1.3: Curvature
 Page ID
 91894
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So far, when we have wanted to approximate a complicated curve by a simple curve near some point, we drew the tangent line to the curve at the point. That's pretty crude. In particular tangent lines are straight — they don't curve. We will get a much better idea of what the complicated curve looks like if we approximate it, locally, by a very simple “curvy curve” rather than by a straight line. Probably the simplest “curvy curve” is a circle^{ 1} and that's what we'll use.
 The circle which best approximates a given curve near a given point is called the circle of curvature or the osculating circle^{ 2} at the point.
 The radius of the circle of curvature is called the radius of curvature at the point and is normally denoted \(\rho\text{.}\)
 The curvature at the point is \(\kappa=\frac{1}{\rho}\text{.}\)
 The centre of the circle of curvature is called centre of curvature at the point.
These definitions are illustrated in the figure below. It shows (part of) the osculating circle at the point \(P\text{.}\) The point \(C\) is the centre of curvature.
Note that when the curvature \(\kappa\) is large, the radius of curvature \(\rho\) is small and we have a very curvy curve. On the other hand when the curvature \(\kappa\) is small, the radius of curvature \(\rho\) is large and our curve is almost straight. In particular, straight lines have curvature exactly zero.
We are now going to determine how to find the circle of curvature, starting by figuring out what its radius should be. We'll first look at curves^{ 3} that lie in the \(xy\)plane and then move on to curves in 3d. Consider the black curve in the figure below.
That figure also contains a (portion of a) red circle that fits the curve really well between the two radial lines that are (a very small) angle \(\text{d}\theta\) apart. So the arclength \(\text{d}s\) of the part of the black curve between the two radial lines, should be (essentially) the same as the arc length of the circle between the two radial lines, which is \(\rho\,\text{d}\theta\text{,}\) where \(\rho\) is the radius of the circle. (We put in absolute values to take into account the possibility that \(\text{d}\theta\) could be negative.) Thus \(\text{d}s = \rho\,\text{d}\theta\text{.}\) When \(\text{d}\theta\) is a macroscopic angle, this is of course an approximation. But in the limit as \(\text{d}\theta\rightarrow 0\text{,}\) we should end up with
\[ \rho = \left\dfrac{ds}{d\theta}\right \nonumber \]
We now have a formula for the radius of curvature, but not in a very convenient form, because to evaluate it we would need to know the arc length along the curve as a function of the angle \(\theta\) in the rightmost figure below. We'll now spend some time developing more convenient formulae for \(\rho\text{.}\) First consider the three figures below. They all show the same curve as in the last figure. The leftmost figure just shows
 the curve of interest, which is the black curve, and
 the (blue) point of interest on the black curve. We want to find the curvature at that point.
The middle figure shows the same curve and point of interest and also shows
 the red circle of curvature (i.e. best fitting circle) for the black curve at the blue dot.
 The red dot is the centre of curvature.
The rightmost figure shows the same black curve, blue point of interest and red circle of curvature (at least part of it) somewhat enlarged.
 The angle \(\theta\) is the angle between \(\hat{\pmb{\imath}}\) and the radius vector from the red dot (the centre of curvature) to the blue dot (the point of interest).
 \(\hat{\textbf{T}}\) is the tangent vector to the black curve at the blue dot.
 The angle \(\phi\) is the angle between \(\hat{\pmb{\imath}}\) and \(\hat{\textbf{T}} \text{.}\) The vector \(\hat{\textbf{T}}\) is also tangent to the red circle. As the tangent and radius vectors for circles are perpendicular to each other^{ 4}, we have that \(\phi=\theta+\frac{\pi}{2}\) and hence \(\rho = \big\dfrac{ds}{d\phi}\big\) too.
