Science:Math Exam Resources/Courses/MATH200/April 2012/Question 02 (a)

MATH200 April 2012

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Question 02 (a)
Let $\displaystyle {}z=f(x,y)=\ln(4x^{2}+y^{2})$ Use a linear approximation of the function z=f(x,y) at (0,1) to estimate f(0.1,1.2).

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Hint
The linear approximation to a multivariable function ƒ(x,y) at a point (x,y) = (a,b) is given by L(x,y) where $\displaystyle L(x,y)=f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ where the subscripts denote partial differentiation with respect to the given variable.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The linear approximation to a multivariable function ƒ(x,y) at a point (x,y) = (a,b) is given by L(x,y) where $\displaystyle L(x,y)=f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$ where the subscripts denote partial differentiation with respect to the given variable. In this case the partial derivatives of $f(x,y)=\ln(4x^{2}+y^{2})$ are $f_{x}(x,y)={\frac {8x}{4x^{2}+y^{2}}},\quad f_{y}(x,y)={\frac {2y}{4x^{2}+y^{2}}}.$ and we wish to approximate ƒ near (x,y) = (0,1). Hence the linear approximation we seek is: ${\begin{aligned}\displaystyle L(x,y)&=\ln(1)+0\cdot (x-0)+2(y-1)\\&=2y-2\end{aligned}}$ Using the linear approximation, we can estimate the value of ƒ(0.1,1.2) as ${\color {blue}f(0.1,1.2)\approx L(0.1,1.2)=0.4}.$

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