Answer
The results:
$\begin{array}{*{20}{c}}
{Method}&{f\left( {2.01,1.02} \right)}&{f\left( {1.97,1.01} \right)}\\
{Linearization}&{4.28}&4\\
{Calculator}&{4.28739}&{3.9985}
\end{array}$
Work Step by Step
The linearization $L\left( {x,y} \right)$ of $f\left( {x,y} \right) = {x^2}{y^3}$ is given by
$L\left( {x,y} \right) = f\left( {a,b} \right) + {f_x}\left( {a,b} \right)\left( {x - a} \right) + {f_y}\left( {a,b} \right)\left( {y - b} \right)$
At $\left( {a,b} \right) = \left( {2,1} \right)$, we get
$L\left( {x,y} \right) = f\left( {2,1} \right) + {f_x}\left( {2,1} \right)\left( {x - 2} \right) + {f_y}\left( {2,1} \right)\left( {y - 1} \right)$
$L\left( {x,y} \right) = 4 + 4\left( {x - 2} \right) + 12\left( {y - 1} \right)$
$L\left( {x,y} \right) = 4x + 12y - 16$
1. Estimate $f\left( {2.01,1.02} \right)$
For $\Delta x = 0.01$ and $\Delta y = 0.02$, the desired estimate is
$L\left( {x,y} \right) = 4x + 12y - 16$
$f\left( {2.01,1.02} \right) \approx L\left( {2.01,1.02} \right) = 4\cdot2.01 + 12\cdot1.02 - 16$
$f\left( {2.01,1.02} \right) \approx 4.28$
Using a calculator: $f\left( {2.01,1.02} \right) \approx 4.28739$
2. Estimate $f\left( {1.97,1.01} \right)$
The desired estimate is
$L\left( {x,y} \right) = 4x + 12y - 16$
$f\left( {1.97,1.01} \right) \approx L\left( {1.97,1.01} \right) = 4\cdot1.97 + 12\cdot1.01 - 16$
$f\left( {1.97,1.01} \right) \approx 4$
Using a calculator: $f\left( {1.97,1.01} \right) \approx 3.9985$
In summary:
$\begin{array}{*{20}{c}}
{Method}&{f\left( {2.01,1.02} \right)}&{f\left( {1.97,1.01} \right)}\\
{Linearization}&{4.28}&4\\
{Calculator}&{4.28739}&{3.9985}
\end{array}$
