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ky2345
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Homework Statement
When a plane intersects a sphere at more than two points, it is a circle (given). Let x^2+y^2+z^2=1 be a sphere S, and P be a plane that intersects S to make a circle (called C). Let q:[a,b] -> R^3 be a unit speed parameterization whose trace is C. Prove that the second derivative of q, q''(t), is orthogonal to S for all t iff P passes through the origin.
Homework Equations
- q''(t) is orthogonal to S at a specific point q(t_0) if q''(t) is orthogonal to the tangent plane of S at q(t_0)
- two vectors are orthogonal iff their dot product is zero (I'm letting * denote dot product)
- A curve is parameterized by arclength exactly when it has unit speed
The Attempt at a Solution
let P be the plane that intersects the origin, and let n be a vector that is orthogonal to P emitting from the origin. Then, an equation for the plane is n*x=0. The vector equation for the sphere is |x|^2=1, or x*x=1. Then, I can say that x*x+n*x=1 is an equation for the circle, or q(t)*q(t)+n*q(t)=1 if we plug in the parameterization. Now, differentiating with respect to t two times, I get that 2[q(t)*q''(t)+q'(t)*q'(t)]+n*q''(t)=0. I know that q'(t)*q'(t)=|q'(t)|^2=1 because q is unit speed. I'm stuck here because I don't know how to do the following two things:
(1) prove that q(t)*q''(t)=-1
(2) Then prove that since n and q''(t) are orthogonal for all t, q"(t) is orthogonal to the tangent plane at any given t_0 in [a,b]