Chapter 9 - Multiveriable Calculus - 9.5 Total Differentials and Approximations - 9.5 Exercises - Page 503: 11

To approximate $$1.03 e^{0.04}.$$ Notice that $1.03\approx 1$ and $0.04 \approx 0$ and we know that $1.03 e^{0.04} \approx 1.0720$. We, therefore, let $$z=f(x, y)=x e^{y}$$ we let $x=1$, $d x=0.03, y=0,$ and $d y=0.04$ Then $$ \begin{aligned} d z &=f_{x}(x, y) d x+f_{y}(x, y) d y \\ &=e^{y} d x+x e^{y} d y \\ &=e^{0}(0.03)+1 \cdot e^{0}(0.04) \\ &=0.07, \\ f(1.03,0.04) &=f(1,0)+\Delta z \\ & \approx f(1,0)+d z \\ &=1 \cdot e^{0}+0.07 \\ &=1.07 \\ \text { Thus, } 1.03 e^{0.04} & \approx 1.07 \\ \text { Using a calculator, } & 1.03 e^{0.04} \approx 1.0720 . \end{aligned} $$ The error is approximately $0.0020.$

To approximate $$1.03 e^{0.04}.$$ Notice that $1.03\approx 1$ and $0.04 \approx 0$ and we know that $1.03 e^{0.04} \approx 1.0720$. We, therefore, let $$z=f(x, y)=x e^{y}$$ we let $x=1$, $d x=0.03, y=0,$ and $d y=0.04$ Then $$ \begin{aligned} d z &=f_{x}(x, y) d x+f_{y}(x, y) d y \\ &=e^{y} d x+x e^{y} d y \\ &=e^{0}(0.03)+1 \cdot e^{0}(0.04) \\ &=0.07 \\ f(1.03,0.04) &=f(1,0)+\Delta z \\ & \approx f(1,0)+d z \\ &=1 \cdot e^{0}+0.07 \\ &=1.07 \\ \text { Thus, } 1.03 e^{0.04} & \approx 1.07 \\ \text { Using a calculator, } & 1.03 e^{0.04} \approx 1.0720 . \end{aligned} $$ The error is approximately $0.0020.$