Question

Expansion of  \(f (x,y) = tan^{-1} \frac{⁡y}{x}\) upto first degree containing (x+1) & (y-1) is __________

(a) \(\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!}  \frac{-1}{2}  + \frac{(y-1)^2}{2!}  \frac{1}{2}\)

(b) \(\frac{π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!}  \frac{1}{4} +  \frac{(y-1)^2}{2!} \frac{1}{4}\)

(c) \(\frac{5π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!}   \frac{-1}{4} + \frac{(y-1)^2}{2!}  \frac{1}{4}\)

(d) \(\frac{3π}{4}  + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!}   \frac{-1}{4}  + \frac{(y-1)^2}{2!} \frac{1}{4}\)

The question was asked during an internship interview.

My question is based upon Taylor’s Theorem Two Variables topic in division Maxima and Minima of Engineering Mathematics

Answer 1

The correct answer is (a) \(\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!}  \frac{-1}{2}  + \frac{(y-1)^2}{2!}  \frac{1}{2}\)

The best explanation: We can expand the given function according to Taylor’s theorem

\(f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\

+2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)\)

 Given a=-1 & b=1, f(-1,1)=tan^-1⁡-1 = \(\frac{3π}{4}\)

\(f_x = \frac{-y}{x^2+y^2} \,at\, (-1,1) = \frac{-1}{2}\)

\(f_y = \frac{x}{x^2+y^2} \,at\, (-1,1) = \frac{-1}{2}\)

\(f_{xy} = \frac{(x^2+y^2)-2x^2}{(x^2+y^2)^2}\)  at (-1,1)=0

\(f_{xx} = \frac{2yx}{(x^2+y^2)^2} \,at\, (-1,1)=\frac{-2}{4} = \frac{-1}{2}\)

\(f_{yy} = \frac{-2yx}{(x^2+y^2)^2} \,at\, (-1,1)=\frac{2}{4} = \frac{1}{2}\) thus the series is given by

\(\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!}  \frac{-1}{2} + \frac{(x+1)^2}{2!}  \frac{-1}{2}  + \frac{(y-1)^2}{2!}  \frac{1}{2}\).