5.9: Kinematic analysis and coordinate systems

The kinematic analysis presented in the previous sections is valid for any coordinate system or constraint curve \(\overrightarrow{\boldsymbol{r}}_{s}\). We now show how to obtain the 3 dimensional motion in a specific coordinate system or a specific path curve \(\overrightarrow{\boldsymbol{r}}_{s}(s)\) that is expressed in Cartesian or cylindrical coordinates. An important case is motion along a straight line.

5.9.1 Rectilinear motion

Motion along a straight \(1 \mathrm{D}\) line is the most simple to analyse. If motion occurs along the \(x\)-coordinate axis in Cartesian coordinates, the path curve pointing in the positive \(x\)-coordinate direction is:

\[\overrightarrow{\boldsymbol{r}}_{s}(s)=s \hat{\boldsymbol{\imath}} \tag{5.53} \label{5.53}\]

A certain point mass \(i\) with path coordinate \(s_{i}\) is located at position vector \(\overrightarrow{\boldsymbol{r}}_{i}=\overrightarrow{\boldsymbol{r}}_{s}\left(s_{i}\right)=s_{i} \hat{\boldsymbol{\imath}}\). Projecting its position on the \(x\)-coordinate axis, we find that its \(x\)-coordinate is identical to its path coordinate \(x_{i}=\overrightarrow{\boldsymbol{r}}_{i} \cdot \hat{\boldsymbol{\imath}}=s_{i}\). So, essentially everything that we have derived for \(s_{i}(t)\) also holds for the \(x\)-coordinate \(x_{i}(t)\) of a point mass that moves on a rectilinear path along the \(x\)-coordinate axis, with \(s_{i}=x_{i}, v_{s, i}=\dot{x}_{i}\) and \(a_{s, i}=\ddot{x}_{i}\), from Eqs. (5.30)-(5.32).

5.9.2 Kinematics in Cartesian coordinates

Cartesian coordinates have the advantage that the direction of the unit vectors \(\hat{\boldsymbol{\imath}}, \hat{\boldsymbol{\jmath}}\) and \(\hat{\boldsymbol{k}}\) is independent of the coordinates in the system. If the coordinate system is not rotating (see also Sec. 6.12), the direction of the unit vectors is also independent of time \(\frac{\mathrm{d} \hat{\imath}}{\mathrm{d} t}=\frac{\mathrm{d} \hat{\jmath}}{\mathrm{d} t}=\frac{\mathrm{d} \hat{\boldsymbol{k}}}{\mathrm{d} t}=\overrightarrow{\mathbf{0}}\). This greatly simplifies the kinematic analysis and allows describing the motion of a point mass \(i\) using its 3 time-dependent coordinates \(x_{i}, y_{i}\) and \(z_{i}\) and Eqs. (5.28)-(5.29) as follows:

\[\begin{align} \overrightarrow{\boldsymbol{r}}_{i}(t) & =x_{i}(t) \hat{\boldsymbol{\imath}}+y_{i}(t) \hat{\boldsymbol{\jmath}}+z_{i}(t) \hat{\boldsymbol{k}} \tag{5.54} \label{5.54}\\[4pt] \overrightarrow{\boldsymbol{v}}_{i}(t) & =\dot{x}_{i}(t) \hat{\boldsymbol{\imath}}+\dot{y}_{i}(t) \hat{\boldsymbol{\jmath}}+\dot{z}_{i}(t) \hat{\boldsymbol{k}} \tag{5.55} \label{5.55}\\[4pt] \overrightarrow{\boldsymbol{a}}_{i}(t) & =\ddot{x}_{i}(t) \hat{\boldsymbol{\imath}}+\ddot{y}_{i}(t) \hat{\boldsymbol{\jmath}}+\ddot{z}_{i}(t) \hat{\boldsymbol{k}} \tag{5.56} \label{5.56}\end{align}\]

