Higher Maths Paper 1

River Song

The integration ones are my most favourite of all of them

I hope that the S/S question is easy, I always do it as I'm trying to avoid the 4 Diff. proofs.

I've been told I'm more than able for an A, but it's ALWAYS paper 2 that screws me over...fun weekend ahead.

hunii07

Generally integration is one of my better questions .. I don't know I still don't like those proofs though

Maybe_Memories

Michael_E wrote: »

I've been told I'm more than able for an A, but it's ALWAYS paper 2 that screws me over...fun weekend ahead.

Me too. I did my LC last year, got 97% on Paper 1, did crap on paper 2, and in the end I got an A2, but I was 0.12% from an A1, a single mark... :rolleyes:

Marking scheme crucified me. I made two tiny mistakes on paper 2. Should have lost 3 marks for each, so 6 in total. But, the way the marking scheme was changed around I lost a total of 30 marks for two tiny addition mistakes... :rolleyes:

River Song

Maybe_Memories wrote: »

Me too. I did my LC last year, got 97% on Paper 1, did crap on paper 2, and in the end I got an A2, but I was 0.12% from an A1, a single mark... :rolleyes:

Marking scheme crucified me. I made two tiny mistakes on paper 2. Should have lost 3 marks for each, so 6 in total. But, the way the marking scheme was changed around I lost a total of 30 marks for two tiny addition mistakes... :rolleyes:

Oh God, that sucks ><

I just did a mock for my teacher the other week. Got an A1 in paper 1, got a B3 in Paper 2 :rolleyes: I am just going balls-out this weekend for paper 2 ><

kpac

Obviously most people here prefer paper 1? I prefer paper 2 to be honest...

RHunce

kpac wrote: »

They're simple! As long as you can do basic intergration you can do these proofs.

Care to explain how they work for somebody who never did them?

River Song

RHunce wrote: »

Care to explain how they work for somebody who never did them?

Sphere

• Draw a circle of equation x^2+y^2=r^2. Radius = r and centre = (0,0)
• Then say that the points it cuts the x-axis are (0,r) and (0,-r) .... they're where we rotate through the x-axis to make our sphere. (i.e. limits)
• Re-arrange equation to get y^2=r^2-x^2
• Intergrate (pi)(y^2) dx between r and -r.

Cone

• Draw a line y=mx (it passes straight through the origin and is diagonal).
• Label a point on this line (r,h)
• So the vertical distance = r (radius) and the horizontal height = h (height)
• If you turn it 90dgs clockwise, it looks more like a cone that way, if you're having trouble visualising it.
• So you can say the slope [m] of this line is the rise over the run so it's r/h
• Replace r/h for m in -> y=mx
• So you now intergrate (pi)(y^2)dx between h and 0.

Maybe_Memories

RHunce wrote: »

Care to explain how they work for somebody who never did them?

You've got a formula in the tables, V = pi (integral of y dx between b and a)
you need to get y in terms of x, so if its a cone or cylinder, just put in the equation of the line, the distance between the x axis and the highest point on y is your base radius, distance between the origin and this x point is your h.

Same with a sphere, except you're dealing with y squared

River Song

RHunce wrote: »

Care to explain how they work for somebody who never did them?

I can write both out quickly and post them here if you like?

kpac

RHunce wrote: »

Care to explain how they work for somebody who never did them?

Basically, the integral of pi times y-squared with respect to x between r and -r will give you 4/3(pi)(r^3)

The cone is similar enough. The integral of pi times y-squared with respect to x between h and 0. It's hard to explain in words.


EDIT: we're not quick enough...

partyatmygaff

Michael_E wrote: »

Sphere

• Draw a circle of equation x^2+y^2=r^2. Radius = r and centre = (0,0)
• Then say that the points it cuts the x-axis are (0,r) and (0,-r) .... they're where we rotate through the x-axis to make our sphere. (i.e. limits)
• Re-arrange equation to get y^2=r^2-x^2
• Intergrate (pi)(y^2) dx between r and -r.

Cone

• Draw a line y=mx (it passes straight through the origin and is diagonal).
• Label a point on this line (r,h)
• So the vertical distance = r (radius) and the horizontal height = h (height)
• If you turn it 90dgs clockwise, it looks more like a cone that way, if you're having trouble visualising it.
• So you can say the slope [m] of this line is the rise over the run so it's r/h
• Replace r/h for m in -> y=mx
• So you now intergrate (pi)(y^2)dx between h and 0.

Proofs are far easier to remember in this format. I do this for all my proofs.

Maybe_Memories

Okay, guy above did a way better job at explaining

in case anyone wants to know why you multiply by pi, pi is the area of a unit circle, so when you have a the area of a triangle, multiplying it by the area of a circle spreads the triangle out making a cone.

Okay that was a terrible explanation, but hey, it makes sense to me

Digits

Area of a circle by integration methods. Likely to come up guys? I cant find the proof if anyone would care to tell me?

aranciata

anybody know how to find the roots of a cubic? (1999 paper is hard, grr)

aranciata

aranciata wrote: »

anybody know how to find the roots of a cubic? (1999 paper is hard, grr)

without guessing, cause I have a kx^2

partyatmygaff

aranciata wrote: »

anybody know how to find the roots of a cubic? (1999 paper is hard, grr)

If you have a Casio FX-83...

