Tangent Planes & Linear Approximation - Multivariable Calculus

1. The function $g(x,y) = \frac{x}{y} + \frac{y}{x}$ is approximated linearly near $(2,1)$ to estimate $g(1.98, 1.01)$. What is the error in this linear approximation compared to the actual value?
   Answer: The error is approximately $0.001$ First, compute the linear approximation. $g(2,1) = \frac{2}{1} + \frac{1}{2} = 2.5$. The partial derivatives are $g_x = \frac{1}{y} - \frac{y}{x^2}$ and $g_y = -\frac{x}{y^2} + \frac{1}{x}$. At $(2,1)$: $g_x(2,1) = 1 - \frac{1}{4} = 0.75$ and $g_y(2,1) = -2 + 0.5 = -1.5$. The linear approximation is $L(1.98,1.01) = 2.5 + 0.75(-0.02) + (-1.5)(0.01) = 2.5 - 0.015 - 0.015 = 2.47$. Now compute the actual value: $g(1.98,1.01) = \frac{1.98}{1.01} + \frac{1.01}{1.98} \approx 1.9604 + 0.5101 = 2.4705$. The error is $2.4705 - 2.47 = 0.0005 \approx 0.001$.
2. Consider the function $w = f(x,y) = \ln(x^2 + y^2)$. If the linear approximation at point $(3,4)$ is used to estimate $f(3.05, 3.95)$, what is the approximated value?
   Answer: $\ln(25) - 0.004$ The linear approximation is $L(x,y) = f(3,4) + f_x(3,4)(x-3) + f_y(3,4)(y-4)$. First, $f(3,4) = \ln(9+16) = \ln(25)$. Next, $f_x = \frac{2x}{x^2+y^2}$, so $f_x(3,4) = \frac{6}{25}$. Similarly, $f_y = \frac{2y}{x^2+y^2}$, so $f_y(3,4) = \frac{8}{25}$. For the point $(3.05, 3.95)$, $\Delta x = 0.05$ and $\Delta y = -0.05$. Therefore, $L(3.05,3.95) = \ln(25) + \frac{6}{25}(0.05) + \frac{8}{25}(-0.05) = \ln(25) + \frac{6 \cdot 0.05 - 8 \cdot 0.05}{25} = \ln(25) + \frac{-0.1}{25} = \ln(25) - 0.004$.

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