it appears we need
1+t = 1+sin s,so
t = sin s
t must be in [0,1] if s is in [0,pi)
also,
2t^2+1 = 2-cos(2s)
2t^2+1 = 2 - (1-2sin^2 s)
2t^2+1 = 2sin^2 s + 1
same domain.
To what interval I must we restrict the parameter t if the graph of the parametric equations
x=t+1
y = 2t^2 +1
for t in the set I
is identical to the graph of the parametric equations
x = 1+ sin s
y = 2- cos(2s)
for s in [0,pi)?
5 answers
Thx!
its not same domain
maybe you misunderstood
@steveiswrong by same domain, he was talking about the domain of t, [0,1], and you misunderstood him. Also, please do not insult people by adding a wrong/right to their name. It's rude. The kind sir took his time to write that solution. You should be thanking him.