Q 65. - Calculus

Question

Prove Theorem 12.33. That is, show that if $z=f(x, y), x=u(s, t)$, and $y=v(s, t)$, then, for all values of $s$ and  $t$ at which $u$ and $v$ are differentiable, and if $f$ is differentiable at $u(s, t), v(s, t))$, it follows that

$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$ and $\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$

Step-by-Step Solution

Answer

Use the definition of differentiability to solve the above equation.

Step 1: Given Information

It is given that

$z=f(x,y),x=u(s,t)\text{ and }y=v(s,t)$

$u,v$ are differentiable for all values of $s,t$

Step 2: Definition of differentiability

Let $z=f(x, y)$ be two variable function, $f(x, y)$ is said to be differentiable at $\left(x_0,y_0\right)$ if partial derivative of $f_x\left(x_0,y_0\right)\text{ and }{f}_y\left(x_0,y_0\right)$ both exists and

$\Delta z=f_x\left(x_0,y_0\right)\Delta x+f_y\left(x_0,y_0\right)\Delta y+{\epsilon }_1\Delta x+{\epsilon }_2\Delta y$

${\epsilon }_1\text{ and }{\epsilon }_2$ are functions of $\Delta x\text{ and }\Delta y$ as $(\Delta x,\Delta y)\to (0,0)$

Step 3: Applying the definition of differentiability

Since $z$ is differentiable

$\Delta z=z_x\Delta x+z_y\Delta y+{\epsilon }_1\Delta x+{\epsilon }_2\Delta y$

$=\left(z_x+{\epsilon }_1\right)\Delta x+\left(z_y+{\epsilon }_2\right)\Delta y$

Also $x,y$ are differentiable for all $s,t$

$\Delta x=x_s\Delta s+x_1\Delta t+{\epsilon }_3\Delta s+{\epsilon }_4\Delta t$

${\epsilon }_3\to 0,{\epsilon }_4\to 0\text{ as }(\Delta s,\Delta t)\to (0,0)$

Also

$\Delta y=y_s\Delta s+y_t\Delta t+{\epsilon }_5\Delta s+{\epsilon }_6\Delta t$

${\epsilon }_5\to 0,{\epsilon }_6\to 0\text{ as }(\Delta s,\Delta t)\to (0,0)$

Hence, $\Delta z=\left(z_x+{\epsilon }_1\right)\Delta x+\left(z_y+{\epsilon }_2\right)\Delta y$

$=\left(z_x+{\epsilon }_1\right)\left(x_s\Delta s+x_t\Delta t+{\epsilon }_3\Delta s+{\epsilon }_4\Delta t\right)+\left(z_y+{\epsilon }_2\right)\left(y_s\Delta s+y_t\Delta t+{\epsilon }_5\Delta s+{\epsilon }_6\Delta t\right)$

Solving, we get

$=\left(z_x{x}_s+z_y{y}_s\right)\Delta s+\left(z_x{x}_t+z_y{y}_t\right)\Delta t+{\epsilon }_7\Delta s+{\epsilon }_8\Delta t$

where ${\epsilon }_7=z_x{\epsilon }_3+z_y{\epsilon }_5+{\epsilon }_1{x}_s+{\epsilon }_2{y}_s+{\epsilon }_1{\epsilon }_3+{\epsilon }_2{\epsilon }_5$

$=\left(z_x+{\epsilon }_1\right){\epsilon }_3+{\epsilon }_1{x}_s+{\epsilon }_2{y}_s+\left(z_y+{\epsilon }_2\right){\epsilon }_5$

and

${\epsilon }_8=z_x{\epsilon }_4+z_y{\epsilon }_6+{\epsilon }_1{x}_t+{\epsilon }_2{y}_t+{\epsilon }_1{\epsilon }_4+{\epsilon }_2{\epsilon }_6$

$=\left(z_x+{\epsilon }_1\right){\epsilon }_4+{\epsilon }_1{x}_t+{\epsilon }_2{y}_t+\left(z_y+{\epsilon }_2\right){\epsilon }_6$

Step 4: Solving for ∂ z ∂ s

If $(\Delta s,\Delta t)\to (0,0)$, ${\epsilon }_7\to 0\text{ and }{\epsilon }_8\to 0$

Solving further,

$\frac{\Delta z}{\Delta s}=\frac{1}{\Delta s}\left\{\left(z_x{x}_s+z_y{y}_s\right)\Delta s+\left(z_x{x}_t+z_y{y}_t\right)·0+{\epsilon }_7\Delta s+{\epsilon }_8·0\right\}$

$=z_x{x}_s+z_y{y}_s+{\epsilon }_7$

Taking limit on both sides

$\underset{\Delta x\to 0}{\lim }\frac{\Delta z}{\Delta s}=\underset{\Delta x\to 0}{\lim }\left(z_x{x}_s+z_y{y}_s+{\epsilon }_7\right)$

$=z_x{x}_s+z_y{y}_s$

$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$

Step 5: Solving for lim Δ y → 0 Δ z Δ t

Taking partial differentiation of $z$ wrt $y$, 

$\frac{\Delta z}{\Delta t}=\frac{1}{\Delta t}\left\{\left(z_x{x}_s+z_y{y}_s\right)·0+\left(z_x{x}_t+z_y{y}_t\right)·\Delta t+{\epsilon }_7·0+{\epsilon }_8·\Delta t\right\}$

$=z_x{x}_t+z_y{y}_t+{\epsilon }_8$

Taking limit on both sides

$\underset{\Delta y\to 0}{\lim }\frac{\Delta z}{\Delta t}=\underset{\Delta y\to 0}{\lim }\left(z_x{x}_t+z_y{y}_t+{\epsilon }_8\right)$

$=z_x{x}_t+z_y{y}_t$

$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$

Hence proved.