It is also possible to integrate the individual functions \(\ddot{x}_{i}(t), \ddot{y}_{i}(t), \ddot{z}_{i}(t)\) to obtain the velocity and position of the point mass when the acceleration is known, similar to the integrals shown in Eqs. (5.35), (5.36), for the velocity and acceleration component along the \(x, y\) and \(z\)-axis:

\[\begin{align} & \overrightarrow{\boldsymbol{r}}_{i}\left(t_{2}\right)=\overrightarrow{\boldsymbol{r}}_{i}\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} \dot{x}_{i}(t) \mathrm{d} t \hat{\boldsymbol{\imath}}+\int_{t_{1}}^{t_{2}} \dot{y}_{i}(t) \mathrm{d} t \hat{\boldsymbol{\jmath}}+\int_{t_{1}}^{t_{2}} \dot{z}_{i}(t) \mathrm{d} t \hat{\boldsymbol{k}} \tag{5.57} \label{5.57}\\[4pt] & \overrightarrow{\boldsymbol{v}}_{i}\left(t_{2}\right)=\overrightarrow{\boldsymbol{v}}_{i}\left(t_{1}\right)+\int_{t_{1}}^{t_{2}} \ddot{x}_{i}(t) \mathrm{d} t \hat{\boldsymbol{\imath}}+\int_{t_{1}}^{t_{2}} \ddot{y}_{i}(t) \mathrm{d} t \hat{\boldsymbol{\jmath}}+\int_{t_{1}}^{t_{2}} \ddot{z}_{i}(t) \mathrm{d} t \hat{\boldsymbol{k}} \tag{5.58} \label{5.58}\end{align}\]

Example 5.3

Let us illustrate the kinematic analysis method in Cartesian coordinates by determining the acceleration and velocity vector of the balloon \(B\) in Figure 5.9, that moves along a parabolic path curve with \(y_{B}=c_{1} x_{B}^{2}\) in the \(x y\)-plane.

Figure 5.9: Example 5.3, determining the velocity and acceleration of a balloon along a parabolic path curve with \(c_{2}=1 \mathrm{~m} / \mathrm{s}\) and \(c_{1}=\frac{1}{20} \mathrm{~m}^{-1}\).

The motion of the balloon is described by:

\[\begin{align} x_{B}(t) & =c_{2} t \\[4pt] y_{B}(t) & =c_{1} c_{2}^{2} t^{2} \end{align}\]

The position vector of the balloon can be differentiated twice, like in Equation 5.54 and Equation 5.56, to obtain the velocity and acceleration vectors as follows:

\[\begin{align} \overrightarrow{\boldsymbol{r}}_{B}(t) & =c_{2} t \hat{\boldsymbol{\imath}}+c_{1} t^{2} \hat{\boldsymbol{\jmath}} \\[4pt] \overrightarrow{\boldsymbol{v}}_{B}(t) & =c_{2} \hat{\boldsymbol{\imath}}+2 c_{1} t \hat{\boldsymbol{\jmath}} \\[4pt] \overrightarrow{\boldsymbol{a}}_{B}(t) & =2 c_{1} \hat{\boldsymbol{j}} \end{align}\]

The speed can be obtained by taking the absolute value of the velocity vector:

\[v_{B}=\left|\overrightarrow{\boldsymbol{v}}_{B}\right|=\sqrt{\overrightarrow{\boldsymbol{v}}_{B} \cdot \overrightarrow{\boldsymbol{v}}_{B}}=\sqrt{c_{2}^{2}+\left(2 c_{1} t\right)^{2}} \tag{5.59} \label{5.59}\]

