Chapter 4.5, Problem 8P

Prove the statement just after (5.2), that $dz$ given ny (3.6) is a good approximation to to $\Delta z$ even though dx and dy are not independent. Hint: let x and y be function of t; then (5.2) is correct, but $\Delta x\ne dx$ and $\Delta y\ne dy$ (because $x$ and $y$ are not independent variables). However, $\Delta x/\Delta t$ is nearly $dx/dt$ for small dt and $dt=\Delta t$ since $t$ is the independent variable. You can then show that $\Delta x=\left(\frac{dx}{dt}+{ϵ}_x\right)dt=dx+{ϵ}_{x}dt$, and a similar formula for $\Delta y,$ and get $\Delta z=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy+\left(\text{ terms containing }{ϵ}^{\prime }s\right)\cdot dt=dz+ϵdt,$ where $ϵ\to 0$ as $\Delta t\to 0$.