Answer
To approximate
$$
2.03 \ln 1.02.$$
Notice that $2.03\approx 2 $ and $1.02 \approx 2 $ and we know that $2.03 \ln 1.02 \approx 0.0402$.
We, therefore, let
$$z=f(x, y)=x \ln y$$
Then
$$
\begin{aligned}
d z &=f_{x}(x, y) d x+f_{y}(x, y) d y \\
&=\ln y d x+\frac{x}{y} d y
\end{aligned}
$$
To approximate $2.03 \ln 1.02,$ we let $x=2$, $d x=0.03, y=1,$ and $d y=0.02$, so we have:
$$
\begin{aligned}
d z &=\ln (1) \cdot(0.03)+\frac{2}{1}(0.02) \\
&=0.04, \\
f(2.03,1.02) &=f(2,1)+\Delta z \\
&\approx f(2,1)+d z \\
&=2 \cdot \ln (1)+0.04 \\
&=0.04
\end{aligned}
$$
Thus, $2.03 \ln 1.02 \approx 0.04$.
Using a calculator, $2.03 \ln 1.02 \approx 0.0402$
The error is approximately $0.0002$.
Work Step by Step
To approximate
$$
2.03 \ln 1.02.$$
Notice that $2.03\approx 2 $ and $1.02 \approx 2 $ and we know that $2.03 \ln 1.02 \approx 0.0402$.
We, therefore, let
$$z=f(x, y)=x \ln y$$
Then
$$
\begin{aligned}
d z &=f_{x}(x, y) d x+f_{y}(x, y) d y \\
&=\ln y d x+\frac{x}{y} d y
\end{aligned}
$$
To approximate $2.03 \ln 1.02,$ we let $x=2$, $d x=0.03, y=1,$ and $d y=0.02$, so we have:
$$
\begin{aligned}
d z &=\ln (1) \cdot(0.03)+\frac{2}{1}(0.02) \\
&=0.04, \\
f(2.03,1.02) &=f(2,1)+\Delta z \\
&\approx f(2,1)+d z \\
&=2 \cdot \ln (1)+0.04 \\
&=0.04
\end{aligned}
$$
Thus, $2.03 \ln 1.02 \approx 0.04$.
Using a calculator, $2.03 \ln 1.02 \approx 0.0402$
The error is approximately $0.0002$.
