Differential Equations

Second Order Equations

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Curvature of Plane Curves

Solved Problems

Example 1.

Determine the radius of curvature of the straight line.

Solution.

Let the line be given by the explicit equation \(y = ax + b,\) where \(a, b\) are some coefficients. We calculate the curvature \(k\) and the radius of curvature \(R\) of this straight line.

The absolute value of the curvature is given by

\[k = \frac{{\left| {y^{\prime\prime}} \right|}}{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}}.\]

In our case

\[y' = \left( {ax + b} \right) = a,\;\; y^{\prime\prime} = a' = 0.\]

This immediately implies that the curvature of the straight line is equal to zero, and the radius of curvature is respectively equal to infinity.

Example 2.

Determine the equation of the railway transition curve.

Solution.

It is known that when a body of mass \(m\) moves with velocity \(v\) along a curve, it experiences a centripetal force, the magnitude of which depends on the radius of curvature \(R:\)

\[F = \frac{{m{v^2}}}{R}.\]

The centripetal force is balanced by a reaction force called the centrifugal force, which acts for example on passengers of a train when it turns. The centrifugal force will remain constant when the body moves along the arc of a circle. To avoid sharp bumps in the transition from linear to circular motion, special transition paths are used, where the curvature increases gradually and uniformly from \(0\) to the final value of \(\frac{1}{R}.\)

Suppose that the transition curve corresponds to the arc \(OP\) (Figure \(3\)), whose length is equal to \(L.\)

Railway transition curve
Figure 3.

As the point \(M\) moves along this curve, the radius of curvature varies in proportion to the way \(s:\)

\[\frac{1}{r} = ms,\]

where \(m\) is the coefficient of proportionality.

This coefficient can be easily found from the boundary conditions: at \(s = OP = L,\) the curvature becomes equal to \(\frac{1} {R}:\)

\[\frac{1}{R} = mL,\;\; \Rightarrow m = \frac{1}{{LR}}.\]

Then the condition for the transition curve can be written as the following equation:

\[k = \frac{1}{r} = \frac{s}{{LR}},\;\; \Rightarrow \frac{{y^{\prime\prime}}}{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{s}{{LR}}.\]

The solution of the problem is simplified by setting \(s = x,\) where \(x\) is the projection of \(M\) on the \(x\)-axis. In this case the derivative \(y'\) will also be small and we can neglect it in the formula for calculating the curvature. As a result, we obtain the following differential equation of the transition curve:

\[y^{\prime\prime} = \frac{x}{{LR}}.\]

Integrating twice, we find the general solution of the equation:

\[y' = \frac{{{x^2}}}{{2LR}} + {C_1},\;\; y = \frac{{{x^3}}}{{6LR}} + {C_1}x + {C_2}.\]

Given the initial conditions \(y\left( {x = 0} \right) = 0\) and \(y'\left( {x = 0} \right) = 0,\) we obtain the final equation of the transition curve:

\[y = \frac{{{x^3}}}{{6LR}},\]

which is a cubic parabola.

Example 3.

Find the curve whose radius of curvature is constant.

Solution.

Let the curve be given by the equation \(y = y\left( x \right).\) Its radius of curvature is defined by the formula

\[R = \frac{1}{{\left| k \right|}} = \frac{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\left| {y^{\prime\prime}} \right|}}.\]

Since, by the condition of the problem, the radius \(R = \text{const}\) is a constant, we have the following differential equation:

\[\frac{{\left| {y^{\prime\prime}} \right|}}{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{1}{R}.\]

We make the substitution \(y' = p\) to reduce the order of the equation:

\[\frac{{\left| {p'} \right|}}{{{{\left[ {1 + {p^2}} \right]}^{\frac{3}{2}}}}} = \frac{1}{R}.\]

Separating the variables and taking the absolute value, we obtain

\[\int {\frac{{dp}}{{{{\left( {1 + {p^2}} \right)}^{\frac{3}{2}}}}}} = \pm \int {\frac{{dx}}{R}} .\]

Apply the new substitution:

\[p = \tan z,\;\;\Rightarrow z = \arctan p,\;\; \Rightarrow dz = \frac{{dp}}{{1 + {p^2}}}.\]

