# Calculus

Applications of the Derivative# Osculating Curves

Problem 1

Problems 2-5

### Order of Contact of Plane Curves

Let \(y = f\left( x \right)\) and \(y = g\left( x \right)\) be two plane curves, which osculate at the point \({M_0}\left( {{x_0},{y_0}} \right)\) (Figure \(1\)) and have derivatives up to the \(\left( {n + 1} \right)\)th order inclusively.

It is said that the curves \(y = f\left( x \right)\) and \(y = g\left( x \right)\) have a contact of order \(n\) at the point \({M_0}\left( {{x_0},{y_0}} \right)\) if the following conditions hold:

{f\left( {{x_0}} \right) = g\left( {{x_0}} \right),}\;\;\;\kern-0.3pt

{f’\left( {{x_0}} \right) = g’\left( {{x_0}} \right),}\;\;\;\kern-0.3pt

{f^{\prime\prime}\left( {{x_0}} \right) = g^{\prime\prime}\left( {{x_0}} \right),\ldots,}\;\;\;\kern-0.3pt

{{f^{\left( n \right)}}\left( {{x_0}} \right) = {g^{\left( n \right)}}\left( {{x_0}} \right),}\;\;\;\kern-0.3pt

{{f^{\left( {n + 1} \right)}}\left( {{x_0}} \right) \ne {g^{\left( {n + 1} \right)}}\left( {{x_0}} \right).}

\]

In particular, if \(n = 1,\) the curves \(y = f\left( x \right)\) and \(y = g\left( x \right)\) have a common tangent line.

Figure 1.

The case \(n = 0\) means that the curves have a common point \({M_0}\left( {{x_0},{y_0}} \right):\) \(f\left( {{x_0}} \right) = g\left( {{x_0}} \right),\) but their first derivatives do not coincide: \(f’\left( {{x_0}} \right) \ne g’\left( {{x_0}} \right).\) In this case, the curves simply intersect at the point \({M_0}.\)

We can consider the difference between the functions \(\varphi \left( x \right) = g\left( x \right) – f\left( x \right)\) in a neighborhood of the point \({x_0}\) and expand it in a Taylor series with Peano’s form of remainder. If the curves \(g\left( x \right)\) and \(f\left( x \right)\) have the \(n\)th order contact, then the first \(n\) terms of the series are zero and the difference \(\varphi \left( x \right)\) is represented as

{\varphi \left( x \right) = \frac{{{\varphi ^{\left( {n + 1} \right)}}\left( {{x_0}} \right) + \alpha }}{{\left( {n + 1} \right)!}}{\left( {x – {x_0}} \right)^{n + 1}} }

= {\frac{{{g^{\left( {n + 1} \right)}}\left( {{x_0}} \right) – {f^{\left( {n + 1} \right)}}\left( {{x_0}} \right) + \alpha }}{{\left( {n + 1} \right)!}}\cdot}\kern0pt{{\left( {x – {x_0}} \right)^{n + 1},}}

\]

i.e. proportional to \({\left( {x – {x_0}} \right)^{n + 1}}.\) Consequently, for even values of \(n,\) the difference \(\varphi \left( x \right)\) has opposite signs to the left and right of the point of contact \({M_0}.\) The particular case \(n = 0\) was considered above.

For odd \(n,\) the curves \(y = f\left( x \right)\) and \(y = g\left( x \right)\) osculate each other at the point \({M_0}\) without mutual intersection.

### Osculating Curve

Consider the following problem. Given the equation of a curve \(y = f\left( x \right)\) and a family of curves

with \(n + 1\) parameters \({a,b, \ldots ,\ell}.\) By changing the values of the parameters, choose a curve from the given family that has the highest possible order of contact with the curve \(y = f\left( x \right)\) at the point \({M_0}\left( {{x_0},{y_0}} \right).\) Such a curve is called the osculating curve.