We are now in a position to develop a bunch of formulae for the radius of curvature \(\rho\) and the curvature \(\kappa=\frac{1}{\rho}\text{,}\) that are more convenient than \(\kappa = \big\dfrac{ds}{d\phi}\big^{1}\text{.}\) These formulae will use the
If \(\vecs{r} (t)\) is a parametrized curve, then
 \(\vecs{v} (t) = \dfrac{d\vecs{r} }{dt}(t)\) is the velocity vector at \(\vecs{r} (t)\)
 \(\textbf{a}(t) = \frac{\mathrm{d}^{2}\vecs{r}}{\mathrm{d}t^{2}}(t) \) is the acceleration vector at \(\vecs{r} (t)\)
 \(\hat{\textbf{T}}(t)\) is the unit tangent vector to the curve at \(\vecs{r} (t)\) that points in the direction of increasing \(t\text{.}\)
 \(\hat{\textbf{N}}(t)\) is the unit normal vector to the curve at \(\vecs{r} (t)\) that points toward the centre of curvature.
 \(\kappa (t)\) is the curvature at \(\vecs{r} (t)\)
 \(\rho(t)\) is the radius of curvature at \(\vecs{r} (t)\)
 Given^{ 5} \(s(\phi)\text{,}\) i.e. if we know the arc length along the curve as a function of the angle^{ 6} \(\phi=\measuredangle(\hat{\pmb{\imath}},\hat{\textbf{T}} )\text{,}\) then
\[ \rho = \left\dfrac{ds}{d\phi}\right\qquad \kappa = \left\dfrac{ds}{d\phi}\right^{1}\qquad \kappa = \left\dfrac{d\phi}{ds}\right \nonumber \]
 Given \(\vecs{r} (s)\text{,}\) i.e. if we have a parametrization of the curve in terms of arc length, then
\[ \dfrac{d\hat{\textbf{T}}}{ds}(s) = \kappa(s)\,\hat{\textbf{N}}(s) \nonumber \]
where \(\hat{\textbf{N}}(s)\) is the unit normal vector to the curve at \(\vecs{r} (s)\) that points toward the centre of curvature.  Given \(\vecs{r} (t)\text{,}\) i.e. if we have a general parametrized curve, then
\[\begin{gather*} \dfrac{d\hat{\textbf{T}} }{dt} = \kappa \dfrac{ds}{dt} \hat{\textbf{N}}\qquad \vecs{v} (t) = \dfrac{ds}{dt}(t)\,\hat{\textbf{T}} (t) \qquad \textbf{a}(t) = \frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}\hat{\textbf{T}} + \kappa\left(\dfrac{ds}{dt}\right)^2\hat{\textbf{N}} \end{gather*}\]
 Given \(\big(x(t)\,,\,y(t)\big)\text{,}\) (for curves in the \(xy\)plane)
\[ \kappa = \left\frac{\vecs{v} (t)\times\textbf{a}(t)}{\big(\dfrac{ds}{dt}\big)^3}\right =\frac{\big\dfrac{dx}{dt}\frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}\dfrac{dy}{dt}\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}\big} { {\big[\big(\dfrac{dx}{dt}\big)^2+\big(\dfrac{dy}{dt}\big)^2\big]}^{3/2} } \nonumber \]
 Given \(y(x)\text{,}\) (again for curves in the \(xy\)plane)
\[ \kappa =\frac{\big\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}\big} { {\big[1+\big(\dfrac{dy}{dx}\big)^2\big]}^{3/2} } \nonumber \]

(a) Given \(s(\phi)\text{,}\) then
\[ \rho = \Big\dfrac{ds}{d\phi}\Big\qquad \kappa = \Big\dfrac{ds}{d\phi}\Big^{1} \nonumber \]
As we are assuming that \(0 \lt \rho=\Big\dfrac{ds}{d\phi}\Big \lt \infty\text{,}\) the inverse function theorem says that we can invert the function \(s(\phi)\) (at least locally) to get \(\phi\) as a function of \(s\text{,}\) and that