Hence, the equation takes the form

\[\int {\frac{{dz}}{{\sqrt {1 + {{\tan }^2}z} }}} = \pm \int {\frac{{dx}}{R}} .\]

Recalling the trigonometric identity

\[1 + {\tan ^2}z = \frac{1}{{{{\cos }^2}z}},\]

this equation can be simplified and then integrated:

\[\int {\cos zdz} = \pm \int {\frac{{dx}}{R}} ,\;\; \Rightarrow \sin z = \pm \frac{1}{R}\left( {x + {C_1}} \right),\]

where \({C_1}\) is a constant of integration.

Let us turn back to the variable \(p\) using the relationship

\[\sin z = \sqrt {\frac{{{{\tan }^2}z}}{{{{\tan }^2}z + 1}}} .\]

Here we consider only the positive square root, since both signs are contained in the right side of the differential equation. Then the equation can be written as

\[{\left[ {\frac{{{{\tan }^2}\left( {\arctan p} \right)}}{{{{\tan }^2}\left( {\arctan p} \right) + 1}}} \right]^{\frac{1}{2}}} = \pm \frac{1}{R}\left( {x + {C_1}} \right),\;\;\Rightarrow \frac{{{p^2}}}{{{p^2} + 1}} = \frac{1}{{{R^2}}}{\left( {x + {C_1}} \right)^2}.\]

Now return to the original variable \(y:\)

\[\frac{{{{\left( {y'} \right)}^2}}}{{{{\left( {y'} \right)}^2} + 1}} = \frac{1}{{{R^2}}}{\left( {x + {C_1}} \right)^2},\;\; \Rightarrow {\left( {y'} \right)^2} = \frac{1}{{{R^2}}}{\left( {x + {C_1}} \right)^2} \left( {{{\left( {y'} \right)}^2} + 1} \right),\;\; \Rightarrow {\left( {y'} \right)^2}\left[ {{R^2} - {{\left( {x + {C_1}} \right)}^2}} \right] = {\left( {x + {C_1}} \right)^2},\;\; \Rightarrow {\left( {y'} \right)^2} = \frac{{{{\left( {x + {C_1}} \right)}^2}}}{{{R^2} - {{\left( {x + {C_1}} \right)}^2}}},\;\; \Rightarrow y' = \pm \sqrt {\frac{{{{\left( {x + {C_1}} \right)}^2}}}{{{R^2} - {{\left( {x + {C_1}} \right)}^2}}}} ,\;\; \Rightarrow y = \pm \int {\sqrt {\frac{{{{\left( {x + {C_1}} \right)}^2}}}{{{R^2} - {{\left( {x + {C_1}} \right)}^2}}}} dx} .\]

To calculate the resulting integral we change the variable (hope it will be the last change of variable in this problem):

\[x + {C_1} = R\sin t,\;\; \Rightarrow dx = R\cos tdt.\]

The integral is equal to

\[y = \pm \int {\sqrt {\frac{{{{\left( {x + {C_1}} \right)}^2}}}{{{R^2} - {{\left( {x + {C_1}} \right)}^2}}}} dx} = \pm \int {\sqrt {\frac{{{R^2}\,{{\sin }^2}t}}{{{R^2} - {R^2}\,{{\sin }^2}t}}} R\cos tdt} = \pm R\int {\tan t\cos tdt} = \pm R\int {\sin t dt} = \mp R\cos t + {C_2}.\]

The resulting expression can be rewritten as

\[y + {C_2} = \pm R\cos t.\]

Eliminate the intermediate variable \(t:\)

\[{\sin ^2}t + {\cos ^2}t = 1,\;\; \Rightarrow \frac{1}{{{R^2}}}{\left( {x + {C_1}} \right)^2} + \frac{1}{{{R^2}}}{\left( {y + {C_2}} \right)^2} = 1,\;\; \Rightarrow {\left( {x + {C_1}} \right)^2} + {\left( {y + {C_2}} \right)^2} = {R^2}.\]

This shows that the curves with a constant radius of curvature \(R\) is a set of circles with an arbitrary center and the same radius \(R.\)

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