Given the position of the particle, we can find its velocity:
v ( t ) = r ′ ( t ) = ( cos t , − sin t , 1 ) \mathbf{v}(t)=\mathbf{r}'(t)=(\cos t, -\sin t , 1) v ( t ) = r ′ ( t ) = ( cos t , − sin t , 1 ) L ( t ) = r × v = ( sin t , cos t , t ) × ( cos t , − sin t , 1 ) . \mathbf{L}(t)=\mathbf{r} \times \mathbf{v}=(\sin t , \cos t , t) \times (\cos t, -\sin t , 1). L ( t ) = r × v = ( sin t , cos t , t ) × ( cos t , − sin t , 1 ) .
Solution 1:
L ( t ) = ( cos t + t sin t ) ı ^ + ( t cos t − sin t ) ȷ ^ − k ^ L ′ ( t ) = t cos t ı ^ − t sin t ȷ ^ ∣ L ′ ( t ) ∣ = t 2 ( sin 2 t + cos 2 t ) = t 2 = ∣ t ∣ \begin{aligned}
\mathbf{L}(t)& = (\cos t + t\sin t)\hat{\boldsymbol{\imath}} + (t\cos t - \sin t)\hat{\boldsymbol{\jmath}}-\hat{\mathbf{k}}\\
\mathbf{L}'(t)&=t\cos t\,\hat{\boldsymbol{\imath}} -t\sin t\, \hat{\boldsymbol{\jmath}}\\
|\mathbf{L}'(t)|&=\sqrt{t^2(\sin^2 t + \cos^2 t)}=\sqrt{t^2}=|t|
\end{aligned} L ( t ) L ′ ( t ) ∣ L ′ ( t ) ∣ = ( cos t + t sin t ) ^ + ( t cos t − sin t ) ^ − k ^ = t cos t ^ − t sin t ^ = t 2 ( sin 2 t + cos 2 t ) = t 2 = ∣ t ∣
Solution 2:
L ′ ( t ) = r ′ ( t ) × v ( t ) + r ( t ) × v ′ ( t ) = r ′ ( t ) × r ′ ( t ) ⏟ 0 + r ( t ) × v ′ ( t ) = ( sin t , cos t , t ) × ( − sin t , − cos t , 0 ) = t cos t ı ^ − t sin t ȷ ^ ∣ L ′ ( t ) ∣ = t 2 cos 2 t + t 2 sin t 2 = ∣ t ∣ \begin{aligned}
\mathbf{L}'(t)&=\mathbf{r}'(t)\times \mathbf{v}(t) + \mathbf{r}(t) \times \mathbf{v}'(t)\\
&=\underbrace{\mathbf{r}'(t)\times \mathbf{r}'(t)}_{0} + \mathbf{r}(t) \times \mathbf{v}'(t)\\
&= (\sin t ,\cos t,t)\times(-\sin t , -\cos t, 0)\\
&=t\cos t\,\hat{\boldsymbol{\imath}}-t\sin t\,\hat{\boldsymbol{\jmath}}\\
|\mathbf{L}'(t)|&=\sqrt{t^2\cos^2 t + t^2\sin t^2 }=|t|
\end{aligned} L ′ ( t ) ∣ L ′ ( t ) ∣ = r ′ ( t ) × v ( t ) + r ( t ) × v ′ ( t ) = 0 r ′ ( t ) × r ′ ( t ) + r ( t ) × v ′ ( t ) = ( sin t , cos t , t ) × ( − sin t , − cos t , 0 ) = t cos t ^ − t sin t ^ = t 2 cos 2 t + t 2 sin t 2 = ∣ t ∣