\[ \kappa = \Big\dfrac{d\phi}{ds}\Big \nonumber \]
(b) Given \(\vecs{r} (s)\text{,}\) then, by Lemma 1.1.4.c, \(\hat{\textbf{T}}(s) = \vecs{r} '(s)\) is a unit tangent to the curve at \(\vecs{r} (s)\) and
\[ \dfrac{d\hat{\textbf{T}}}{ds} = \dfrac{d\hat{\textbf{T}}}{d\phi} \dfrac{d\phi}{ds} \tag{$*$} \nonumber \]
Now up to a sign \(\dfrac{d\phi}{ds}\) is \(\kappa\text{,}\) and just because \(\phi=\measuredangle(\hat{\pmb{\imath}},\hat{\textbf{T}})\text{,}\) with \(\hat{\textbf{T}}\) a unit vector,
\[ \begin{split} \hat{\textbf{T}} &=\cos\phi\,\hat{\pmb{\imath}} + \sin\phi\,\hat{\pmb{\jmath}} \\ \implies \dfrac{d\hat{\textbf{T}} }{d\phi}&= \sin\phi\,\hat{\pmb{\imath}} + \cos\phi\,\hat{\pmb{\jmath}} \end{split} \tag{$**$} \nonumber \]
So \(\dfrac{d\hat{\textbf{T}} }{d\phi}\) is a unit vector that is perpendicular^{ 7} to \(\hat{\textbf{T}}\text{,}\) and hence to the curve at \(\vecs{r} (s)\text{,}\) and
\[ \dfrac{d\hat{\textbf{T}}}{ds}(s) = \kappa(s)\,\hat{\textbf{N}}(s) \tag{$\dagger$} \nonumber \]
with \(\hat{\textbf{N}}(s)\) a unit normal vector to the curve at \(\vecs{r} (s)\text{.}\) In fact, \(\hat{\textbf{N}}(s)\) is the unit normal vector to the curve at \(\vecs{r} (s)\) that points toward the centre of curvature.
To see that, look at the figures below^{ 8}, and note that substituting the sign information from each figure into (\(*\)) gives (\(\dagger\)). For example,
consider the figure on the lower left. In that figure,
 the \(x\) component of \(\hat{\textbf{T}}\) is negative (\(\hat{\textbf{T}}\) is leftward pointing in the figure),
 which makes \(\cos\phi\) negative (see (\(**\))),
 which makes the \(y\) component of \(\dfrac{d\hat{\textbf{T}}}{d\phi}\) negative (see (\(**\)) again),
 so \(\dfrac{d\hat{\textbf{T}}}{d\phi}\) is downward pointing,
so \(\dfrac{d\hat{\textbf{T}}}{d\phi}=\hat{\textbf{N}}\) (the centre of curvature is the red dot above the curve) and
 as \(s\) increases (i.e. as you move in the direction of the arrow on the curve), \(\phi\) decreases (on the far right hand part of the curve \(\phi\approx\frac{3\pi}{2}\text{,}\) while on the far left hand part of the curve \(\phi\approx\pi\)), so \(\dfrac{d\phi}{ds} \lt 0\) and \(\kappa = \big\dfrac{d\phi}{ds}\big =  \dfrac{d\phi}{ds}\text{.}\)
 So by (\(*\)), \(\dfrac{d\hat{\textbf{T}}}{ds} = \dfrac{d\hat{\textbf{T}}}{d\phi} \dfrac{d\phi}{ds} =\big(\hat{\textbf{N}})(\kappa) = \kappa\hat{\textbf{N}}\text{.}\)
In each of the three other figures we also end up with \(\dfrac{d\hat{\textbf{T}}}{ds} = \kappa(s)\hat{\textbf{N}}(s)\text{.}\)
Note that if \(\kappa(s)=0\text{,}\) then \(\hat{\textbf{N}}(s)\) is not defined. This makes sense: if the curve is (locally) a straight line, there is no “best fitting circle”.
