Find d²y/dx² in terms of t if x= (1-t²)/(1+t²) and y=t/(1+t²)?

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Answer 1 · 2015-08-25T07:53:02

$(-4y^2-x^2)/(16y^3)$

First find $dx/dt= ((1+t^2)(-2t)- (1-t^2)(2t))/(1+t^2)^2$ = $(-4t)/(1+t^2)^2$

$dy/dt=(1(1+t^2)- t (2t))/(1+t^2)^2$ = $(1-t^2)/(1+t^2)^2$

This would give $dy/dx=(1-t^2)/ (-4t)$

Since $y(1+t^2) = t$ and $x(1+t^2) = 1-t^2$ , you get:

$dy/dx=(xcancel((1+t^2)))/ (-4ycancel((1+t^2))) = x/(-4y)$

$(d^2y)/dx^2= (-4y -x(-4)dy/dx)/(16y^2)$ = $(-4y+4xdy/dx)/(16y^2)$

= $(-4y^2-x^2)/(16y^3)$

Answer 2 · 2015-08-28T05:29:37

$dy/dx= x/(-4y)$

It is like this:

It was derived that $dy/dx=(1-t^2)/(-4t)$ .
Now, from the given equations, it is evident that $1-t^2= x(1+t^2)$ and
$t= y(1+t^2)$ . Hence,

$dy/dx=(1-t^2)/(-4t)= (x(1+t^2))/(-4y(1+t^2))=x/(-4y)$