求证曲线R(t)=(sec t, sec t tan t)的曲率恰在两点处为零

Alright, let's walk through this proof step by step. I'll break it down into clear, manageable parts so you can follow along easily.

First, let's start with the curvature formula for parametric curves:
For a curve defined by ( \mathbf{R}(t) = (x(t), y(t)) ), the curvature ( k(t) ) is given by:

k(t) = |x'(t)y''(t) - x''(t)y'(t)| / (x'(t)^2 + y'(t)^2)^{3/2}

Curvature is zero if and only if the numerator is zero (the denominator is always positive in our domain, since ( \sec t = 1/\cos t ) is defined and non-zero here, so the derivatives ( x'(t) ) and ( y'(t) ) can't both be zero). So we just need to solve ( x'(t)y''(t) - x''(t)y'(t) = 0 ).

Step 1: Compute all first and second derivatives #

Let's calculate each derivative one by one using trigonometric rules and the product rule:

• For ( x(t) = \sec t ):
  • First derivative: ( x'(t) = \sec t \tan t )
  • Second derivative:

    x''(t) = \sec t \cdot \sec^2 t + \tan t \cdot \sec t \tan t = \sec^3 t + \sec t \tan^2 t = \sec t (\sec^2 t + \tan^2 t)
    

• For ( y(t) = \sec t \tan t ):
  • First derivative:

    y'(t) = \sec t \cdot \sec^2 t + \tan t \cdot \sec t \tan t = \sec^3 t + \sec t \tan^2 t = \sec t (\sec^2 t + \tan^2 t)
    

  • Second derivative (differentiate term by term):

    d/dt [sec^3 t] = 3 \sec^2 t \cdot \sec t \tan t = 3 \sec^3 t \tan t
    d/dt [sec t \tan^2 t] = \sec t \cdot 2 \tan t \sec^2 t + \tan^2 t \cdot \sec t \tan t = 2 \sec^3 t \tan t + \sec t \tan^3 t
    y''(t) = 3 \sec^3 t \tan t + 2 \sec^3 t \tan t + \sec t \tan^3 t = \sec t \tan t (5 \sec^2 t + \tan^2 t)
    

Step 2: Simplify the numerator expression #

Substitute the derivatives into ( x'(t)y''(t) - x''(t)y'(t) ):

x'(t)y''(t) - x''(t)y'(t) = [\sec t \tan t \cdot \sec t \tan t (5 \sec^2 t + \tan^2 t)] - [\sec t (\sec^2 t + \tan^2 t) \cdot \sec t (\sec^2 t + \tan^2 t)]

Factor out the common ( \sec^2 t ) term:

= \sec^2 t \left[ \tan^2 t (5 \sec^2 t + \tan^2 t) - (\sec^2 t + \tan^2 t)^2 \right]

Now expand and simplify the bracket:

1. Expand the first term: ( 5 \sec^2 t \tan^2 t + \tan^4 t )
2. Expand the second term: ( (\sec^2 t + \tan^2 t)^2 = \sec^4 t + 2 \sec^2 t \tan^2 t + \tan^4 t )
3. Subtract the second from the first:

   5 \sec^2 t \tan^2 t + \tan^4 t - \sec^4 t - 2 \sec^2 t \tan^2 t - \tan^4 t = 3 \sec^2 t \tan^2 t - \sec^4 t
   

Factor out ( \sec^2 t ) again:

= \sec^2 t (3 \tan^2 t - \sec^2 t)

The full numerator becomes:

\sec^2 t \cdot \sec^2 t (3 \tan^2 t - \sec^2 t) = \sec^4 t (3 \tan^2 t - \sec^2 t)

Step 3: Solve for zero curvature #

In ( -\pi/2 < t < \pi/2 ), ( \sec t \neq 0 ), so ( \sec^4 t \neq 0 ). We only need to solve:

3 \tan^2 t - \sec^2 t = 0

Use the Pythagorean identity ( \sec^2 t = 1 + \tan^2 t ):

3 \tan^2 t - (1 + \tan^2 t) = 0 \\
2 \tan^2 t - 1 = 0 \\
\tan^2 t = 1/2 \\
\tan t = \pm \frac{\sqrt{2}}{2}

Step 4: Find the corresponding curve points #

In ( (-\pi/2, \pi/2) ), each ( \tan t ) value maps to exactly one ( t ):

• When ( \tan t = \sqrt{2}/2 ):
  ( \cos t = \frac{2}{\sqrt{2^2 + (\sqrt{2})^2}} = \frac{\sqrt{6}}{3} ), so ( \sec t = \frac{\sqrt{6}}{2} ).
  ( y = \sec t \tan t = \frac{\sqrt{6}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{3}}{2} ).
  This gives the point ( \left( \frac{\sqrt{6}}{2}, \frac{\sqrt{3}}{2} \right) ).
• When ( \tan t = -\sqrt{2}/2 ):
  ( \sec t ) remains ( \frac{\sqrt{6}}{2} ) (secant is an even function), and ( y = -\frac{\sqrt{3}}{2} ).
  This gives the point ( \left( \frac{\sqrt{6}}{2}, -\frac{\sqrt{3}}{2} \right) ).

These are the only two points in the domain where curvature is zero—since ( \tan t ) is one-to-one on ( (-\pi/2, \pi/2) ), there are no other solutions.

内容的提问来源于stack exchange，提问作者user380605