(c) Given \(\vecs{r} (t)\text{,}\) i.e. if we have a general parametrized curve, we can determine a unit tangent vector by using Lemma 1.1.4:
\[ \vecs{v} (t) = \dfrac{d\vecs{r} }{dt}(t) = \dfrac{ds}{dt}(t)\,\hat{\textbf{T}} (t) \quad\implies\quad \hat{\textbf{T}} (t) = \frac{\vecs{r} '(t)}{\vecs{r} '(t)} \nonumber \]
Then we can determine \(\kappa\) and \(\hat{\textbf{N}}\) by differentiating \(\hat{\textbf{T}} (t)\) and using the chain rule:
\[ \dfrac{d\hat{\textbf{T}}}{dt} = \dfrac{d\hat{\textbf{T}}}{ds}\dfrac{ds}{dt} = \kappa \dfrac{ds}{dt} \hat{\textbf{N}} \quad\implies\quad \kappa(t) = \frac{\hat{\textbf{T}} '(t)}{\vecs{r} '(t)} \nonumber \]
Also, if we differentiate \(\vecs{v} (t) = \dfrac{ds}{dt}\hat{\textbf{T}} (t)\text{,}\) we get that the acceleration
\[ \textbf{a}(t) = \frac{\mathrm{d}^{2}\vecs{r}}{\mathrm{d}t^{2}}{t} = \frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}\hat{\textbf{T}} + \dfrac{ds}{dt}\,\dfrac{d\hat{\textbf{T}}}{dt} = \frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}\hat{\textbf{T}}+ \kappa\Big(\dfrac{ds}{dt}\Big)^2\hat{\textbf{N}} \nonumber \]
(d) Given \(\big(x(t)\,,\,y(t)\big)\text{,}\) (for curves in the \(xy\)plane), we can read off the curvature from
\[\begin{align*} \vecs{v} (t)\times\textbf{a}(t) &=\Big(\dfrac{ds}{dt}(t)\,\hat{\textbf{T}}(t)\Big)\times \Big(\frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}\hat{\textbf{T}} + \kappa\Big(\dfrac{ds}{dt}\Big)^2\hat{\textbf{N}}\Big)\\ &= \kappa\Big(\dfrac{ds}{dt}\Big)^3\hat{\textbf{T}}\times\hat{\textbf{N}} \qquad\text{(since }\hat{\textbf{T}} \times\hat{\textbf{T}} =\vecs{0} \end{align*}\]
Think of \(\hat{\textbf{T}} \) and \(\hat{\textbf{N}}\) as 3d vectors that whose \(z\)components happen to be zero. As \(\hat{\textbf{T}} \) and \(\hat{\textbf{N}}\) are mutually perpendicular unit vectors in the \(xy\)plane, the crossproduct \(\hat{\textbf{T}} \times\hat{\textbf{N}}\) will be either \(+\hat{\mathbf{k}}\) or \(\hat{\mathbf{k}}\text{.}\) In both cases, \(\vecs{v} (t)\times\textbf{a}(t)\big = \kappa\big\dfrac{ds}{dt}\big^3\text{.}\) So
\[\begin{align*} \kappa &= \left\frac{\vecs{v} (t)\times\textbf{a}(t)}{\big(\dfrac{ds}{dt}\big)^3}\right =\left\frac{\big[\dfrac{dx}{dt}\hat{\pmb{\imath}}+\dfrac{dy}{dt}\hat{\pmb{\jmath}}\big]\times \big[\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}\hat{\pmb{\imath}}+\frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}\hat{\pmb{\jmath}}\big]} {\big(\dfrac{ds}{dt}\big)^3}\right\\ &=\left\frac{\big[\dfrac{dx}{dt}\frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}\dfrac{dy}{dt}\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}\big]\hat{\mathbf{k}}} {\big(\dfrac{ds}{dt}\big)^3}\right\\ &=\frac{\big\dfrac{dx}{dt}\frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}\dfrac{dy}{dt}\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}\big} { {\big[\big(\dfrac{dx}{dt}\big)^2+\big(\dfrac{dy}{dt}\big)^2\big]}^{3/2} } \end{align*}\]
