Final answer:
The parametrization of the curve formed by the intersection of the plane y = 6 with the sphere x² + y² + z² = 136 is x(t) = 10 cos(t), y(t) = 6, z(t) = 10 sin(t), with t ranging from 0 to 2π.
Step-by-step explanation:
The student's question asks for a parametrization of the curve formed by the intersection of the plane y = 6 with the sphere x² + y² + z² = 136.
First, we notice that any point on this curve will have a y-coordinate of 6. Plugging y = 6 into the equation of the sphere, we get:
x² + 6² + z² = 136
x² + 36 + z² = 136
x² + z² = 100
This describes a circle of radius 10 in the xz-plane. To parametrize this circle, we use cosine and sine functions which describe a unit circle, and simply scale them by the radius of 10.
The parametric equations are:
- x(t) = 10 × cos(t)
- y(t) = 6
- z(t) = 10 × sin(t)
where t represents the parameter, usually the angle in radians, running from 0 to 2π to describe the full circle.