Osculating Curves
Solved Problems
Example 3.
Find the equation of the curve \[y = g\left( x \right) = \frac{a}{{x + b}},\] which osculates with the graph of the logarithmic function \[f\left( x \right) = \ln x + 1\] at the point \({x_0} = 1.\)
Solution.
The family of curves \(g\left( x \right)\) has two parameters \(a\) and \(b.\) Therefore, these curves will have the first order of contact. The conditions of osculation are written as follows:
\[\left\{ \begin{array}{l}
f\left( {{x_0}} \right) = g\left( {{x_0}} \right)\\
f'\left( {{x_0}} \right) = g'\left( {{x_0}} \right)
\end{array} \right..\]
Here the derivatives are given by
\[f'\left( x \right) = \left( {\ln x + 1} \right)^\prime = \frac{1}{x},\;\;\;g'\left( x \right) = \left( {\frac{a}{{x + b}}} \right)^\prime = - \frac{a}{{{{\left( {x + b} \right)}^2}}}.\]
Substituting the expressions for the functions and their derivatives, we obtain:
\[\left\{ \begin{array}{l}
\ln {x_0} + 1 = \frac{a}{{{x_0} + b}}\\
\frac{1}{{{x_0}}} = - \frac{a}{{{{\left( {x + b} \right)}^2}}}
\end{array} \right..\]
Given that \({x_0} = 1,\) we find the values of \(a\) and \(b:\)
\[
\left\{ \begin{array}{l}
\ln 1 + 1 = \frac{a}{{1 + b}}\\
\frac{1}{1} = - \frac{a}{{{{\left( {1 + b} \right)}^2}}}
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
a = 1 + b\\
a = - \left( {1 + b} \right)
\end{array} \right.,\;\; \Rightarrow
a = - {a^2},\;\; \Rightarrow
a + {a^2} = 0,\;\; \Rightarrow
a\left( {1 + a} \right) = 0.\]
The meaningful solution is the root \(a = -1.\) It corresponds to \(b = -2.\) Thus, the rational function, that has the \(1\)st order of contact (i.e. a common tangent) with the curve \(f\left( x \right) = \ln x + 1\) at \({x_0} = 1\) is described by the equation
\[y = g\left( x \right) = - \frac{1}{{x - 2}} = \frac{1}{{2 - x}}.\]
Example 4.
Write the equation of a cubic function \[y = g\left( x \right) = a{x^3} + b{x^2} + cx + d,\] the graph of which osculates with the curve \[f\left( x \right) = \tan x\] at the point \({x_0} = 0.\)
Solution.
We deal here with a family of functions containing \(4\) parameters. Therefore, the highest possible order of contact of the curves is equal to \(3.\) To determine the coefficients \(a, b, c,\) and \(d,\) we write the following conditions of osculation:
\[\left\{ \begin{array}{l}
f\left( {{x_0}} \right) = g\left( {{x_0}} \right)\\
f'\left( {{x_0}} \right) = g'\left( {{x_0}} \right)\\
f^{\prime\prime}\left( {{x_0}} \right) = g^{\prime\prime}\left( {{x_0}} \right)\\
f^{\prime\prime\prime}\left( {{x_0}} \right) = g^{\prime\prime\prime}\left( {{x_0}} \right)
\end{array} \right..\]
The derivatives of the cubic functions are given by
\[g'\left( x \right) = \left( {a{x^3} + b{x^2} + cx + d} \right)^\prime = 3a{x^2} + 2bx + c,\]
\[g^{\prime\prime}\left( x \right) = \left( {3a{x^2} + 2bx + c} \right)^\prime = 6ax + 2b,\]
\[g^{\prime\prime\prime}\left( x \right) = \left( {6ax + 2b} \right)^\prime = 6a.\]
Calculate the derivatives of the tangent function:
\[f'\left( x \right) = \left( {\tan x} \right)^\prime = \frac{1}{{{{\cos }^2}x}},\]
\[f^{\prime\prime}\left( x \right) = \left( {\frac{1}{{{{\cos }^2}x}}} \right)^\prime = \left[ {{{\left( {\cos x} \right)}^{ - 2}}} \right]^\prime = - 2{\left( {\cos x} \right)^{ - 3}} \cdot \left( { - \sin x} \right) = \frac{{2\sin x}}{{{{\cos }^3}x}},\]
