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Help with C2 question

3.
A curve has equation: Y=1/[(x^2)+1]
The shaded region R is bounded by the curve, the coordinate axis and the line x=2.

a) Use the trapezium rule with 4 strips of equal width to estimate the area R.
[DONE this and got 1.10 to 3sfg]

The cross-section of a support for a bookshelf is modelled by R with 1 unit on each axis representing 8 cms. Given that the support is 2 cm thick,
b) Find an estimate for the volume of the support

4.
Amy plans to join a savings scheme in which she will pay in £500 at the start of each year.
One scheme that she is considering pays 6% interest on the amount in the account at the end of each year.
For this scheme,

a) find the amount of interest paid into the account at the end of the second year,
b) show that after interest is piad at the end of the 8th year, the amount in the account will be £5246 to the nearest pound.

Another scheme that she us considering pays 0.5% interest on the amount at the end of the each month.
c) find, to the nearest pound, how much more or less will be in the account at the end of the 8th year under this scheme.

Answers
a) £61.8
b) £5246
These are correct

for c) here is what I did

96 months in 8 years
Start Month 1 = 500
End month 1 = 500 * 1.005
Start month 2 = (500 * 1.005)+500
End month 2 = [(500*1.005)+500]*1.005

SO end month 2 = 500 * 1.005^2 + 500 * 1.005
= 500(1.005 + 1.005^2)

Therefore:

Sn = [a(r^n-1)] / (r-1)
Sn = 500* [1.005(1.005^96 - 1) / (1.005 - 1)
Sn = 61721.342

BUT
the answer is 5285.71

Therefore I cant calculate the difference

What they did
They said (1.005)^12 = 1.0617
and then said first term = 1.0617, n = 8, r = 1.0617

I don't understand their method.

(edited 9 years ago)

Reply 1

ps1265A

The curve
http://www.thestudentroom.co.uk/showthread.php?t=213215

Reply 2

m4ths/maths247

For 3b could you use a GCSE topic about length and area?

Reply 3

ps1265A

Original post by m4ths/maths247

For 3b could you use a GCSE topic about length and area?

Yup, but the mark scheme states that the area of the cross section is 8^2 x 1.10...

That's the only part of the working I don't understand.

Reply 4

Matureb

You are assuming that £500 is paid in monthly. Not just the interest.

Reply 5

m4ths/maths247

Original post by ps1265A

Yup, but the mark scheme states that the area of the cross section is 8^2 x 1.10...

That's the only part of the working I don't understand.

That is my point though. :smile:

If you understand Linear/Area Scale factors from GCSE it will make sense.

I did this a couple of weeks back for my y11s

https://www.youtube.com/watch?v=101Lneukck0

Reply 6

TenOfThem

Original post by ps1265A

Yup, but the mark scheme states that the area of the cross section is 8^2 x 1.10...

That's the only part of the working I don't understand.

Your area assumes units on the axes are 1, they are actually 8

Reply 7

ps1265A

Original post by TenOfThem

Your area assumes units on the axes are 1, they are actually 8

There are 2 units on the x axis, doesn't this make it 16cm on the x axis and 8 cm on the y axis?

Reply 8

TenOfThem

Original post by ps1265A

There are 2 units on the x axis, doesn't this make it 16cm on the x axis and 8 cm on the y axis?

I am not sure what your point is

A square measuring 1cm by 1cm has area = ?

A square measuring 8cm by 8cm has area = ?

So each units squared in your original area now has an area of ?

Reply 9

MP2205

Original post by ps1265A

Yup, but the mark scheme states that the area of the cross section is 8^2 x 1.10...

That's the only part of the working I don't understand.

you do 8^2 because you are working out the area and not a length so the scale factor has to be squared since you're supposedly including the width*height in your answer