Answer:
The parametric equation is
[tex]x = \frac{1}{2}[/tex]
[tex]y = 1[/tex]
[tex]z = \frac{1}{2} + t[/tex]
Step-by-step explanation:
From the question we are told that
The equation given are
[tex]x^2 + y^2 = 4[/tex]
and y = 1
Thus
Let
[tex]f(x,y,z) = x^2 + y^2 -4 = 0[/tex]
And [tex]g(x,y,z ) = y- 1[/tex]
Now we will differentiate f(x, y , z ) and g(x,y,z) to obtain value of f(x,y,z ) and g(x,y,z) where the line passed through
So
[tex]\Delta f(x,y,z) = 2x + 2y + 0z[/tex]
At Point: (1/2, 1, 1/2)
[tex]\Delta f(1/2,1 ,1/2) = 2 (1/2) + 2(1) + 0(1/2)[/tex]
[tex]\Delta f(1/2,1 ,1/2) = 1 + 2 + 0[/tex]
converting to coordinate form
[tex]\Delta f(1/2,1 ,1/2) = <1 \ , 2 \ ,0>[/tex]
And
[tex]\Delta g(x,y,z) = 0x + 1 + 0z[/tex]
At Point: (1/2, 1, 1/2)
[tex]\Delta g(1/2,1 ,1/2) = 0 (1/2) + 1 + 0(1/2)[/tex]
converting to coordinate form
[tex]\Delta g(1/2,1 ,1/2) \ = \ <0 \ , 1 \ , 0 >[/tex]
At point of intersection we have that
[tex]G = \Delta f \ X \ \Delta g = | det \left[\begin{array}{ccc}i&j&k\\1&2&0\\0&1&0\end{array}\right] |[/tex]
=> [tex]G = i((2 * 0 ) - (0 * 1))- j ((1 * 0 ) - (0*0)) + k((1 * 1 ) - (2 * 0))[/tex]
=> [tex]G = 0i - j (0) + k[/tex]
Thus the parametric equation is
[tex]x = \frac{1}{ 2} - 0 t[/tex]
=> [tex]x = \frac{1}{2}[/tex]
[tex]y = 1 - 0(t)[/tex]
=> [tex]y = 1[/tex]
and
[tex]z = \frac{1}{2} + t[/tex]
