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the parametric equations x(t)=1/sqrt(t+1) and y(t)=t/(t+1) where t is not -1, represent what curve?
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solve the first equation for t(x) and sub that into the y equation in place of the ts and you should get a normal looking y(x) equation to compare with
how do i solve the first one?
you cannot solve\[x=\frac1{\sqrt{t+1}}\]for t?
x^2-1?
t=x^2-1, which when i sub it into the other gives me y=2. im i doing this right?
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\[x=\frac1{\sqrt{t+1}}\]\[x^2=\frac1{t+1}\]\[t+1=\frac1{x^2}\]\[t=\frac1{x^2}-1\]
still gives me y=2?
wait i got it, and i simplified it to y=1-x^2. how would i express this as a polar?
Show that for any metric space X, the set X \ {x} is open in X.
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