Question

Eliminate arbitrary constants from z=(x-a)²+(y-b)² to form the partial differential equation.

Answer 1

$$z=(x-a)^2+(y-b)^2$$

Differentiating partially with respect to $x$ and $y,$ we get

$$\dfrac{\partial z}{\partial x}=2(x-a)$$

$$\dfrac{\partial z}{\partial y}=2(y-b)$$

Squaring and adding these equations, we have

$$(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2=(2(x-a))^2+(2(y-b))^2$$

$$(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2=4((x-a)^2+(y-b)^2)$$

Since $(x-a)^2+(y-b)^2=z,$ we get

$$(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2=4z$$