Answer
\begin{array}{}
a & L(x, y)=1+0.5(x-1)+0.3\left(x_{0}, y_{0}\right)(y-1) \\
b & \text { Magnitude of approximation error with the distance b/w } \mathrm{P} \text { and } \mathrm{Q}=0.01073
\end{array}
Work Step by Step
\[
f(x, y)=x^{0.5} y^{0.3} \quad P \equiv(1,1), Q \equiv(1.05,0.97)
\]
The linear approximation theorem is
\[
L(x, y)=f\left(x_{0}, y_{0}\right)+f_{x}\left(x_{0}, y_{0}\right)\left(x-x_{0}\right)+f_{y}\left(x_{0}, y_{0}\right)\left(y-y_{0}\right)
\]
Now obtaining $\left.f_{(} x\right)$ and $f_{y}$
\[
\begin{aligned}
&\frac{\partial f}{\partial x} =f_{x} \\
&\frac{\partial}{\partial x}\left(x^{0.5} y^{0.3}\right)=f_{x} \\
&y^{0.3} \frac{\partial}{\partial x}\left(x^{0.5}\right) =f_{x}\\
&0.5 x^{-0.5} y^{0.3}=f_{x} \\
&0.5 x^{-0.5} y^{0.3}=0.5=f_{x}(1,1)
\end{aligned}
\]
Similarly $f_{y}$
\[
\begin{aligned}
&\frac{\partial f}{\partial y}=f_{y} \\
&\frac{\partial}{\partial y}\left(x^{0.5} y^{0.3}\right) =f_{v}\\
&x^{0.5} \frac{\partial}{\partial y}\left(y^{0.3}\right)=f_{y} \\
&0.3 x^{0.5} y^{-0.7}=0.3=f_{y}(1,1)
\end{aligned}
\]
$\because f(1,1)=1$, so substituting values of $f(1,1), f_{x}, f_{y}$ in the equation
\[
L(x, y)=1+0.5(x-1)+0.3(y-1)
\]
b) Here, $\Delta x=0.05$ and $\Delta y=-0.03$:
\[
\begin{aligned}
\text { Distance }=&|P Q|=\sqrt{(x+\Delta x-x)^{2}+(y+\Delta y-y)^{2}} \\
&|P Q|=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}} \\
&|P Q|=\sqrt{(0.05)^{2}+(-0.03)^{2}} \\
&|P Q|=0.05831
\end{aligned}
\]
Obtaining the approximated value at Point $\mathrm{Q}$
\[
\begin{array}{l}
L(1.05,0.97)=1+0.5(1.05-1)+0.3(0.97-1) \\
L(1.05,0.97)=1+0.5(1.05-1)+0.3(0.97-1) \\
L(1.05,0.97)=1.016
\end{array}
\]
Error is defined by $\mathrm{E}$
\[
E=|L(1.05,0.97)-f(1.05,0.97)|
\]
\[
E=1.016-1.01537=0.0006257
\]
The change in error with respect to distance between $\mathrm{P}$ and $\mathrm{Q}$ is
\[
\begin{aligned}
\frac{E}{|P Q|} &=\frac{0.0006257}{0.05831} \\
\frac{E}{|P Q|} &=0.01073
\end{aligned}
\]