5.9.3 Kinematics in cylindrical coordinates

Cylindrical coordinates, which were introduced in section 3.3.2 are often convenient when the kinematics involves rotation around an axis, like in Figure 5.10. The position, velocity and acceleration vector in cylindrical coordinates can be written as:

\[\begin{align} \overrightarrow{\boldsymbol{r}}_{i} & =\rho_{i} \hat{\boldsymbol{\rho}}+z_{i} \hat{\boldsymbol{k}} \tag{5.60} \label{5.60}\\[4pt] \overrightarrow{\boldsymbol{v}}_{i} & =v_{i, \rho} \hat{\boldsymbol{\rho}}+v_{i, \phi} \hat{\boldsymbol{\phi}}+v_{i, z} \hat{\boldsymbol{k}} \tag{5.61} \label{5.61}\\[4pt] \overrightarrow{\boldsymbol{a}}_{i} & =a_{i, \rho} \hat{\boldsymbol{\rho}}+a_{i, \phi} \hat{\boldsymbol{\phi}}+a_{i, z} \hat{\boldsymbol{k}} \tag{5.62} \label{5.62}\end{align}\]

Figure 5.10: Point mass \(i\) moving in a circular orbit in the \(x y\)-plane. Its position is indicated by cylindrical coordinates \(\rho_{i}\) and \(\phi_{i}\). Its velocity vector (orange) points in the \(\hat{\phi}\) direction, tangential to its path curve, its acceleration vector (purple) is in the negative \(\hat{\rho}\) direction.

Concept. Kinematics in cylindrical coordinates

We will show that by taking the time derivatives of the position vector, the components of the velocity vector and acceleration vector in Eqs. (5.60-5.62) are found to be:

\[\begin{align} v_{i, \rho} & =\dot{\rho}_{i} \tag{5.63} \label{5.63}\\[4pt] v_{i, \phi} & =\rho_{i} \dot{\phi}_{i} \tag{5.64} \label{5.64}\\[4pt] v_{i, z} & =\dot{z}_{i} \tag{5.65} \label{5.65}\\[4pt] a_{i, \rho} & =\ddot{\rho}_{i}-\rho_{i} \dot{\phi}_{i}^{2} \tag{5.66} \label{5.66}\\[4pt] a_{i, \phi} & =2 \dot{\rho}_{i} \dot{\phi}_{i}+\rho_{i} \ddot{\phi}_{i} \tag{5.67} \label{5.67}\\[4pt] a_{i, z} & =\ddot{z}_{i} \tag{5.68} \label{5.68}\end{align}\]

Derivation.

The determination of these velocity and acceleration vectors in cylindrical coordinates is more difficult because the direction of the unit vectors \(\hat{\rho}\) and \(\hat{\boldsymbol{\phi}}\) is not constant. To determine their time derivatives, we use that the position vector can also be expressed in Cartesian unit vectors as:

\[\overrightarrow{\boldsymbol{r}}_{i}=\rho_{i} \cos \phi_{i} \hat{\boldsymbol{\imath}}+\rho_{i} \sin \phi_{i} \hat{\boldsymbol{\jmath}}+z_{i} \hat{\boldsymbol{k}} \tag{5.69} \label{5.69}\]

From the directions of the unit vectors, as shown in Figure 5.10, the cylindrical unit vectors can be expressed in Cartesian unit vectors as:

\[\begin{align} & \hat{\boldsymbol{\rho}}=\cos \phi_{i} \hat{\boldsymbol{\imath}}+\sin \phi_{i} \hat{\boldsymbol{\jmath}} \tag{5.70} \label{5.70}\\[4pt] & \hat{\boldsymbol{\phi}}=-\sin \phi_{i} \hat{\boldsymbol{\imath}}+\cos \phi_{i} \hat{\boldsymbol{\jmath}} \tag{5.71} \label{5.71}\end{align}\]

Now it can be seen that the time derivatives of the unit vectors obey (chain rule):

\[\begin{align} \frac{\mathrm{d} \hat{\boldsymbol{\rho}}}{\mathrm{d} t} & =\left(-\sin \phi_{i} \hat{\boldsymbol{i}}+\cos \phi_{i} \hat{\boldsymbol{j}}\right) \dot{\phi}_{i}=\dot{\phi}_{i} \hat{\boldsymbol{\phi}}=\overrightarrow{\boldsymbol{\omega}} \times \hat{\boldsymbol{\rho}} \tag{5.72} \label{5.72}\\[4pt] \frac{\mathrm{d} \hat{\boldsymbol{\phi}}}{\mathrm{d} t} & =\left(-\cos \phi_{i} \hat{\boldsymbol{i}}-\sin \phi_{i} \hat{\boldsymbol{j}}\right) \dot{\phi}_{i}=-\dot{\phi}_{i} \hat{\boldsymbol{\rho}}=\overrightarrow{\boldsymbol{\omega}} \times \hat{\boldsymbol{\phi}} \tag{5.73} \label{5.73}\end{align}\]

