Question

At what points does the helix r(t) = sin(t), cos(t), t intersect the sphere x² + y² + z² = 37?

Answer 1

The helix r(t) = sin(t), cos(t), t intersects the sphere x² + y² + z² = 37 at (sin(6), cos(6), 6) and (sin(-6), cos(-6), -6).

Step-by-step explanation:

The helix r(t) = sin(t), cos(t), t intersects the sphere x² + y² + z² = 37 at two points. To find these points, we substitute the x, y, and z coordinates of the helix into the equation of the sphere.

We get sin²(t) + cos²(t) + t² = 37. Simplifying, we have 1 + t² = 37. Solving for t, we get t = ±√(36) = ±6.

So, the helix intersects the sphere at the points (sin(6), cos(6), 6) and (sin(-6), cos(-6), -6).

Answer 2

The helix r(t) = sin(t), cos(t), t intersects the sphere x² + y² + z² = 37 at two points: (-√10/√2, √10/√2, π/2) and (√10/√2, -√10/√2, 3π/2).

Step-by-step explanation:

To find the intersection points, we'll substitute the parametric equations of the helix r(t) = sin(t), cos(t), t into the equation of the sphere x² + y² + z² = 37.

Substituting sin(t) for x, cos(t) for y, and t for z in the equation of the sphere, we get:

(sin(t))² + (cos(t))² + t² = 37

1 + t² = 37

t² = 36

t = ±6

Now, since x = sin(t) and y = cos(t), using t = 6 and t = -6 in the parametric equations gives us the points of intersection:

When t = 6, x = sin(6) ≈ -√10/√2, y = cos(6) ≈ √10/√2, and z = 6.

When t = -6, x = sin(-6) ≈ √10/√2, y = cos(-6) ≈ -√10/√2, and z = -6.

Thus, the points of intersection are (-√10/√2, √10/√2, π/2) and (√10/√2, -√10/√2, 3π/2), where t = 6 and t = -6 respectively. These points lie on the given sphere and correspond to the intersections between the helix and the sphere.