Week 5 Discussion Notes
Table of Contents
Notation
Here's a quick table with the main definitions/notation used:
quantity description r ⃗ ( t ) position r ⃗ ′ ( t ) velocity ∥ r ⃗ ′ ( t ) ∥ speed s ( t ) = ∫ a t ∥ r ⃗ ′ ( u ) ∥ d u arc length T ⃗ ( t ) = r ⃗ ′ ( t ) ∥ r ⃗ ′ ( t ) ∥ = d r ⃗ d s unit tangent N ⃗ ( t ) = T ⃗ ′ ( t ) ∥ T ⃗ ′ ( t ) ∥ unit normal \def\arraystretch{1.5}
\begin{array}{cll}
\hline
\text{quantity} &\hspace{1em}& \text{description} \\\hline
\vec{r}\p{t}
&& \text{position} \\
\vec{r}'\p{t}
&& \text{velocity} \\
\norm{\vec{r}'\p{t}}
&& \text{speed} \\
\displaystyle s\p{t} = \int_a^t \norm{\vec{r}'\p{u}} \,\diff{u}
&& \text{arc length} \\[2ex]
\displaystyle \vec{T}\p{t} = \frac{\vec{r}'\p{t}}{\norm{\vec{r}'\p{t}}} = \deriv{\vec{r}}{s}
&& \text{unit tangent} \\[2ex]
\displaystyle \vec{N}\p{t} = \frac{\vec{T}'\p{t}}{\norm{\vec{T}'\p{t}}}
&& \text{unit normal} \\[3.5ex]
\hline
\end{array} quantity r ( t ) r ′ ( t ) ∥ r ′ ( t ) ∥ s ( t ) = ∫ a t ∥ r ′ ( u ) ∥ d u T ( t ) = ∥ r ′ ( t ) ∥ r ′ ( t ) = d s d r N ( t ) = ∥ ∥ T ′ ( t ) ∥ ∥ T ′ ( t ) description position velocity speed arc length unit tangent unit normal Normally, to calculate the curvature κ \kappa κ d T ⃗ d s \deriv{\vec{T}}{s} d s d T r ⃗ ( t ) \vec{r}\p{t} r ( t )
κ ( t ) = ∥ d T ⃗ d s ⃗ ∥ = ∥ d T ⃗ d t ⃗ d t d s ∥ = 1 ∥ r ⃗ ′ ( t ) ∥ ∥ d T ⃗ d t ⃗ ∥ κ ( t ) = ∥ r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) ∥ ∥ r ⃗ ′ ( t ) ∥ 3 \begin{gathered}
\kappa\p{t}
= \norm{\deriv{\vec{T}}{\vec{s}}}
= \norm{\deriv{\vec{T}}{\vec{t}} \deriv{t}{s}}
= \frac{1}{\norm{\vec{r}'\p{t}}} \norm{\deriv{\vec{T}}{\vec{t}}} \\
\kappa\p{t}
= \frac{\norm{\vec{r}'\p{t} \times \vec{r}''\p{t}}}{\norm{\vec{r}'\p{t}}^3}
\end{gathered} κ ( t ) = ∥ ∥ d s d T ∥ ∥ = ∥ ∥ d t d T d s d t ∥ ∥ = ∥ r ′ ( t ) ∥ 1 ∥ ∥ d t d T ∥ ∥ κ ( t ) = ∥ r ′ ( t ) ∥ 3 ∥ r ′ ( t ) × r ′′ ( t ) ∥ Generally, the second one is more useful, but sometimes the calculation using the first one is easier. Either way, you'll get the same answer.
If you read the book, there are other formulas for curvature, but they're all special cases of the second formula above. For example, if you have a parametrization ⟨ x ( t ) , y ( t ) ⟩ \ang{x\p{t}, y\p{t}} ⟨ x ( t ) , y ( t ) ⟩
κ ( t ) = ∣ x ′ ( t ) y ′ ′ ( t ) − y ′ ( t ) x ′ ′ ( t ) ∣ ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3 / 2 . \kappa\p{t} = \frac{\abs{x'\p{t}y''\p{t} - y'\p{t}x''\p{t}}}{\p{x'\p{t}^2 + y'\p{t}^2}^{3/2}}. κ ( t ) = ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3/2 ∣ x ′ ( t ) y ′′ ( t ) − y ′ ( t ) x ′′ ( t ) ∣ . To derive it, all you have to do is set r ⃗ ( t ) = ⟨ x ( t ) , y ( t ) , 0 ⟩ \vec{r}\p{t} = \ang{x\p{t}, y\p{t}, 0} r ( t ) = ⟨ x ( t ) , y ( t ) , 0 ⟩
r ⃗ ′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t ) ⟩ r ⃗ ′ ′ ( t ) = ⟨ x ′ ′ ( t ) , y ′ ′ ( t ) ⟩ , \begin{aligned}
\vec{r}'\p{t} &= \ang{x'\p{t}, y'\p{t}} \\
\vec{r}''\p{t} &= \ang{x''\p{t}, y''\p{t}},
\end{aligned} r ′ ( t ) r ′′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t ) ⟩ = ⟨ x ′′ ( t ) , y ′′ ( t ) ⟩ , and we get
∥ r ⃗ ′ ( t ) ∥ = ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 1 / 2 ∥ r ⃗ ′ ( t ) × r ′ ′ ⃗ ( t ) ∥ = ∥ ⟨ 0 , 0 , x ′ ( t ) y ′ ′ ( t ) − y ′ ( t ) x ′ ′ ( t ) ⟩ ∥ = ∣ x ′ ( t ) y ′ ′ ( t ) − y ′ ( t ) x ′ ′ ( t ) ∣ . \begin{aligned}
\norm{\vec{r}'\p{t}}
&= \p{x'\p{t}^2 + y'\p{t}^2}^{1/2} \\
\norm{\vec{r}'\p{t} \times \vec{r''}\p{t}}
&= \norm{\ang{0, 0, x'\p{t}y''\p{t} - y'\p{t}x''\p{t}}} \\
&= \abs{x'\p{t}y''\p{t} - y'\p{t}x''\p{t}}.
\end{aligned} ∥ r ′ ( t ) ∥ ∥ ∥ r ′ ( t ) × r ′′ ( t ) ∥ ∥ = ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 1/2 = ∥ ⟨ 0 , 0 , x ′ ( t ) y ′′ ( t ) − y ′ ( t ) x ′′ ( t ) ⟩ ∥ = ∣ x ′ ( t ) y ′′ ( t ) − y ′ ( t ) x ′′ ( t ) ∣ . Putting it all together,
κ ( t ) = ∥ r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) ∥ ∥ r ⃗ ′ ( t ) ∥ 3 = ∣ x ′ ( t ) y ′ ′ ( t ) − y ′ ( t ) x ′ ′ ( t ) ∣ ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3 / 2 . \kappa\p{t}
= \frac{\norm{\vec{r}'\p{t} \times \vec{r}''\p{t}}}{\norm{\vec{r}'\p{t}}^3}
= \frac{\abs{x'\p{t}y''\p{t} - y'\p{t}x''\p{t}}}{\p{x'\p{t}^2 + y'\p{t}^2}^{3/2}}. κ ( t ) = ∥ r ′ ( t ) ∥ 3 ∥ r ′ ( t ) × r ′′ ( t ) ∥ = ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3/2 ∣ x ′ ( t ) y ′′ ( t ) − y ′ ( t ) x ′′ ( t ) ∣ .