where the terms in brackets were replaced by \(\hat{\boldsymbol{\phi}}\) and \(\hat{\boldsymbol{\rho}}\) using Eqs. (5.70) and (5.71), and the equations were further simplified using the vector \(\overrightarrow{\boldsymbol{\omega}}=\dot{\phi}_{i} \hat{\boldsymbol{k}}\). We use these time derivatives of the unit vectors to find the velocity and acceleration in cylindrical coordinates, and the components given in Eqs. (5.63-5.68) by differentiating twice:

\[\begin{align} \overrightarrow{\boldsymbol{r}}_{i} & =\rho_{i} \hat{\boldsymbol{\rho}}+z_{i} \hat{\boldsymbol{k}} \tag{5.74} \label{5.74}\\[4pt] \overrightarrow{\boldsymbol{v}}_{i} & =\frac{\mathrm{d} \overrightarrow{\boldsymbol{r}}_{i}}{\mathrm{~d} t}=\dot{\rho}_{i} \hat{\boldsymbol{\rho}}+\rho_{i} \frac{\mathrm{d} \hat{\boldsymbol{\rho}}}{\mathrm{d} t}+\dot{z}_{i} \hat{\boldsymbol{k}} \tag{5.75} \label{5.75}\\[4pt] & =\dot{\rho}_{i} \hat{\boldsymbol{\rho}}+\rho_{i} \dot{\phi}_{i} \hat{\boldsymbol{\phi}}+\dot{z}_{i} \hat{\boldsymbol{k}} \tag{5.76} \label{5.76}\\[4pt] \overrightarrow{\boldsymbol{a}}_{i} & =\frac{\mathrm{d} \overrightarrow{\boldsymbol{v}}_{i}}{\mathrm{~d} t}=\ddot{\rho}_{i} \hat{\boldsymbol{\rho}}+\dot{\rho}_{i} \frac{\mathrm{d} \hat{\boldsymbol{\rho}}}{\mathrm{d} t}+\dot{\rho}_{i} \dot{\phi}_{i} \hat{\boldsymbol{\phi}}+\rho_{i} \ddot{\phi}_{i} \hat{\boldsymbol{\phi}}+\rho_{i} \dot{\phi}_{i} \frac{\mathrm{d} \hat{\boldsymbol{\phi}}}{\mathrm{d} t}+\ddot{z}_{i} \hat{\boldsymbol{k}} \tag{5.77} \label{5.77}\\[4pt] & =\left(\ddot{\rho}_{i}-\rho_{i} \dot{\phi}_{i}^{2}\right) \hat{\boldsymbol{\rho}}+\left(2 \dot{\rho}_{i} \dot{\phi}_{i}+\rho_{i} \ddot{\phi}_{i}\right) \hat{\boldsymbol{\phi}}+\ddot{z}_{i} \hat{\boldsymbol{k}} \tag{5.78} \label{5.78}\end{align}\]

The components of these vectors are identical to the equations (5.63-5.68) thus proving the correctness of those equations.