(e) Given \(y(x)\text{,}\) again for curves in the \(xy\)plane, we can parametrize the curve using \(x\) as the parameter:
\[ \vecs{r} (t) = \big(X(t)\,,\,Y(t)\big) \qquad\text{with $X(t)=t$ and $Y(t) =y(t)$} \nonumber \]
Then
\[ \dfrac{dX}{dt} = 1 \qquad \frac{\mathrm{d}^{2}X}{\mathrm{d}t^{2}} = 0 \qquad \dfrac{dY}{dt} = \dfrac{dy}{dx} \qquad \frac{\mathrm{d}^{2}Y}{\mathrm{d}t^{2}} = \frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}} \nonumber \]
and
\[ \kappa =\frac{\big\dfrac{dX}{dt}\frac{\mathrm{d}^{2}Y}{\mathrm{d}t^{2}}\dfrac{dY}{dt}\frac{\mathrm{d}^{2}X}{\mathrm{d}t^{2}}\big} { {\big[\big(\dfrac{dX}{dt}\big)^2+\big(\dfrac{dY}{dt}\big)^2\big]}^{3/2} } =\frac{\big\frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}\big} { {\big[1+\big(\dfrac{dy}{dx}\big)^2\big]}^{3/2} } \nonumber \]
Take another look at Theorem 1.3.3 and note that
 the tangential component of acceleration, i.e. \(\frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}\text{,}\) arises purely from change in speed while
 the normal component of acceleration, i.e. \(\kappa\big(\dfrac{ds}{dt}\big)^2\text{,}\) arises from curvature and is proportional to the square of the speed \(\dfrac{ds}{dt}\text{.}\) Think about what you feel when you are driving. That's why velodromes and (car) race tracks often have banked corners.
 the \(x\) component of \(\hat{\textbf{T}}\) is negative (\(\hat{\textbf{T}}\) is leftward pointing in the figure),
As a warm up example, and also a check that our formulae make sense, we'll find the curvature \(\kappa\text{,}\) radius of curvature, \(\rho\text{,}\) unit tangent vector, \(\hat{\textbf{T}} \text{,}\) unit normal vector, \(\hat{\textbf{N}}\text{,}\) and centre of curvature of the parametrized curve
\[ \vecs{r} (t) = a\cos t\,\hat{\pmb{\imath}} + a\sin t\,\hat{\pmb{\jmath}} \nonumber \]
with the constant \(a \gt 0\text{.}\) This is, of course, the circle of radius \(a\) centred on the origin. As
\[ \vecs{v} (t)=\dfrac{d\vecs{r} }{dt}(t)= a\sin t\,\hat{\pmb{\imath}} + a\cos t\,\hat{\pmb{\jmath}} \implies \dfrac{ds}{dt}(t) = \vecs{v} (t) = a \nonumber \]
we have that the unit tangent vector
\[ \vecs{T} (t) = \frac{\vecs{v} (t)}{\vecs{v} (t)} = \sin t\,\hat{\pmb{\imath}} + \cos t\,\hat{\pmb{\jmath}} \nonumber \]
Note, as a check, that this is indeed a vector of length one and is perpendicular to the radius vector (as expected — the curve is a circle). As
\[ \dfrac{d\hat{\textbf{T}} }{dt}(t)= \cos t\,\hat{\pmb{\imath}} \sin t\,\hat{\pmb{\jmath}} \nonumber \]
we have that
\[\begin{align*} \hat{\textbf{N}}(t) &= \frac{\dfrac{d\hat{\textbf{T}} }{dt}(t)}{\big\dfrac{d\hat{\textbf{T}} }{dt}(t)\big} = \cos t\,\hat{\pmb{\imath}} \sin t\,\hat{\pmb{\jmath}}\\ \kappa(t) &= \frac{\big\dfrac{d\hat{\textbf{T}} }{dt}(t)\big}{\dfrac{ds}{dt}(t)} =\frac{1}{a}\\ \rho(t)&=\frac{1}{\kappa(t)} =a \end{align*}\]
Now look at the figure.
