Differential and Integral Calculus Questions and Answers – Taylor’s Theorem Two Variables

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This set of Differential and Integral Calculus Multiple Choice Questions & Answers (MCQs) focuses on “Taylor’s Theorem Two Variables”.

1. Among the following which is the correct expression for Taylor’s theorem in two variables for the function f (x, y) near (a, b) where h=x-a & k=y-b upto second degree?
a) \(f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\
+2 \frac{(x-a)(y-b)}{4!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)\)
b) \(f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\
+ \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)\)
c) \(f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\
+ \frac{(y-b)^2}{2!} f_{yy}(a, b)\)
d) \(f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\
+2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)\)
View Answer

Answer: d
Explanation: By definition
\(f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\
+2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)\)
here we can observe that second degree is of the form (p+q)2 similarly Taylor’s theorem is expanded to third degree which is of the form (p+q)3 & f (a+ h, b+ k) = f (x, y)
where\((f_x=\frac{∂f (x,y)}{∂x}, f_y=\frac{∂f (x,y)}{∂y}, f_{xx}=\frac{∂}{∂x}(\frac{∂f(x,y)}{∂x}), f_{yy}=\frac{∂}{∂y} (\frac{∂f (x,y)}{∂y}), \\
f_{xy}=\frac{∂}{∂x}(\frac{∂f (x,y)}{∂y})).\)
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2. Given f (x,y)=ex cos⁡y, what is the value of the fifth term in Taylor’s series near (1,\(\frac{π}{4}\)) where it is expanded in increasing order of degree & by following algebraic identity rule?
a) \(\frac{-e(x-1)(y-\frac{π}{4})}{\sqrt{2}}\)
b) \(-\sqrt{2} e(x-1)(y-\frac{π}{4})\)
c) \(\frac{e(x-1)^2}{\sqrt{2}}\)
d) \(\frac{e(y-\frac{π}{4})^2}{\sqrt{2}}\)
View Answer

Answer: a
Explanation: Taylor’s series expansion is given by
\(f (a+ h, b+ k) = f (a, b) + \frac{x-a}{1!} f_x(a, b) + \frac{y-b}{1!} f_y(a, b) + \frac{(x-a)^2}{2!} f_{xx}(a, b)\\
+2 \frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + \frac{(y-b)^2}{2!} f_{yy}(a, b)\)
Thus fifth term is given by \(2 \frac{(x-a)(x-b)}{2!} f_{xy} (a,b)\)..(1) where a=1, b=π/4 & \(f_{xy}=\frac{∂}{∂x}(\frac{∂f(x,y)}{∂x}) = \frac{∂}{∂x}(\frac{∂e^x cos⁡y}{∂y}) =- e^x \,sin⁡y \) at (1, \(\frac{π}{4}), f_{xy}=\frac{-e}{\sqrt{2}}\) substituting in (1)
We get fifth term as \(2 \frac{(x-1)(x-π/4)}{2!} \frac{-e}{\sqrt{2}} = \frac{-e(x-1)(y-\frac{π}{4})}{\sqrt{2}}\).

3. Given f (x,y)=sin⁡xy, what is the value of the third degree first term in Taylor’s series near (1,-\(\frac{π}{2}\)) where it is expanded in increasing order of degree & by following algebraic identity rule?
a) \(\frac{π^3}{8}\)
b) \(\frac{π^3}{8} \frac{(x-1)(y+\frac{π}{2})}{3!}\)
c) 0
d) \(-\frac{π^3}{8} \frac{(x-1)^3}{3!}\)
View Answer

Answer: c
Explanation: Third degree first term in Taylor’s series is given by \(\frac{(x-a)^3 f_{xxx} (x,y)}{3!}\) Where a=1 \(b=-\frac{π}{2}, f_{xxx} (x,y)=\frac{∂^3 f(x,y)}{∂x^3} \,i.e\, \frac{∂^3 sin⁡xy}{∂x^3} = -y^3 cos⁡xy\)…… (partial differentiating f (x,y) w.r.t x only)
at \(a=1, b=-\frac{π}{2}, \frac{∂^3 sin⁡xy}{∂x^3} = -\frac{π^3 cos-\frac{⁡π}{2}}{8}=0\) hence third degree first term is given by \(-\frac{π^3}{8} \frac{(x-1)^3}{3!}.0 = 0.\)

4. Taylor’s theorem is mainly used in expressing the function as sum with infinite terms.
a) True
b) False
View Answer

Answer: a
Explanation: Taylor’s theorem helps in expanding a function into infinite terms however, it can be applied to functions that can be expressed finitely.
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5. 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) \(\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}\)
View Answer

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn