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 \(-\frac{π^3}{8} \frac{(x-1)^3}{3!}.0 = 0.\)
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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.

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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