The Student Room Group

Fp1 exam question help

question 7 jan 09 fp1 aqa
fp1 jan 09.png

(link if wanted) http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MFP1-W-QP-JAN09.PDF

i understand the theory i.e i have to prove it is linear interpolation however i can't seem to obtain the rearranged version for r - mainly the c-d part why is that needed and not just d for the similar triangles?, could someone walk it through please.

Also, on q3 am i right in thinking i have too minus the (pi divided by 2) then divide by -3?
FP1 09 q3.png
(edited 11 years ago)

Scroll to see replies

Reply 1
Oh yeah ... horrible question
Original post by TenOfThem
Oh yeah ... horrible question


so i shouldn't feel that annoyed that i was stuck and spent a lot of time on this?
Reply 3
I seem to remember that I spent ages doing convoluted adding of stuff to make it work

Then when I was showing the class I saw a really obvious way
(edited 11 years ago)
Reply 4
Original post by coolstorybrother
so i shouldn't feel that annoyed that i was stuck and spent a lot of time on this?


I spent time using the little triangles .... then in class I thought ... hang on and used this

Using For Working Out Stuff P15 (1).png
Original post by TenOfThem
I seem to remember that I spent ages doing convoluted adding of stuff to make it work

Then when I was showing the case I saw a really obvious way


im not sure but is the really obvious way using similar triangles (r-a)/c = (b-r)/d ? could you help on q3?
Reply 6
Original post by coolstorybrother
im not sure but is the really obvious way using similar triangles (r-a)/c = (b-r)/d ? could you help on q3?


It is about using the correct similar triangles

see ^^^^^^
Reply 7
What is arctan(rt3)
Reply 8
Original post by TenOfThem
I seem to remember that I spent ages doing convoluted adding of stuff to make it work

Then when I was showing the case I saw a really obvious way


Which part of the question is difficult?
Reply 9
Original post by raheem94
Which part of the question is difficult?


none of it
Reply 10
Original post by TenOfThem
none of it


On which part did you spent ages?
I may like to have a go at it.
Original post by raheem94
On which part did you spent ages?
I may like to have a go at it.


The initial proof

As I said I used the 2 small triangles and then had to do manipulation to get the format required

But ... as my diagram shows it was very easy
Reply 12
Original post by TenOfThem
The initial proof

As I said I used the 2 small triangles and then had to do manipulation to get the format required

But ... as my diagram shows it was very easy


I did the 1st part in this way,

The gradient of the line(using points P(a,c) and Q(b,d) ) is, m=dcba \displaystyle m = \frac{d-c}{b-a}

Now form the equation of the line, using the point R(r,0),
y0=dcba(xr)    y(ba)=(dc)(xr)    y(ba)=dxdrcx+cr    y(ba)=x(dc)+r(cd)    y(ba)=x(cd)+r(cd)    y(bacd)=x+r    r=x+y(bacd) \displaystyle y - 0 = \frac{d-c}{b-a} (x-r) \implies y(b-a) = (d-c)(x-r) \\ \implies y(b-a) = dx - dr -cx +cr \implies y(b-a) = x(d-c) + r(c-d) \\ \implies y(b-a) = -x(c-d) + r(c-d) \implies y \left( \frac{b-a}{c-d} \right) = -x + r \\ \implies r = x + y \left( \frac{b-a}{c-d} \right)

Now sub in P(a,c), this gives,

r=a+c(bacd) \boxed{ r = a + c \left( \frac{b-a}{c-d} \right) }

How is this solution?
I did Purple ratio = Green ratio

rac=bacd\dfrac{r-a}{c} = \dfrac{b-a}{c-d}

so

ra=c(bacd)r-a = c(\frac{b-a}{c-d})

so

r=a+c(bacd)r = a + c(\frac{b-a}{c-d})
Original post by raheem94


How is this solution?



I did similar to this before I realised the easy way I have shown above

That is what I meant when I said it took too much manipulation
Reply 15
Original post by TenOfThem
I did similar to this before I realised the easy way I have shown above

That is what I meant when I said it took too much manipulation


Your way is better, but it is difficult to realize that technique. Similar triangles aren't much used at A-Level, so the students will probably not spot the easy technique.

Though my way is also not very long, it only takes around a minute.
Original post by raheem94
Your way is better, but it is difficult to realize that technique. Similar triangles aren't much used at A-Level, so the students will probably not spot the easy technique.

Though my way is also not very long, it only takes around a minute.


Yeah ... from experience I just knew it should be "see-able" rather than "workable"


Once I "saw" it I was like "OMG how obvious"
Original post by TenOfThem
It is about using the correct similar triangles

see ^^^^^^

i kind of understand this... everything except why it is c-d?

Original post by TenOfThem
What is arctan(rt3)


umm i really dont know, should i? :frown:
Original post by coolstorybrother
i kind of understand this... everything except why it is c-d?



umm i really dont know, should i? :frown:


c-d ... yeah I had to think ... but d is already a negative number


Yes you should :frown:
Reply 19
Original post by raheem94
Your way is better, but it is difficult to realize that technique. Similar triangles aren't much used at A-Level, so the students will probably not spot the easy technique.

Though my way is also not very long, it only takes around a minute.


Similar triangles is widely used as a method for linear-interpolation.

Quick Reply