We introduce the notation

The osculation conditions are written as

\Phi \left( {{x_0},a,b, \ldots ,\ell} \right) = 0\\

{\Phi’_x}\left( {{x_0},a,b, \ldots ,\ell} \right) = 0\\

{\Phi^{\prime\prime}_{xx}}\left( {{x_0},a,b, \ldots ,\ell} \right) = 0\\

\cdots \cdots \cdots \cdots \cdots \cdots \cdots \\

\Phi _{{x^n}}^{\left( n \right)}\left( {{x_0},a,b, \ldots ,\ell} \right) = 0

\end{array} \right..\]

As a result, we have the system of \(n + 1\) equations with \(n + 1\) unknown values of the parameters. By solving this system, we find the parameters \({a,b, \ldots ,\ell}\) and the equation of the osculating curve. Usually its order of contact is not lower than \(n\) (in case of \(n + 1\) parameters). Thus, the order of contact of an osculating curve is usually one less than the number of parameters.

### Osculating Circle

In this section we derive the equation of osculating circle. Let be given a function \(y = f\left( x \right),\) which is at least twice differentiable. The family of circles is described by the equation

As can be seen, we are dealing here with three parameters: the coordinates of the center of circle \(a, b\) and its radius \(R.\) It is clear that in this case the highest possible order of contact is equal to \(2.\)

Denoting

we write the derivatives of the function \(\Phi:\)

{{{\Phi’_x}\left( {{x_0},a,b,R} \right)} = {2\left( {x – a} \right) + 2\left( {y – b} \right)y’,}}\;\;\;\kern-0.3pt

{{{\Phi^{\prime\prime}_{xx}}\left( {{x_0},a,b,R} \right)} = {2 + 2{\left( {y’} \right)^2} + 2\left( {y – b} \right)y^{\prime\prime}.}}

\]

Assuming that the curves osculate at the point \(\left( {{x_0},{y_0}} \right),\) we obtain the following system of three equations for finding the osculating circle.

{\left\{ \begin{array}{l}

\Phi \left( {{x_0},a,b,R} \right) = 0\\

{\Phi’_x}\left( {{x_0},a,b,R} \right) = 0\\

{\Phi^{\prime\prime}_{xx}}\left( {{x_0},a,b,R} \right) = 0

\end{array} \right.,\;\;}\Rightarrow

{\left\{ \begin{array}{l}

{\left( {{x_0} – a} \right)^2} + {\left( {{y_0} – b} \right)^2} – {R^2} = 0\\

2\left( {{x_0} – a} \right) + 2\left( {{y_0} – b} \right){y’_0} = 0\\

2 + 2\left( {{y’_0}} \right)^2 + 2\left( {{y_0} – b} \right){y^{\prime\prime}_0} = 0

\end{array} \right.}

\]

From the last equation we find the value of \(b:\)

{2 + 2{\left( {{y’_0}} \right)^2} + 2\left( {{y_0} – b} \right){y^{\prime\prime}_0} = 0,\;\;}\Rightarrow

{\left( {{y_0} – b} \right){y^{\prime\prime}_0} = – 1 – {\left( {{y’_0}} \right)^2},\;\;}\Rightarrow

{{y_0} – b = – \frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}},\;\;}\Rightarrow

{b = {y_0} + \frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}.}

\]

Substituting \({{y_0} – b}\) into the second equation, we get the coordinate \(a\) of the center of circle:

{2\left( {{x_0} – a} \right) + 2\left( {{y_0} – b} \right){y’_0} = 0,\;\;}\Rightarrow

{{x_0} – a = – \left( {{y_0} – b} \right){y’_0},\;\;}\Rightarrow

{{x_0} – a = \frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}{y’_0},\;\;}\Rightarrow

{a = {x_0} – \frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}{y’_0}.}

\]

The radius of the osculating circle is determined from the first equation:

{{\left( {{x_0} – a} \right)^2} + {\left( {{y_0} – b} \right)^2} – {R^2} = 0,\;\;}\Rightarrow

{{R^2} = {\left( {{x_0} – a} \right)^2} + {\left( {{y_0} – b} \right)^2},\;\;}\Rightarrow

{{R^2} = {\left( {\frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}{y’_0}} \right)^2} + {\left( {\frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}} \right)^2},\;\;}\Rightarrow

