(b) A simple way would be as follows. Let z=t, where 2
Advanced Engineering Mathematics
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Chapter2: Second-order Linear Odes
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hello, the previous expert have yet to explain where did the sin and cos functions come from and how do we go about solving it.

Transcribed Image Text:(b) A simple way would be as follows. Let z=t, where 2<t<9. It follows that
x + y = t-1. We can then choose x= \t-1cos(u) and y= Vt-1sin(1) , where u
is another free parameter. Thus, a parametric representation for the surface is given by
x= Vt-1 cos(u),
y= \t-1sin(u) '} for 2<t<9.
Z= t

Transcribed Image Text:(b) 1+ x² + y –z = 0 for 2< z<9.
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