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

Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: loggerhead on October 07, 2009, 12:13:16 am

Title: CIE A2 Maths Question Help Me!
Post by: loggerhead on October 07, 2009, 12:13:16 am: this one is similar to me last question.

it from May/June 2008 paper 3 which iv attached and its question number 8 part ii.

8) In the diagram the tangent to a curve at a general point P with coordinates (x, y) meets the x-axis at T.
The point N on the x-axis is such that PN is perpendicular to the x-axis. The curve is such that, for all
values of x in the interval 0 < x < 1/2(Pi), the area of triangle PTN is equal to tan x, where x is in radians.

(i) Using the fact that the gradient of the curve at P is PN/TN, show that

dy/dx = 1/2 y^2 cot x. (i can do part i. i just put it in to help with part ii)

(ii) Given that y = 2 when x = 1/6(pi), solve this differential equation to find the equation of the curve,
expressing y in terms of x.

the diagram is in the attachment

please help
Title: Re: CIE A2 Maths Question Help Me!
Post by: astarmathsandphysics on October 07, 2009, 09:21:41 am: $\frac {dy}{dx} =0.5y^2 cotx$
separate vaiable and integrate
$\int 2y^{-2}dy =\int cotx$
$-2y=ln(sinx) +c$
y(|pi /6)=2 so $-4=ln(sin(\pi /6))+c=-ln2+c$
c=ln2-4 hence $y=\frac {ln(sinx)+ln2-4}{-2} =\frac {4-ln(2sinx)}{2}$
Title: Re: CIE A2 Maths Question Help Me!
Post by: loggerhead on October 07, 2009, 11:05:35 pm: your left hand integration was wrong.

should be -2/y not -2y.

never the less this still helped a lot so thank you.
Title: Re: CIE A2 Maths Question Help Me!
Post by: astarmathsandphysics on October 09, 2009, 03:23:30 pm: Even the best people at maths, late at night or early in morning might think that 1+1=3