5.9.4 Circular motion

A special important case of constrained motion on a path curve is motion on a circle with constant radius \(\rho\). So, in Figure 5.10 this means \(\rho_{i}=\) constant. The components of the velocity and acceleration vector can be found by substituting \(\dot{\rho}_{i}=0\) and \(\ddot{\rho}_{i}=0\) into Eqs. (5.75-5.78) to obtain:

\[\begin{align} v_{i, \rho} & =0 \tag{5.79} \label{5.79}\\[4pt] v_{i, \phi} & =\rho \dot{\phi}_{i}=\dot{s}_{i} \tag{5.80} \label{5.80}\\[4pt] a_{i, \rho} & =-\rho \dot{\phi}_{i}^{2}=-\frac{v_{s, i}^{2}}{\rho} \tag{5.81} \label{5.81}\\[4pt] a_{i, \phi} & =\rho \ddot{\phi}_{i}=\dot{v}_{s, i}=\ddot{s}_{i} \tag{5.82} \label{5.82}\end{align}\]

It can be seen that the velocity vector only has a component in the \(\hat{\phi}\) direction, tangential to the circle. The acceleration has both a tangential component \(a_{i, \phi}\),
and a radial or centripetal component \(a_{i, \rho}\) that points towards the centre of the circle. The time derivatives of the angle \(\omega_{o, i}=\dot{\phi}_{i}\) and \(\alpha_{o, i}=\ddot{\phi}_{i}\) are called the orbital angular velocity and orbital angular acceleration of the point mass. The word orbital is used to distinguish this angular velocity from the angular velocity of a rigid body, which is sometimes called the spin angular velocity and will be discussed later in Ch. 9.

5.9.5 Natural \(t, n, b\) coordinates for curvilinear motion

Natural coordinates make use of the fact that even if a curved trajectory is not really a circle, it is possible to approximate it by a circle with radius \(\rho\) at every point of the trajectory, as shown in Figure 5.11. Like in the previous subsection, a cylindrical coordinate system can be chosen such that the centre of that circle (the centre of curvature \(O_{i}\) ) is at the origin and the point mass position is determined by a path coordinate \(s_{i}=\rho \phi_{i}\), where \(\rho\) is called the radius of curvature. The plane in which the instantaneous circle is positioned is called the osculating plane. For an instantaneous moment, the motion of the point mass can be described using natural coordinates or \(t, n, b\) coordinates, that are defined by a unit vector \(\hat{\boldsymbol{t}}\) that is tangential to the curvilinear path \({ }^{2}\), a normal unit vector \(\hat{\boldsymbol{n}}\) that points towards the centre of the instantaneous circle and a binormal unit vector \(\hat{\boldsymbol{b}}=\hat{\boldsymbol{t}} \times \hat{\boldsymbol{n}}\), that points perpendicular to the osculating plane. At every instant of time the path curve can have a different origin \(O\) and radius \(\rho\). Note that in Figure 5.11 the origin changes from point \(O_{1}\) to \(O_{2}\). As a consequence, at times \(t_{1}\) and times \(t_{3}\), the directions of both the \(\hat{\boldsymbol{n}}\) and \(\hat{\boldsymbol{b}}\) unit vectors have flipped because the motion changed from anti-clockwise to clockwise, even though the tangential unit vector \(\hat{\boldsymbol{t}}\) is the same. In \(t, n, b\) coordinates, the components of the instantaneous velocity vector are \(\overrightarrow{\boldsymbol{v}}_{i}=v_{i, t} \hat{\boldsymbol{t}}+v_{i, n} \hat{\boldsymbol{n}}\) and the acceleration vector is \(\overrightarrow{\boldsymbol{a}}_{i}=a_{i, t} \hat{\boldsymbol{t}}+a_{i, n} \hat{\boldsymbol{n}}\). Note the sign change \(a_{i, n}=-a_{i, \rho}\) compared to Equation 5.81, because \(\hat{\boldsymbol{n}}\) points inward, while \(\hat{\boldsymbol{\rho}}\) points away from the rotation axis. The velocity and acceleration components are:

\[\begin{align} v_{i, t} & =\dot{s}_{i} \tag{5.83} \label{5.83}\\[4pt] v_{i, n} & =0 \tag{5.84} \label{5.84}\\[4pt] a_{i, t} & =\dot{v}_{s, i}=\ddot{s}_{i} \tag{5.85} \label{5.85}\\[4pt] a_{i, n} & =v_{s, i}^{2} / \rho \tag{5.86} \label{5.86}\end{align}\]

Figure 5.11: Natural \(t, n, b\) coordinates that define the local unit vectors depending on the shape of the path curve.