\[f^{\prime\prime\prime}\left( x \right) = \left( {\frac{{2\sin x}}{{{{\cos }^3}x}}} \right)^\prime = \frac{{{{\left( {2\sin x} \right)}^\prime }{{\cos }^3}x - 2\sin x{{\left( {{{\cos }^3}x} \right)}^\prime }}}{{{{\cos }^6}x}} = \frac{{2\,{{\cos }^3}x + 6\,{{\sin }^2}x\,{{\cos }^2}x}}{{{{\cos }^6}x}} = \frac{{2 + 4\,{{\sin }^2}x}}{{{{\cos }^4}x}}.\]
Then the system of equations takes the form:
\[\left\{ \begin{array}{l}
\tan {x_0} = ax_0^3 + bx_0^2 + c{x_0} + d\\
\frac{1}{{{{\cos }^2}{x_0}}} = 3ax_0^2 + 2b{x_0} + c\\
\frac{{2\sin {x_0}}}{{{{\cos }^3}{x_0}}} = 6a{x_0} + 2b\\
\frac{{2 + 4{{\sin }^2}{x_0}}}{{{{\cos }^4}{x_0}}} = 6a
\end{array} \right.\]
Substituting the value \({x_0} = 0,\) we have
\[\left\{ \begin{array}{l}
d = 0\\
c = 1\\
2b = 0\\
6a = 2
\end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l}
a = \frac{1}{3}\\
b = 0\\
c = 1\\
d = 0
\end{array} \right..\]
Thus, the osculating cubic function is written as
\[y = \frac{{{x^3}}}{3} + x.\]
This curve has the third order of contact with the tangent curve at the origin.
Note that the resulting cubic function is the third order Maclaurin polynomial of the tangent function.
Example 5.
Write the equation of a circle osculating with the curve \[f\left( x \right) = \arctan x\] at the point \({x_0} = 1.\)
Solution.
Obviously, the point of contact of the two curves has the coordinates
\[\left( {{x_0},{y_0}} \right) = \left( {1,\frac{\pi }{4}} \right).\]
The coordinates of the center of the osculating circle and its radius are given by
\[a = {x_0} - \frac{{1 + {{\left( {{y'_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}{y'_0},\;\;\;b = {y_0} + \frac{{1 + {{\left( {{y'_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}},\;\;\;R = \frac{{{{\left[ {1 + {{\left( {{y'_0}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\left| {{y^{\prime\prime}_0}} \right|}}.\]
Find the derivatives used in these expressions:
\[y' = \left( {\arctan x} \right)^\prime = \frac{1}{{1 + {x^2}}},\;\;\;y^{\prime\prime} = \left( {\frac{1}{{1 + {x^2}}}} \right)^\prime = - \frac{{2x}}{{{{\left( {1 + {x^2}} \right)}^2}}}.\]
Their values at \({x_0} = 1\) are
\[{y'_0} = y'\left( 1 \right) = \frac{1}{2},\;\;\;{y^{\prime\prime}_0} = y^{\prime\prime}\left( 1 \right) = - \frac{1}{2}.\]
Then the coordinates of the center of the osculating circle are equal to
\[a = {x_0} - \frac{{1 + {{\left( {{y'_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}{y'_0} = 1 - \frac{{1 + {{\left( {\frac{1}{2}} \right)}^2}}}{{\left( { - \frac{1}{2}} \right)}} \cdot \frac{1}{2} = \frac{9}{4} = 2,25;\]
\[b = {y_0} + \frac{{1 + {{\left( {{y'_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}} = \frac{\pi }{4} + \frac{{1 + {{\left( {\frac{1}{2}} \right)}^2}}}{{\left( { - \frac{1}{2}} \right)}} = \frac{\pi }{4} - \frac{5}{2} \approx - 1,71.\]
We also calculate the radius of the osculating circle:
\[R = \frac{{{{\left[ {1 + {{\left( {{y'_0}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\left| {{y^{\prime\prime}_0}} \right|}} = \frac{{{{\left[ {1 + {{\left( {\frac{1}{2}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\left| { - \frac{1}{2}} \right|}} = \frac{{{{\left( {1 + \frac{1}{4}} \right)}^{\frac{3}{2}}}}}{{\frac{1}{2}}} = 2{\left( {\frac{5}{4}} \right)^{\frac{3}{2}}} = \frac{{\sqrt {125} }}{4} \approx 2,80.\]
Thus, the center of the osculating circle is at the point \(\left( {{\frac{9}{4}},{\frac{\pi }{4}} - {\frac{5}{2}}} \right)\) (Figure \(3\)).
Figure 3.