To get to the centre of curvature we should start from \(\vecs{r} (t)\) and walk a distance \(\rho(t)\text{,}\) which after all is the radius of curvature, in the direction \(\hat{\textbf{N}}(T)\text{,}\) which is pointing towards the centre of curvature. So the centre of curvature is
\[ \vecs{r} (t)+\rho(t)\hat{\textbf{N}}(t) =\big[a\cos t\,\hat{\pmb{\imath}} +a\sin t\,\hat{\pmb{\jmath}}\big] +a\big[\cos t\,\hat{\pmb{\imath}} \sin t\,\hat{\pmb{\jmath}}\big] = \vecs{0} \nonumber \]
This makes perfectly good sense — the radius of curvature is the radius of the original circle and the centre of curvature is the centre of the original circle.
One alternative calculation of the curvature, using \(x(t) = a\cos t \text{,}\) \(y(t)=a\sin t\text{,}\) is
\[\begin{align*} \kappa(t) &=\frac{\big \dfrac{dx}{dt}(t)\frac{\mathrm{d}^{2}y}{\mathrm{d}t^{2}}(t)\dfrac{dy}{dt}(t)\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}(t) \big}{\Big[\big(\dfrac{dx}{dt}(t)\big)^2 +\big(\dfrac{dy}{dt}(t)\big)^2\Big]^{3/2}}\\ &=\frac{\big a\sin t\big(a\sin t\big)a\cos t\big(a\cos t\big) \big}{\big[\big(a\sin t\big)^2 +\big(a\cos t\big)^2\big]^{3/2}}\\ &=\frac{1}{a} \end{align*}\]
Another alternative calculation of the curvature, using \(y(x) =\sqrt{a^2x^2}\) (for the part of the circle with \(y \gt 0\)),
\[\begin{align*} y'(x) &= \frac{x}{\sqrt{a^2x^2}}=\frac{x}{y(x)} \\ y''(x) &= \frac{y(x)  xy'(x)}{y(x)^2} = \frac{y(x)^2 + x^2}{y(x)^3} = \frac{a^2}{y(x)^3} \end{align*}\]
is
\[\begin{align*} \kappa(x) &=\frac{\big\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}(x)\big} {\Big[1+\big(\dfrac{dy}{dx}(x)\big)^2\Big]^{3/2}} =\frac{\frac{a^2}{y(x)^3}} {\Big[1+\frac{x^2}{y(x)^2}\Big]^{3/2}} =\frac{a^2}{\big[y(x)^2+x^2\big]^{3/2}}\\ &=\frac{1}{a} \end{align*}\]
As a more computationally involved example, we'll analyze
\[\begin{align*} \vecs{r} (t) &= \big(\cos t + t\sin t\big)\hat{\pmb{\imath}} +\big(\sin tt\cos t\big)\hat{\pmb{\jmath}} \qquad t \gt 0\\ \vecs{v} (t) &= t\cos t\,\hat{\pmb{\imath}} + t\sin t\,\hat{\pmb{\jmath}}\\ \textbf{a}(t) &= \big(\cos t  t\sin t\big)\hat{\pmb{\imath}} +\big(\sin t+t\cos t\big)\hat{\pmb{\jmath}} \end{align*}\]
We can read off from \(\vecs{v} (t)\text{,}\) that
\[\begin{align*} \dfrac{ds}{dt}(t) &= \vecs{v} (t) =t\\ \frac{\mathrm{d}^{2}s}{\mathrm{d}t^{2}}(t) &= 1\\ \vecs{T} (t)&=\frac{\vecs{v} (t)}{\vecs{v} (t)} = \cos t\,\hat{\pmb{\imath}} + \sin t\,\hat{\pmb{\jmath}} \end{align*}\]
Next, from \(\textbf{a}(t)\text{,}\) we read off that
(by Theorem 1.3.3 .c)
\begin{align*} &=\cos t\,\hat{\pmb{\imath}} + \sin t\,\hat{\pmb{\jmath}} +t^2\kappa(t) \hat{\textbf{N}}(t)\\ \implies t^2\kappa(t) \hat{\textbf{N}}(t)&=  t\sin t\,\hat{\pmb{\imath}} + t\cos t\,\hat{\pmb{\jmath}} \end{align*}so that \(t^2\kappa(t)\) is the length of \( t\sin t\,\hat{\pmb{\imath}} + t\cos t\,\hat{\pmb{\jmath}}\text{,}\) which is \(t\text{.}\) Thus
