Consider a plane curve defined by the equation y = f (x). Suppose that the tangent line is drawn to the curve at a point M(x, y). The tangent forms an angle α with the horizontal axis (Figure 1).
Figure 1.
At the displacement \(\Delta s\) along the arc of the curve, the point \(M\) moves to the point \({M_1}.\) The position of the tangent line also changes: the angle of inclination of the tangent to the positive \(x-\text{axis}\) at the point \({M_1}\) will be \(\alpha + \Delta\alpha.\) Thus, as the point moves by the distance \(\Delta s,\) the tangent rotates by the angle \(\Delta\alpha.\) (The angle \(\alpha\) is supposed to be increasing when rotating counterclockwise.)
The absolute value of the ratio \(\frac{{\Delta \alpha }}{{\Delta s}}\) is called the mean curvature of the arc \(M{M_1}.\) In the limit as \(\Delta s \to 0,\) we obtain the curvature of the curve at the point \(M:\)
From this definition it follows that the curvature at a point of a curve characterizes the speed of rotation of the tangent of the curve at this point.
For a plane curve given by the equation \(y = f\left( x \right),\) the curvature at a point \(M\left( {x,y} \right)\) is expressed in terms of the first and second derivatives of the function \(f\left( x \right)\) by the formula
\[K = \frac{{\left| {y^{\prime\prime}\left( x \right)} \right|}}{{{{\left[ {1 + {{\left( {y'\left( x \right)} \right)}^2}} \right]}^{\frac{3}{2}}}}}.\]
If a curve is defined in parametric form by the equations \(x = x\left( t \right),\) \(y = y\left( t \right),\) then its curvature at any point \(M\left( {x,y} \right)\) is given by
The radius of curvature of a curve at a point \(M\left( {x,y} \right)\) is called the inverse of the curvature \(K\) of the curve at this point:
\[R = \frac{1}{K}.\]
Hence for plane curves given by the explicit equation \(y = f\left( x \right),\) the radius of curvature at a point \(M\left( {x,y} \right)\) is given by the following expression:
\[R = \frac{{{{\left[ {1 + {{\left( {y'\left( x \right)} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\left| {y^{\prime\prime}\left( x \right)} \right|}}.\]
Solved Problems
Example 1.
Calculate the curvature of the ellipse \[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\]
at its vertices.
Solution.
Obviously, it suffices to find the curvature of the ellipse at points \(A\left( {a,0} \right)\) and \(B\left( {0,b} \right)\) (Figure \(2\)), because due to the symmetry of the curve, the curvature at the two opposite vertices of the ellipse will be the same.
Figure 2.
To calculate the curvature, it is convenient to pass from the canonical equation of the ellipse to the equation in parametric form:
\[x = a\cos t,\;\;\;y = b\sin t,\]
where \(t\) is a parameter. The parameter has the value \(t = 0\) at the point \(A\left( {a,0} \right)\) and is equal to \(t = \frac{\pi }{2}\) at the point \(B\left( {0,b} \right).\)
Find the curvature and radius of curvature of the curve \[y = \cos mx\] at a maximum point.
Solution.
This function reaches a maximum at the points \(x = {\frac{{2\pi n}}{m}},\;n \in Z.\) By the periodicity, the curvature at all maximum points is the same, so it is sufficient to consider only the point \(x = 0.\)