{{R^2} = {\left( {\frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}} \right)^2}\cdot\kern0pt{\left[ {{{\left( {{y’_0}} \right)}^2} + 1} \right],}\;\;}\Rightarrow

{{R^2} = \frac{{{{\left[ {1 + {{\left( {{y’_0}} \right)}^2}} \right]}^3}}}{{{{\left( {{y^{\prime\prime}_0}} \right)}^2}}},\;\;}\Rightarrow

{R = \frac{{{{\left[ {1 + {{\left( {{y’_0}} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}{{\left| {{y^{\prime\prime}_0}} \right|}}.}

\]

As can be seen, the coordinates of the center of circle \(a, b\) are the coordinates of the center of curvature for the curve \(y = f\left( x \right)\) at \({x_0},\) and the radius of the osculating circle coincides with the radius of curvature of the curve at the point of contact.

## Solved Problems

Click on problem description to see solution.

### ✓ Example 1

Find the equation of the parabola osculating with the exponential function \(f\left( x \right) = {e^x}\) at the point \({x_0} = 0.\)

### ✓ Example 2

Find the equation of the parabola osculating with the function \(f\left( x \right) = \cos x\) at \({x_0} = 0.\)

### ✓ Example 3

Find the equation of the curve

which osculates with the graph of the logarithmic function \(f\left( x \right) = \ln x + 1\) at the point \({x_0} = 1.\)

### ✓ Example 4

Write the equation of a cubic function

the graph of which osculates with the curve \(f\left( x \right) = \tan x\) at the point \({x_0} = 0.\)

### ✓ Example 5

Write the equation of a circle osculating with the curve \(f\left( x \right) = \arctan x\) at the point \({x_0} = 1.\)

### Example 1.

Find the equation of the parabola osculating with the exponential function \(f\left( x \right) = {e^x}\) at the point \({x_0} = 0.\)

*Solution.*

We assume that the parabola is defined by the equation \(y = g\left( x \right) = a{x^2} + bx + c.\) This function has \(3\) parameters. Therefore, we can suppose that the order of contact of the curves is equal to \(2.\) Then the coefficients \(a, b, c\) are found from the following conditions:

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)

\end{array} \right..\]

The derivatives of the functions \(f\left( x \right) = {e^x}\) and \(g\left( x \right) = a{x^2} + bx + c\) are given by the formulas

{f’\left( x \right) = {\left( {{e^x}} \right)^\prime } = {e^x},}\;\;\;\kern-0.3pt

{f^{\prime\prime}\left( x \right) = {\left( {{e^x}} \right)^\prime } = {e^x};}

\]

{g’\left( x \right) = {\left( {a{x^2} + bx + c} \right)^\prime } = 2ax + b,}\;\;\;\kern-0.3pt

{g^{\prime\prime}\left( x \right) = {\left( {2ax + b} \right)^\prime } = 2a.}

\]

Then the system of equations takes the following form:

{e^{{x_0}}} = ax_0^2 + b{x_0} + c\\

{e^{{x_0}}} = 2a{x_0} + b\\

{e^{{x_0}}} = 2a

\end{array} \right..\]

Substituting \({x_0} = 0,\) we have

\left\{ \begin{array}{l}

c = 1\\

b = 1\\

2a = 1

\end{array} \right.

\;\;\kern-0.3pt{\text{or}\;\;

\left\{ \begin{array}{l}

a = \frac{1}{2}\\

b = 1\\

c = 1

\end{array} \right..}

\]

So the parabola osculating with the exponential function at the point \({x_0} = 0\) has the second order of contact and is determined by the formula

If we write its equation in the form

{y = \frac{{{x^2}}}{2} + x + 1 }

= {\frac{1}{2}\left( {{x^2} + 2x} \right) + 1 }

= {\frac{1}{2}\left( {{x^2} + 2x + 1 – 1} \right) + 1 }

= {\frac{1}{2}{\left( {x + 1} \right)^2} + \frac{1}{2},}

\]

Figure 2.

we see that the vertex of the parabola is at the point \(\left( { – 1,{\large\frac{1}{2}\normalsize}} \right).\) Schematically, both osculating curves are shown in Figure \(2.\)

Problem 1

Problems 2-5