\[\begin{gather*} \kappa(t) = \frac{1}{t} \quad\text{and}\quad \hat{\textbf{N}}(t)= \frac{ t\sin t\,\hat{\pmb{\imath}} + t\cos t\,\hat{\pmb{\jmath}}}{t^2\kappa(t)} = \sin t\,\hat{\pmb{\imath}} + \cos t\,\hat{\pmb{\jmath}} \end{gather*}\]
As an alternative calculation of the curvature, we have
\[\begin{align*} \kappa(t) &=\frac{\vecs{v} (t)\times\textbf{a}(t)}{(\dfrac{ds}{dt}(t))^3}\\ &=\frac{\big\big[t\cos t\,\hat{\pmb{\imath}} + t\sin t\,\hat{\pmb{\jmath}}\big]\times \big[\big(\cos t  t\sin t\big)\hat{\pmb{\imath}} +\big(\sin t+t\cos t\big)\hat{\pmb{\jmath}}\big]\big} {(\dfrac{ds}{dt}(t))^3}\\ &=\frac{\big\big[t\cos t\big(\sin t+t\cos t\big) t\sin t\big(\cos t  t\sin t\big)\big]\hat{\mathbf{k}}\big} {(\dfrac{ds}{dt}(t))^3}\\ &=\frac{t^2\hat{\mathbf{k}}}{t^3} =\frac{1}{t} \end{align*}\]
It pays to think before you calculate!
Exercises
Stage 1
There are a lot of constants in this chapter that might be new to you. They can take a little getting used to. Questions 1.3.1.11.3.1.5 provide practice working with and interpreting these constants and their relations to each other.
Sketch the curve \(\vecs{r} (t)=(3\sin t,3\cos t)\text{.}\) At the point \((0,3)\text{,}\) label \(\hat{\textbf{T}}\) and \(\hat{\textbf{N}}\text{.}\) Give the values of \(\kappa\) and \(\rho\) at this point as well.
Consider the circle \(\vecs{r} (t)=(3\sin t,3\cos t)\text{.}\) Find \(\hat{\textbf{T}}(t)\) and \(\hat{\textbf{T}}(s)\text{.}\) Then, use parts (b) and (c) of Theorem 1.3.3 to find \(\hat{\textbf{N}}(t)\) and \(\hat{\textbf{N}}(s)\text{.}\)
The functon \(\vecs{r} (t)=(t\cos t, t\sin t)\text{,}\) \(t \ge 0\text{,}\) defines a spiral centred at the origin. Using only geometric intuition (no calculation), predict \(\displaystyle\lim_{t \to \infty}\kappa(t)\text{.}\)
Let \(\vecs{r} (t)=(e^t,3t,\sin t)\text{.}\) What is \(\dfrac{ds}{dt}\text{?}\)
In Question 1.2.1.5 of Section 1.2, we found that the spiral
\[ \vecs{r} (t) = e^t (\cos t, \sin t) \nonumber \]
parametrized in terms of arclength is
\[ \textbf{R}(s)=\frac{s}{\sqrt{2}}\left(\cos\Big(\ln\Big(\frac{s}{\sqrt{2}}\Big)\Big)\,,\, \sin\Big(\ln\Big(\frac{s}{\sqrt{2}}\Big)\Big)\right). \nonumber \]
Find \(\dfrac{d\hat{\textbf{T}}}{ds}\) and \(\dfrac{d\hat{\textbf{T}}}{dt}\) for this curve.
In this exercise, we make more precise the sense in which the osculating circle is the circle which best approximates a plane curve at a point.
 By translating and rotating our coordinate system, we can always arrange that the point is \((0,0)\) and that the curve is \(y=f(x)\) with \(f'(0)=0\) and \(f''(0) \gt 0\text{.}\) (We are assuming that the curvature at the point is nonzero.)
