don't know if my decent results this year were luck or my maths skills deteriorate incredibly quickly without sufficient practice...I have absolutely no clue how to approach the last question given, or would not have had any without the ideas of others (still probably can't work towards a solution even given that). Starting to feel a maths degree is not for me..

Original post by fatart123

lol, I found this after thinking about matrices, but I'm not going to use/adapt it because it's kinda cheating now:
http://www.emathhelp.net/notes/calculus-2/applications-of-integrals/volume-of-solid-of-revolution-about-slant-line/

May have to resort to that...although it looks somewhat ugly

Reply 1258

Original post by A Slice of Pi

My method was a little more complicated I'd say (making it more likely incorrect). I drew a diagram showing the curve, the tangent and the required area. What I then did was introduce a second pair of axes (w and z, as shown) at (1.5, 0) and started trying to express w and z in terms of x and y. I came up with

w = \frac{2x-3}{\sqrt{5}}

and

z = \frac{\sqrt{5}y}{2}

. All I did then was make substitutions in the formula

V = \int_{0}^{\frac{3\sqrt{5}}{2}}w^2 dz

to get it in terms of x and dy. The limits etc. were explained in my full method. Excuse the poor diagram...

Hah, I tried to do this. Instead of introducing new axis, I converted the translated equations of the line and curve to polar form, rotated them in that system (quite a bit easier) and then reconverted to cartesian. Sadly, it's seems actually impossible (Wolfram could only approximate) to explicitly express y in terms of x after doing so, so that didn't work for me.

I like how you've tried it though, so I'll have a go at doing the question that way tbh.

Original post by 1 8 13 20 42

I don't know if my decent results this year were luck or my maths skills deteriorate incredibly quickly without sufficient practice...I have absolutely no clue how to approach the last question given, or would not have had any without the ideas of others (still probably can't work towards a solution even given that). Starting to feel a maths degree is not for me..

May have to resort to that...although it looks somewhat ugly

It's just lots and lots of boring algebra. I was too confused as to how to get it to work from underneath, so I just worked it out from the 'inverse' of the quadratic above the tangent

(edited 8 years ago)

Reply 1259

OP