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CIE A2 Maths Question Help Me!

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loggerhead:
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

astarmathsandphysics:

separate vaiable and integrate


y(|pi /6)=2 so
c=ln2-4 hence

loggerhead:
your left hand integration was wrong.

should be -2/y not -2y.

never the less this still helped a lot so thank you.

astarmathsandphysics:
Even the best people at maths, late at night or early in morning might think that 1+1=3

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