Question Number 188017 by Rupesh123 last updated on 24/Feb/23
Answered by aleks041103 last updated on 24/Feb/23
$${f}\left({x},{y}\right)={ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{2}{ln}\left({r}\right) \\ $$$${r}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$\Rightarrow{f}\left({x},{y}\right)={f}\left({r}\right) \\ $$$$\Rightarrow\bigtriangledown{f}=\frac{\mathrm{2}}{{r}}\overset{\rightarrow} {{e}_{{r}} } \\ $$$$\Rightarrow{max}\:{rate}\:{of}\:{change}\:=\:\frac{\mathrm{2}}{{r}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{4}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} }}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${direction}\:=\:\overset{\rightarrow} {{e}_{{r}} }=\left(\frac{\mathrm{4}}{\mathrm{5}},\frac{−\mathrm{3}}{\mathrm{5}}\right) \\ $$$$\Rightarrow{Ans}: \\ $$$${Max}\:{rate}\:{of}\:{change}\:=\:\mathrm{0}.\mathrm{4}\:=\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${Direction}\:=\:\left(\mathrm{0}.\mathrm{8},−\mathrm{0}.\mathrm{6}\right)=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{4},−\mathrm{3}\right) \\ $$
Commented by Rupesh123 last updated on 24/Feb/23
Perfect ��