Answered: Write the equation of a circle…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Write the equation of a circle osculating with the curve f(x)=arctanx at the point x₀=1.

Expert Solution

Step 1

The curve $f (x) = a r c \tan x$ .

We have to find the equation of an osculating circle at the point $x_{0} = 1$ .

Step 2

Formula used:

Let $f (x)$ be any curve, the equation of an osculating circle at the point $(x_{0}, y_{0})$ is ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ .

Where,

$h = x_{0} - \frac{1 + {(y' (x_{0}))}^{2}}{y'' (x_{0})} . y' (x_{0}) k = y_{0} + \frac{1 + {(y' (x_{0}))}^{2}}{y'' (x_{0})} r = \frac{1}{|κ|}, κ = \frac{y'' (x_{0})}{{[1 + {(y' (x_{0}))}^{2}]}^{\frac{3}{2}}} .$

Step 3

$f (x) = a r c \tan x$

The point $x_{0} = 1$ .

$\begin{array}{rcl} y_{0} & = & f (x_{0}) \\ = & f (1) \\ = & a r c \tan (1) \\ y_{0} & = & \frac{π}{4} \end{array}$

Differentiate $y = f (x) = a r c \tan x$ with respect to $x$ ,

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

$y' (1) = \frac{1}{1 + 1^{2}} = \frac{1}{2}$

Again differentiate $y' = \frac{1}{1 + x^{2}}$ with respect to $x$ ,

$\begin{array}{rcl} y'' & = & - \frac{2 x}{{(1 + x^{2})}^{2}} \\ y'' (1) & = & - \frac{1}{2} . \end{array}$

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