Velocity and acceleration along the \(\hat{\boldsymbol{b}}\) direction are always zero. Note that these equations are very similar to those for circular motion, except for the minus sign in Equation 5.81. The radius \(\rho\) can change continuously along the curvilinear path and needs to be determined at every point of the curve in order to use these equations to determine the given velocity and acceleration components. Furthermore, because they do not have a fixed origin, natural coordinates are not used for describing positions, and are therefore not a replacement for Cartesian and cylindrical coordinates.

5.9.6 Summary parametrised motion and kinematics

After having discussed kinematics in different coordinate systems we conclude by summarising the different ways these coordinates allow us to parametrise motion in 3D and obtain the kinematics. Depending on the situation, the time dependent motion \(\overrightarrow{\boldsymbol{r}}(t)\) can be fully described either using 3 time dependent coordinate functions \(x(t), y(t), z(t)\), or using the path curve \(\overrightarrow{\boldsymbol{r}}_{s}(s)=x(s) \hat{\boldsymbol{\imath}}+\) \(y(s) \hat{\boldsymbol{j}}+z(s) \hat{\boldsymbol{k}}\) and a single time dependent coordinate function \(s(t):\)

Parametrisations of motion and determination of \(\vec{v}\) and \(\vec{a}\)

The time dependent coordinate functions:

1. Cartesian coordinate functions \(x(t)\) and \(y(t)\).
   • Use methods from Sec. 5.9.2 to obtain \(\overrightarrow{\boldsymbol{v}}\) and \(\overrightarrow{\boldsymbol{a}}\).
2. Cylindrical coordinate functions \(\rho(t)\) and \(\phi(t)\).
   • Use methods from Sec. 5.9.3 to obtain \(\overrightarrow{\boldsymbol{v}}\) and \(\overrightarrow{\boldsymbol{a}}\).
   • First convert to Cartesian coordinates using \(x(t)=\rho(t) \cos \phi(t)\) and \(y(t)=\rho(t) \sin \phi(t)\), and proceed like in 1.

The path curve and one coordinate function of time:

1. Cartesian path curve \(x(s), y(s)\), and path coordinate \(s(t)\).
   • Use methods from Sec. 5.8 to determine the path velocity \(v_{s}\), and tangential component of acceleration \(a_{s}\) from \(s(t)\).
   • Use \(t, n, b\) coordinates (Sec. 5.9.5) to obtain the normal component of acceleration \(a_{n}\) by first determining the local path radius \(\rho\).
   • Alternatively one can determine \(x(t)=x(s(t)), y(t)=y(s(t))\) and proceed like in 1.
2. Cartesian path curve \(y(x)\) and coordinate function \(x(t)\).
   • Determine the function \(y(t)=y(x(t))\) and use method 1 .
3. Cylindrical path curve \(\rho(s), \phi(s)\) and path coordinate \(s(t)\).
   • Proceed using \(s(t)\) and \(t, n, b\) coordinates like in 3 .
   • Use \(x(t)=\rho(s(t)) \cos \phi(s(t))\) and \(y(t)=\rho(s(t)) \sin \phi(s(t))\), and proceed like in 1.
4. Cylindrical path curve \(\rho(\phi)\) and coordinate function \(\phi(t)\).
   • Determine the function \(\rho(t)=\rho(\phi(t))\) and use method 2 .

As becomes clear from this list, there are often multiple methods to determine the velocity and acceleration from given kinematic information. However, some methods might be much more mathematically complex than others. It is therefore useful to become familiar with the different methods, and get experience in choosing the easiest one in a certain situation.

