Question Number 99923 by bemath last updated on 24/Jun/20
$$\mathrm{Eliminate}\:\mathrm{arbitrary}\:\mathrm{constant}\: \\ $$$${a}\:\mathrm{and}\:{b}\:\mathrm{from}\:\mathrm{z}\:=\:\left(\mathrm{x}−{a}\right)^{\mathrm{2}} +\left(\mathrm{y}−{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{to}\:\mathrm{form}\:\mathrm{the}\:\mathrm{partial}\:\mathrm{differential} \\ $$$$\mathrm{equation}.\: \\ $$
Commented by bobhans last updated on 24/Jun/20
$$\frac{\partial\mathrm{z}}{\partial\mathrm{x}}\:=\:\mathrm{2}\left({x}−{a}\right)\:;\:\frac{\partial\mathrm{z}}{\partial\mathrm{y}}\:=\:\mathrm{2}\left(\mathrm{y}−{b}\right) \\ $$$$\left(\frac{\partial\mathrm{z}}{\partial\mathrm{x}}\right)^{\mathrm{2}} +\left(\frac{\partial\mathrm{z}}{\partial\mathrm{y}}\right)^{\mathrm{2}} \:=\:\mathrm{4}\left\{\left({x}−{a}\right)^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} \right\} \\ $$$$\left(\frac{\partial\mathrm{z}}{\partial\mathrm{x}}\right)^{\mathrm{2}} +\left(\frac{\partial\mathrm{z}}{\partial\mathrm{y}}\right)^{\mathrm{2}} =\:\mathrm{4z}\: \\ $$
Commented by bemath last updated on 24/Jun/20
$${thank}\:{you}\: \\ $$