Given the cone z=sqrt(x^2+y^2) and the plane z=5+y. Represent the curve of intersection of the surfaces with a vector function r(t).

babeeb0oL 2021-09-15 Answered
Given the cone z=x2+y2 and the plane z=5+y.
Represent the curve of intersection of the surfaces with a vector function r(t).
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Expert Answer

Fatema Sutton
Answered 2021-09-16 Author has 88 answers
x=f(t),y=g(t),andz=h(t)
The vector:
r(t)=f(t)i+g(t)j+h(t)k
Consider the equation of cone surface:
z=x2+y2 (a)
Substitute z=5+y to (a)
5+y=x2+y2
(5+y)2=(x2+y2)2
25+10y+y2=x2+y2
25+10y=x2
10y=x225
y=110(x225)
y=110x252
Consider the equation for plane surface:
z=5+y (b)
Substitute y=110x252 to (b)
z=5+110x252
z=110x2+52
To define each variable in terms of the parameter t, set x=t
y=110t252
and y=110t252
Thus,
r(t)=(t)i+(110t252)j+(110t2+52)k
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