IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: thecandydoll on May 17, 2011, 09:57:27 am

Title: P3 help :D
Post by: thecandydoll on May 17, 2011, 09:57:27 am: I got three doubts,could y'all help me out?

Oct/Nov 2003 Q4 (ii)
May June 2004 (Q6)
Oct/Nov 2004 Q10 (ii)
Title: Re: P3 help :D
Post by: thecandydoll on May 17, 2011, 12:54:13 pm: BUMPPP
Title: Re: P3 help :D
Post by: elemis on May 17, 2011, 01:25:58 pm: From the equation y=x we can :

Replace x in the original equation (given in the question) and solve for x.

Do the same for y, except replace with x.

You should get the coordinates of intersection as (a/4,a/4)

Input the these into dy/dx and solve :

$\frac{dy}{dx}=-\frac{\sqrt{a/4}}{\sqrt{a/4}}$

Simplify to -1.

Construct an equation :

y-a/4 = -1(x-a/4)

Simplify to answer in markscheme.
Title: Re: P3 help :D
Post by: elemis on May 17, 2011, 01:36:52 pm: Split the variables :

$\frac{1}{\frac{y^3+1}{3y^2}}dy=1dx$

$\frac{3y^2}{y^3+1}dy=1dx$

$\frac{1}{3}ln(y^3+1)=x+c$

When y=1 and x=0, c = \frac{1}{3}ln(2)

$ln(y^3+1)=3x+ln2<br /><br />ln(y^3+1)-ln2=3x$

Can't you continue from here ?