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)#