Author Topic: CIE A2 Maths Question Help Me!  (Read 814 times)

Offline loggerhead

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CIE A2 Maths Question Help Me!
« 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

Offline astarmathsandphysics

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Re: CIE A2 Maths Question Help Me!
« Reply #1 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}

Offline loggerhead

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Re: CIE A2 Maths Question Help Me!
« Reply #2 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.

Offline astarmathsandphysics

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Re: CIE A2 Maths Question Help Me!
« Reply #3 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