証明しようとする式： \[ \frac{\overline{ OP'} - r }{ \Delta t} \ \ \ \longrightarrow \ v\ cos( \theta ) \ \ \ ( \Delta t \longrightarrow 0 ) \\ \ \\ \frac{ 1 - cos( a ) }{\Delta t } \ \ \ \longrightarrow \ 0 \ \ \ ( \Delta t \longrightarrow 0 ) \\ \ \\ \frac{ sin( a ) }{\Delta t } \ \ \longrightarrow \ \frac{ v\ sin(\theta) }{r} \ \ \ \ ( \Delta t \longrightarrow 0 ) \] \( \overline{ OP'} \) は， \[ \overline{ OP'}^2 \\ = ( r + v\ \Delta t\ cos( \theta ) )^2 \\ \quad + ( v\ \Delta t\ sin( \theta ) )^2 \\ = r^2 + 2\ r\ v\ \Delta t\ cos( \theta ) + v^2 \Delta t^2\ cos( \theta )^2 \\ \quad + v^2 \Delta t^2\ sin( \theta )^2 \\ = r^2 + 2\ r\ v\ \Delta t\ cos( \theta ) + v^2 \Delta t^2 \\ \] よって， \[ \frac{\overline{ OP'}^2 - r^2 }{ \Delta t} \\ = 2\ r\ v\ cos( \theta ) + v^2 \Delta t \ \\ \ \\ \longrightarrow \ 2\ r\ v\ cos( \theta ) + 0 = 2\ r\ v\ cos( \theta ) \ \ ( \Delta t \longrightarrow 0 ) \\ \ \\ \ \\ \frac{\overline{ OP'} - r }{ \Delta t} \\ = \frac{\overline{ OP'}^2 - r^2 }{ \Delta t} \ \ \frac{ 1 }{\overline{ OP'} + r } \ \\ \ \\ \longrightarrow \ ( 2\ r\ v\ cos( \theta ) )\ \frac{1}{ 2\ r } = v\ cos( \theta ) \ \ ( \Delta t \longrightarrow 0 ) \] また， \[ \frac{ 1 - cos( a ) }{\Delta t } \\ \ \\ = \frac{1}{ \Delta t } \frac{ \overline{ OP'} - \overline{ OP'} cos( a ) }{ \overline{ OP'} } \\ \ \\ = \frac{1}{\overline{ OP'} \Delta t } \bigl( \overline{ OP'} - ( r + v \Delta t cos( \theta ) ) \bigr) \\ \ \\ = \frac{1}{\overline{ OP'}} \bigl( \frac{ \overline{ OP'} - r }{\Delta t} - v\ cos( \theta ) \bigr) \\ \ \\ \longrightarrow \ \frac{ 1 }{ r }\ ( v\ cos( \theta ) - v cos( \theta ) ) = 0 \ \ ( \Delta t \longrightarrow 0 ) \] \[ \frac{ sin( a ) }{\Delta t } \\ \ \\ = \frac{1}{ \Delta t }\ \frac{ \overline{ OP'}\ sin( a ) }{ \overline{ OP'} } \\ \ \\ = \frac{1}{\overline{ OP'} \Delta t }\ v\ \Delta t\ sin(\theta) \\ \ \\ \longrightarrow \ \frac{ v\ sin(\theta) }{r} \ \ \ \ ( \Delta t \longrightarrow 0 ) \]

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