5.9.7 Determining the path coordinate*

We note that it is also possible to determine the path coordinate \(s(t)\) by going from parametrisation 4 to 3 . For that purpose you use that it holds from Pythagoras’ theorem that the distance \(\mathrm{d} s\) along a straight infinitesimal line element obeys \(\mathrm{d} s^{2}=\mathrm{d} x^{2}+\mathrm{d} y^{2}+\mathrm{d} z^{2}\), such that \(s(t)\) can be determined using the integral:

\[s(t)=\int_{0}^{s(t)} \mathrm{d} s=\int_{0}^{x(t)} \sqrt{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^{2}+\left(\frac{\mathrm{d} z}{\mathrm{~d} x}\right)^{2}} \mathrm{~d} x \tag{5.87} \label{5.87}\]

For example if we deal with motion at constant speed along a straight tilted line, \(y=\alpha x\), and \(x(t)=v_{x} t\) then we have:

\[s(t)=\int_{0}^{v_{x} t} \sqrt{1+\alpha^{2}} \mathrm{~d} x=\sqrt{1+\alpha^{2}} \times v_{x} t \tag{5.88} \label{5.88}\]

Noting that the \(y\) velocity component is \(v_{y}=\dot{y}=\frac{\mathrm{d} y}{\mathrm{~d} x} \dot{x}=\alpha v_{x}\), we see that the path velocity \(v_{s}=\dot{s}=\sqrt{1+\alpha^{2}} \times v_{x}=\sqrt{v_{x}^{2}+v_{y}^{2}}\) as expected.

5.9.8 Determining \(\rho\), the radius of curvature*

When working with natural \(t, n, b\) coordinates, one needs to know the local radius of curvature \(\rho\) at each position to determine the velocity and acceleration components using (5.83-5.86). However, determining the radius of curvature \(\rho\) is not always easy. For a path curve in the \(x y\)-plane that can be written in the form \(y(x)\), the instantaneous radius \(\rho\) can be determined using this convenient function:

\[\rho=\frac{\left[1+\frac{\mathrm{d} y}{\mathrm{~d} x}\right]^{\frac{3}{2}}}{\left|\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\right|} \tag{5.89} \label{5.89}\]

Derivation: Radius of Curvature

This equation can be derived by considering a circle with centre \(\left(x_{0}, y_{0}\right)\) and radius \(\rho\) while ensuring that its first and second derivative with respect to \(x\) match those of the path curve \(y(x)\) at the point \((x, y)\) where they touch:

\[\begin{align} \rho^{2} & =\left(y-y_{0}\right)^{2}+\left(x-x_{0}\right)^{2} \tag{5.90} \label{5.90}\\[4pt] y & =y_{0}+\sqrt{\rho^{2}-\left(x-x_{0}\right)^{2}} \tag{5.91} \label{5.91}\\[4pt] \frac{\mathrm{d} y}{\mathrm{~d} x} & =-\frac{x-x_{0}}{y-y_{0}} \tag{5.92} \label{5.92}\\[4pt] \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} & =\frac{-1}{y-y_{0}}+\frac{\mathrm{d} y}{\mathrm{~d} x} \frac{x-x_{0}}{\left(y-y_{0}\right)^{2}}=-\frac{1+\frac{\mathrm{d} y^{2}}{\mathrm{~d} x}}{y-y_{0}} \tag{5.93} \label{5.93}\\[4pt] y-y_{0} & =-\frac{1+\frac{\mathrm{d} y}{\mathrm{~d} x}}{\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}} \tag{5.94} \label{5.94}\\[4pt] \rho^{2} & =\left(y-y_{0}\right)^{2}\left[1+\frac{\left(x-x_{0}\right)^{2}}{\left(y-y_{0}\right)^{2}}\right](\text { from Equation 5.90) } \tag{5.95} \label{5.95}\\[4pt] & =\left[\frac{1+\frac{\mathrm{d} y}{\mathrm{~d} x}}{\left|\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\right|}\right]^{2}\left[1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^{2}\right](\text { used }(5.92) \&(5.94)) \tag{5.96} \label{5.96}\\[4pt] \rho & =\frac{\left(1+\frac{\mathrm{d} y^{2}}{\mathrm{~d} x}\right)^{3 / 2}}{\left|\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}\right|} \tag{5.97} \label{5.97}\end{align}\]