Expand the function f(x, y) = (1 + 4x² + y²)^(1/2) in Taylor series up to the 2nd degree about (1, 2).
Understand the Problem
The question is asking to expand the function f(x, y) = (1 + 4x² + y²)^(1/2) in a Taylor series up to the 2nd degree around the point (1, 2). This involves finding the partial derivatives of the function at that point and constructing the Taylor series expression.
Answer
The Taylor series expansion is given by: \[ f(x,y) \approx 3 + \frac{4}{3}(x - 1) + \frac{2}{3}(y - 2) + \frac{2}{3}(x - 1)^2 + \frac{8}{9}(x - 1)(y - 2) + \frac{2}{9}(y - 2)^2 \]
Answer for screen readers
The Taylor series expansion of ( f(x, y) ) at ( (1, 2) ) to the second degree is: [ f(x,y) \approx 3 + \frac{4}{3}(x - 1) + \frac{2}{3}(y - 2) + \frac{2}{3}(x - 1)^2 + \frac{8}{9}(x - 1)(y - 2) + \frac{2}{9}(y - 2)^2 ]
Steps to Solve
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Define the function and point of expansion
We are given the function ( f(x, y) = (1 + 4x^2 + y^2)^{1/2} ) and we need to expand it around the point ( (1, 2) ). -
Calculate ( f(1, 2) )
Substituting ( x = 1 ) and ( y = 2 ): [ f(1, 2) = (1 + 4(1)^2 + (2)^2)^{1/2} = (1 + 4 + 4)^{1/2} = (9)^{1/2} = 3 ] -
Find the first partial derivatives ( f_x ) and ( f_y )
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For ( f_x ): [ f_x = \frac{\partial}{\partial x} (1 + 4x^2 + y^2)^{1/2} = \frac{4x}{(1 + 4x^2 + y^2)^{1/2}} ] Evaluating at ( (1, 2) ): [ f_x(1, 2) = \frac{4 \cdot 1}{(1 + 4 \cdot 1^2 + 2^2)^{1/2}} = \frac{4}{3} ]
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For ( f_y ): [ f_y = \frac{\partial}{\partial y} (1 + 4x^2 + y^2)^{1/2} = \frac{y}{(1 + 4x^2 + y^2)^{1/2}} ] Evaluating at ( (1, 2) ): [ f_y(1, 2) = \frac{2}{3} ]
- Calculate the second partial derivatives ( f_{xx}, f_{yy}, f_{xy} )
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For ( f_{xx} ): [ f_{xx} = \frac{\partial}{\partial x} f_x = \frac{4(1 + 4x^2 + y^2)^{1/2} - 4x \cdot \frac{4x}{(1 + 4x^2 + y^2)^{1/2}}}{(1 + 4x^2 + y^2)} = \frac{12}{9} = \frac{4}{3} ] Evaluating at ( (1, 2) ): [ f_{xx}(1, 2) = \frac{4}{3} ]
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For ( f_{yy} ): [ f_{yy} = \frac{\partial}{\partial y} f_y = \frac{(1 + 4x^2 + y^2)^{1/2} - \frac{y^2}{(1 + 4x^2 + y^2)^{1/2}}}{(1 + 4x^2 + y^2)} = \frac{4}{9} ] Evaluating at ( (1, 2) ): [ f_{yy}(1, 2) = \frac{4}{9} ]
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For ( f_{xy} ): [ f_{xy} = \frac{\partial}{\partial y} f_x = \frac{\frac{y}{(1 + 4x^2 + y^2)^{1/2}} \cdot 4}{(1 + 4x^2 + y^2)} = \frac{8}{9} ] Evaluating at ( (1, 2) ): [ f_{xy}(1, 2) = \frac{8}{9} ]
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Construct the Taylor series expansion
The Taylor series expansion of ( f(x, y) ) around ( (1, 2) ) up to the second order is: [ f(x,y) \approx f(1,2) + f_x(1,2) (x-1) + f_y(1,2) (y-2) + \frac{1}{2} f_{xx}(1,2) (x-1)^2 + f_{xy}(1,2) (x-1)(y-2) + \frac{1}{2} f_{yy}(1,2) (y-2)^2 ] Replacing variables: [ = 3 + \frac{4}{3}(x - 1) + \frac{2}{3}(y - 2) + \frac{2}{3}(x - 1)^2 + \frac{8}{9}(x - 1)(y - 2) + \frac{2}{9}(y - 2)^2 ]
The Taylor series expansion of ( f(x, y) ) at ( (1, 2) ) to the second degree is: [ f(x,y) \approx 3 + \frac{4}{3}(x - 1) + \frac{2}{3}(y - 2) + \frac{2}{3}(x - 1)^2 + \frac{8}{9}(x - 1)(y - 2) + \frac{2}{9}(y - 2)^2 ]
More Information
The Taylor series provides a way to approximate functions using polynomials. It's particularly useful in multivariable calculus for approximating functions near a point.
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