Push mode and switch to table mode. Type in the cubic equation and set start and end values and the step value to 1. Look for where f(x)=0 and they are your roots.

Then work backwards to show your steps on the exam paper

Maybe_Memories

Digits wrote: »

Area of a circle by integration methods. Likely to come up guys? I cant find the proof if anyone would care to tell me?

Equation of a circle is y^2 + x^2 = r^2, rearrange and take square roots to get

y = root(x^2 - r^2)

Plug this into formula i mentioned a few posts above

There;s a substitution for this, which I don't remember off the top of my head, someone mentioned it as the thing on the 1999 paper.

Integrate between r and -r, unsubstitute if thats even a word...

Should be okay

River Song

I'm working on deriving the area of a circle proof, I doubt it'll come up. It's not in my book (T&T)

aranciata wrote: »

without guessing, cause I have a kx^2

• Sub in 0, 1, -1, 2, -2 ......... for x. When you find a value for x that makes the equation = 0, you can say f(number) = 0, therefore (x - number) is a factor.
• Divide that (x - number) into the original equation, you'll find a quadratic equation.
• The two roots of that quadratic, and the original (x - number) are your 3 roots of the cubic.

partyatmygaff

Ah... are you on Q3bii of 1999?

You use long division to get a quadratic and you solve that for the remaining two roots.

Divide [LATEX]z^3 - (k)z^2++(22)z - 20 [/LATEX] by [LATEX]z-3-i[/LATEX]

Maybe_Memories

partyatmygaff wrote: »

If you have a Casio FX-83...

Push mode and switch to table mode. Type in the cubic equation and set start and end values and the step value to 1. Look for where f(x)=0 and they are your roots.

Then work backwards to show your steps on the exam paper

This.

Also, to find one of the roots straight away, if you have ax^3 + bx^2 + cx + d, one of the factors has to be d/a, so firstly try that and the negative value of it.

Get a factor from the root, divide into the cubic to get a quadratic.

aranciata

partyatmygaff wrote: »

Ah... are you on Q3bii of 1999?

You use long division to get a quadratic and you solve that for the remaining two roots.

Divide [LATEX]z^3 - (k)z^2+(22)z - 20 [/LATEX] by [LATEX]z-3-i[/LATEX]

question 7(c)(i) from 1999. working from the answer, x has to be greater than the y value of it's local maximum. that doesn't make much sense to me, does the local maximum have to be above the x axis??

Digits

Maybe_Memories wrote: »

Equation of a circle is y^2 + x^2 = r^2, rearrange and take square roots to get

y = root(x^2 - r^2)

Plug this into formula i mentioned a few posts above

There;s a substitution for this, which I don't remember off the top of my head, someone mentioned it as the thing on the 1999 paper.

Integrate between r and -r, unsubstitute if thats even a word...

Should be okay

Yeah I'm getting that far but I cant seem to figure out what the substitution should be.

Maybe_Memories

Digits wrote: »

Yeah I'm getting that far but I cant seem to figure out what the substitution should be.

I'm not 100% certain as to what it is and don't want to post my initial hunch in case it mixes you up.

There should be a section in the integration chapter of your book about integrating functions of the form squareroot(a^2 - x^2)

Maybe_Memories

Let x = rsin(theta)

partyatmygaff

aranciata wrote: »

question 7(c)(i) from 1999. working from the answer, x has to be greater than the y value of it's local maximum. that doesn't make much sense to me, does the local maximum have to be above the x axis??

In a cubic function the local max and local min have to be on opposite sides of the x axis to cut it three times hence three roots. For it to have three roots the local maximum y coordinate must be positive. To show the range where it has three roots simply put the the y coordinate of the maximum value greater than 0 and solve.

For two real roots and one imaginary root the y coordinate of the local max must be equal to 0.

Maybe_Memories

Nevermind

Maybe_Memories

partyatmygaff wrote: »

In a cubic function the local max and local min have to be on opposite sides of the x axis to cut it three times hence three roots. For it to have three roots the local maximum y coordinate must be positive. To show the range where it has three roots simply put the the y coordinate of the maximum value greater than 0 and solve.

For two real roots and one imaginary root the y coordinate of the local max must be equal to 0.

It doesn't say they have to be distinct, so one of the local max/mins could infact be on the x axis.

aranciata

partyatmygaff wrote: »

In a cubic function the local max and local min have to be on opposite sides of the x axis to cut it three times hence three roots. For it to have three roots the local maximum y coordinate must be positive. To show the range where it has three roots simply put the the y coordinate of the maximum value greater than 0 and solve.

For two real roots and one imaginary root the y coordinate of the local max must be equal to 0.

taking this down and committing it to memory! a sincere thanks!

River Song

Okay here are the proofs of the sphere/cone volume. I figured out the area one, but it's on my whiteboard and not on paper, sorry :P

Sphere


Cone

Digits

Maybe_Memories wrote: »

Let x = rsin(theta)

Thanks:)