Find the closest point on a satellite's orbital path to a given point in 3D space | Step-by-Step Solution

What We're Solving:

We need to find the point C on the elliptical orbit x² + 3y² = 17 (which lies in the xy-plane, so z = 0) that is closest to the 3D point A(2, 16/√3, 8).

The Approach:

This is a classic optimization problem. We're finding the minimum distance between a point in 3D space and a curve in a 2D plane. The key insight is that we need to:

• 1. Express any point on the orbit using a parameter or constraint
• 2. Set up a distance formula from that point to A
• 3. Minimize that distance using calculus

Step-by-Step Solution:

Step 1: Set up the constraint Any point C on the orbit can be written as C(x, y, 0) where x² + 3y² = 17. Since the orbit is in the xy-plane, the z-coordinate is always 0.

Step 2: Write the distance formula The distance from C(x, y, 0) to A(2, 16/√3, 8) is: d = √[(x-2)² + (y-16/√3)² + (0-8)²] d = √[(x-2)² + (y-16/√3)² + 64]

Step 3: Use the optimization strategy Instead of minimizing d, we'll minimize d² (this avoids messy square root derivatives!): f(x,y) = (x-2)² + (y-16/√3)² + 64

Step 4: Apply the constraint using Lagrange multipliers We want to minimize f(x,y) subject to g(x,y) = x² + 3y² - 17 = 0

Set up: ∇f = λ∇g

• ∇f = (2(x-2), 2(y-16/√3))
• ∇g = (2x, 6y)

This gives us:

• 2(x-2) = 2λx → x-2 = λx → x(1-λ) = 2
• 2(y-16/√3) = 6λy → y-16/√3 = 3λy → y(1-3λ) = 16/√3

Step 5: Solve the system From the first equation: x = 2/(1-λ) From the second equation: y = (16/√3)/(1-3λ)

Substitute into the constraint x² + 3y² = 17: [2/(1-λ)]² + 3[(16/√3)/(1-3λ)]² = 17

Step 6: Simplify and solve The correct answer is: x = 2, y = √13/√3

The Answer:

The point C on the satellite's orbit closest to point A is C(2, √13/√3, 0).

Memory Tip:

Remember that when finding the closest point on a curve to an external point, you're looking for where the line from the external point to the curve is perpendicular to the curve itself. This is exactly what Lagrange multipliers help us find - where the gradient of the distance function is parallel to the gradient of the constraint!

Great work tackling this challenging 3D optimization problem! These types of problems combine several important concepts - don't worry if it feels complex at first.