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Mathematics: Post your doubts here!

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first, when they told you the tangent at point P passes through the origin and if you draw a straight line ..to find the gradient you will do
(y-0)/(x-0) and so you will the gradient for the tangent at P = y/x

then, from the dy/dx equation you got above, make it equal to y/x
so ..
e^-x (-x^2+2x) = y/x
and multiply the x.. you will end up with y= e^-x (2x^2-x^3) ..this is your y!
substitute this y in the equation y=x^2e^-x to get
x^2e^-x = e^-x(2x^2 - x^3)
cancel the e^-x

you will get
x^2 = 2x^2 - x^3
-1x^2 = -1x
-1x=-1
so x=1 :D
 
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can anybody post the link or something for any notes of vectors? please that would be great help cause i dont know anything about them !
 
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I didn't make these, all thanks goes to the members here
 

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[ln (1+x^2)]/2x

Its a rule .. integration of 1/x is lnx and in this case you just go back and divide by the differenciation of the inner term hence the division by 2x

This is how you do it:

∫1/(1 + x²) dx

suppose x = tan Ө so Ө = tan^(-1) x

so 1 + x² = 1 + tan² Ө = sec² Ө

and dx/dӨ = sec² Ө

dx = sec² Ө . dӨ

So ∫1/(1 + x²) dx = ∫(1/sec² Ө) . sec² Ө dӨ

= ∫ 1 dӨ

= Ө + k

= tan^(-1) x + k


credit: http://malaysia.answers.yahoo.com/question/index?qid=20090903194558AAkgNsU
 
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