Curvature and acceleration

Accelerating an object can change both in the magnitude and direction of the velocity. When driving a car, you can accelerate forwards by stepping on the gas (that's why the gas pedal is called the accelerator!), backwards by stepping on the brake, and left or right by turning the steering wheel. When taking a corner you typically do a combination of these. When flying a plane, you have even more ways to accelerate! In general, we'll soon see that acceleration forwards or backwards has to do with changes in speed, and that acceleration sideways has to do with curvature.

Our intuition tells us that the curvature of a circle of radius $R$ should be $1/R$, i.e., smaller circles have tighter curves and more curvature, while the curvature of a straight line should be 0. But coming up with the appropriate definition takes some care because the same curve can have many different parametrizations. For example, tracing the vector-valued functions $$ R\cos(t) \,{\bf i} + R\sin (t)\, {\bf j}\,, \qquad R\cos (2t) \, {\bf i} + R\cos (2t)\, {\bf j}\,, \qquad R\cos (t^3) \, {\bf i} + R\sin (t^3)\, {\bf j}\,, $$ gives the same circle of radius $R$ centered at the origin. They obviously all have the same curvature, but particles having these as position functions will not have the same velocity or the same acceleration. So the definition of curvature has to be based on some 'standard', unambiguous choice of parametrization. This is done by using arc length as our parameter.

Fix a curve $C$ parametrized by ${\bf r}(t)$, and choose some initial fixed point ${\bf r}(a)$ corresponding to $t = a$. Now define the arc length parameter $s$ by $$s \ = \ s(t) \ = \ \int_a^t \, \|{\bf r}'(u)\|\, du\,.$$ This is the arc length of the portion of graph of ${\bf r}(t)$ between ${\bf r}(a)$ and ${\bf r}(t)$, and by the Fundamental Theorem of calculus, $$ \frac{ds}{dt} \ = \ \|{\bf r}'(t)\|\,;$$ in particular, we can parametrize $C$ by defining a new vector-valued function ${\bf c}(s)$, setting $${\bf c} (s) \ = \ {\bf c} (s(t)) \ = \ {\bf r}(t)\,.$$ In this case, many texts speak of $C$ as being parametrized by arc length. Of course, the unit tangent vectors to $C$ coincide at ${\bf c} (s), \, {\bf r}(t)$, (they are the same point on $C$), i.e., $${\bf T}(t) \ = \ \frac{{\bf r}'(t)}{\|{\bf r}'(t)\|} \ = \ \frac{{\bf c}'(s)}{\|{\bf c}'(s)\|}\,.$$

The Curvature $\kappa$ of the space curve $C$ at ${\bf c} (s)$ is the scalar $$\kappa \ = \ \left \| \frac{d}{ds} \frac{{\bf c}'(s)}{\|{\bf c}'(s)\|}\right \| \ = \ \left \|\frac{d {\bf T}}{ds} \right \|\,, $$ in other words, the curvature of $C$ is the rate at which the direction of the unit tangent vector ${\bf T}$ is changing with respect to arc length.

For a circle ${\bf r}(t) = R\cos t \, {\bf i} + R\sin t \, {\bf j}$, we see that $${\bf r}'(t) \,=\, R(-\sin t \, {\bf i} + \cos t )\, {\bf j}\,, \quad \|{\bf r}'(t)\| \ = \ R(\cos^2 t + \sin^2 t)^{1/2} \ = \ R\,, $$ while $$ s \ = \ \int_0^t\, R\, dt \ = \ Rt\,, \quad \frac{ds}{dt} \ = \ R\,.$$ Thus in terms of arc length, $$\frac{d {\bf T}}{ds} \ = \ \frac{d {\bf T}}{dt} \frac{dt}{ds} \ = \ \frac{1}{R} \Bigl (\frac{d}{dt} \frac{{\bf r}'(t)}{\|{\bf r}'(t)\|} \Big) \ = \ -\frac{1}{R^2}\Big(R\big(\cos t\, {\bf i} +\sin t\, {\bf j}\big)\Big)\,.$$ This means that a circle of radius $R$ has $$ \hbox{curvature} \ \kappa \ = \ \left\|\frac{1}{R^2} \Big(R\big(\cos t\, {\bf i} +\sin t\, {\bf j}\big)\Big)\right\| \ = \ \frac{1}{R}\,,$$ exactly as we suspected all along.

We use the term radius of curvature even when the motion isn't exactly in a circle. For any point on a curve, the radius of curvature is $1/\kappa.$ In other words, the radius of curvature is the radius of a circle with the same instantaneous curvature as the curve.

In practice, parametrizing a curve by arc length often isn't convenient - what we really want are expressions for $\kappa(t)$ and ${\bf a}(t)$ in terms of $t$, avoiding mention of arc length. There are several such results whose proof uses lots of the ideas we've developed so far: since $$\frac{ds}{dt}\ = \ \|{\bf r}'(t)\|\,, \qquad {\bf T}'(t) \ = \ {\bf N}(t) \|{\bf T}'(t)\|\,, \qquad \kappa(t) \ = \ \left\|\frac{d{\bf T}}{ds}\right\|\,, \qquad (*)$$ the Chain Rule gives $${\bf T}'(t) \ = \ \frac{d{\bf T}}{dt} \ = \ \frac{ds}{dt}\frac{d{\bf T}}{ds} \ = \ \|{\bf r}'(t)\|\frac{d{\bf T}}{ds}\,,$$ so that by the second of the results in (*) $${\bf T}'(t) \ = \ {\bf N}(t) \|{\bf T}'(t)\| \ = \ \|{\bf r}'(t)\| \kappa(t) {\bf N}(t)\,.$$ But ${\bf r}'(t) = \|{\bf r}'(t)\| {\bf T}(t)$, so by the Product Rule and this last result, $${\bf r}''(t) \ = \ \frac{d}{dt}\big(\|{\bf r}'(t)\| {\bf T}(t)\big) \ = \ \frac{d \|{\bf r}'(t)\|}{dt} {\bf T}(t) + \|{\bf r}'(t)\| {\bf T}'(t) \ = \ \frac{d \|{\bf r}'(t)\|}{dt} {\bf T}(t) + \|{\bf r}'(t)\|^2 \kappa(t) {\bf N} (t)\,. \qquad (**)$$ Thus

Fundamental Result I: For an object with position function ${\bf r}(t)$,     the acceleration at the point ${\bf r}(t)$ decomposes into tangential and normal components $${\bf a}(t) \ = \ v'(t)\, {\bf T} (t) + \kappa(t) v(t)^2\, {\bf N}(t)$$ where the speed $v(t) = \|{\bf r}'(t)\|$ and acceleration ${\bf a}(t) = {\bf r}''(t)$.

Fundamental Result II: For a space curve $C$ parametrized by ${\bf r} (t)$     the curvature at the point ${\bf r} (t)$ is $$\kappa(t) \ = \ \frac{\|{\bf r}'(t) \times {\bf r}''(t)\|} { \|{\bf r}'(t)\|^3}\,;$$ in other words, the curvature can be expressed in terms of velocity and acceleration.