 Let \(y=g(x)\) be the bottom half of the circle of radius \(r\) which is centred at \((0,r)\text{.}\)
Show that if \(f(x)\) and \(g(x)\) have the same second order Taylor approximation at \(x=0\text{,}\) then \(r\) is the radius of curvature of \(y=f(x)\) at \(x=0\text{.}\)
Stage 2
Given a curve \(\vecs{r} (t)=(e^t,t^2+t)\text{,}\) compute the following quantities:
 \(\displaystyle \vecs{v} (t)\)
 \(\displaystyle \textbf{a}(t)\)
 \(\displaystyle \dfrac{ds}{dt}\)
 \(\displaystyle \hat{\textbf{T}}(t)\)
 \(\displaystyle \kappa(t)\)
Find the curvature \(\kappa(t)\) of \(\vecs{r} (t)=(\cos t+\sin t , \sin t  \cos t)\text{.}\)
Find the minimum and maximum values for the curvature of the ellipse \(x(t)= a \cos t\text{,}\) \(y(t)=b\sin t\text{.}\) Here \(a \gt b \gt 0\text{.}\)
 Find the curvature of \(y=e^x\) at \((0,1)\text{.}\)
 Find the equation of the circle best fitting \(y=e^x\) at \((0,1)\text{.}\)
Consider the motion of a thumbtack stuck in the tread of a tire which is on a bicycle moving at constant speed. This motion is given by the parametrized curve
\[ \vecs{r} (t) = \big(t  \sin t\,,\, 1  \cos t\big) \nonumber \]
with \(t \gt 0\text{.}\)
 Sketch the curve in the \(xy\)plane for \(0 \lt t \lt 4\pi\text{.}\)
 Find and simplify the formula for the curvature \(\kappa(t)\text{.}\)
 Find the radius of curvature of the osculating circle to \(\vecs{r} (t)\) at \(t = \pi\text{.}\)
 Find the equation of the osculating circle to \(\vecs{r} (t)\) at \(t = \pi\text{.}\)
Stage 3
Find the curvature \(\kappa\) as a function of arclength \(s\) (measured from \((0,0)\)) for the curve
\[ x(\theta)=\int_0^\theta \cos\big(\frac{1}{2}\pi t^2\big)dt\quad \quad y(\theta)=\int_0^\theta \sin\big(\frac{1}{2}\pi t^2\big)dt \nonumber \]
Let \(C\) be the curve in \(\mathbb{R}^2\) given by the graph of the function \(y=\frac{x^3}{3}\text{.}\) Let \(\kappa(x)\) be the curvature of \(C\) at the point \((x, x^3/3)\text{.}\) Find all points where \(\kappa(x)\) attains its maximal values, or else explain why such points do not exist. What are the limits of \(\kappa(x)\) as \(x \rightarrow \infty\) and \(x \rightarrow \infty\text{?}\)
 Circles are good for studying “curvature”, because, unlike parabolas for example, the rate at which a circle curves is uniform over the entire circle.
 “Osculare” is the Latin verb “to kiss”. The German mathematician Gottfried Wilhelm (von) Leibniz (16461716) named the circle the “circulus osculans”.
 We'll also assume that the curves of interest are smooth, with no cusps for example, and not straight, so that the radius of curvature \(0 \lt \rho \lt \infty\text{.}\)
 We saw that in Example 1.1.6.

The equation \(s=s(\phi)\) is called the “intrinsic equation of the curve”.

The notation \(\measuredangle(\hat{\pmb{\imath}},\hat{\textbf{T}})\) means “the angle between \(\hat{\pmb{\imath}}\) and \(\hat{\textbf{T}}\)”.

Think about why this should be the case. In particular, sketch \(\hat{\textbf{T}}\) and \(\phi\) and think about what the sketch says about \(\dfrac{d\hat{\textbf{T}} }{d\phi}\text{.}\)

In each of the four figures, the arrow on the curve specifies the direction of increasing arc length \(s\) and the red dot is the centre of curvature for the curve at the blue dot.