# Calculus

## Applications of the Derivative # Osculating Curves

• ### 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 $${M_0}\left( {{x_0},{y_0}} \right)$$ $$\left({\text{Figure }1}\right)$$ 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.

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

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

that is 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

$G\left( {x,y,a,b, \ldots ,\ell} \right) = 0$

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

$\Phi \left( {x,a,b, \ldots ,l} \right) = G\left( {x,f\left( x \right),a,b, \ldots ,l} \right).$

The osculation conditions are written as

$\left\{ \begin{array}{l} \Phi \left( {{x_0},a,b, \ldots ,\ell} \right) = 0\\ {\Phi’_x}\left( {{x_0},a,b, \ldots ,\ell} \right) = 0\\ {\Phi^{\prime\prime}_{xx}}\left( {{x_0},a,b, \ldots ,\ell} \right) = 0\\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \Phi _{{x^n}}^{\left( n \right)}\left( {{x_0},a,b, \ldots ,\ell} \right) = 0 \end{array} \right..$

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

${\left( {x – a} \right)^2} + {\left( {y – b} \right)^2} = {R^2}.$

As it 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

${\Phi \left( {x,a,b,R} \right)} = {{\left( {x – a} \right)^2} + {\left( {y – b} \right)^2} – {R^2},}$

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:

${{\left( {{x_0} – a} \right)^2} + {\left( {{y_0} – b} \right)^2} – {R^2} = 0,\;\;}\Rightarrow {{R^2} = {\left( {{x_0} – a} \right)^2} + {\left( {{y_0} – b} \right)^2},\;\;}\Rightarrow {{R^2} = {\left( {\frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}{y’_0}} \right)^2} }+{ {\left( {\frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}} \right)^2},\;\;}\Rightarrow {{R^2} = {\left( {\frac{{1 + {{\left( {{y’_0}} \right)}^2}}}{{{y^{\prime\prime}_0}}}} \right)^2}\cdot\kern0pt{\left[ {{{\left( {{y’_0}} \right)}^2} + 1} \right],}\;\;}\Rightarrow {{R^2} = \frac{{{{\left[ {1 + {{\left( {{y’_0}} \right)}^2}} \right]}^3}}}{{{{\left( {{y^{\prime\prime}_0}} \right)}^2}}},\;\;}\Rightarrow {R = \frac{{{{\left[ {1 + {{\left( {{y’_0}} \right)}^2}} \right]}^{\large\frac{3}{2}\normalsize}}}}{{\left| {{y^{\prime\prime}_0}} \right|}}.}$

We see that 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 a problem to see the 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
$y = g\left( x \right) = \frac{a}{{x + b}},$
which osculates with the graph of the logarithmic function $$f\left( x \right) = \ln x + 1$$ at the point $${x_0} = 1.$$

### 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 $${x_0} = 0.$$

### 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:

$\left\{ \begin{array}{l} f\left( {{x_0}} \right) = g\left( {{x_0}} \right)\\ f’\left( {{x_0}} \right) = g’\left( {{x_0}} \right)\\ f^{\prime\prime}\left( {{x_0}} \right) = g^{\prime\prime}\left( {{x_0}} \right) \end{array} \right..$

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:

$\left\{ \begin{array}{l} {e^{{x_0}}} = ax_0^2 + b{x_0} + c\\ {e^{{x_0}}} = 2a{x_0} + b\\ {e^{{x_0}}} = 2a \end{array} \right..$

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

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

If we write its equation in the form

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

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 $$2.$$