Question Number 228203 by Kassista last updated on 02/Apr/26
$${let}\:{f}\left({x},{y}\right)={x}^{\mathrm{2}} +{y} \\ $$$${find},\:{from}\:{first}\:{principles},\:{the}\:{directional}\:{derivative} \\ $$$${of}\:{f}\:{along}\:{the}\:{vector}\:\left(\mathrm{2},\mathrm{9}\right) \\ $$
Answered by TonyCWX last updated on 02/Apr/26
![Vector, a = ⟨2, 9⟩ Unit Vector, u = ⟨(2/( (√(85)))), (9/( (√(85))))⟩ _u f(x_0 , y_0 ) = lim_(h→0) [(((x + ((2h)/( (√(85)))))^2 + (y + ((9h)/( (√(85))))) − (x^2 + y))/h)] = lim_(h→0) [((x^2 + ((4xh)/( (√(85)))) + ((4h^2 )/(85)) + y + ((9h)/( (√(85)))) − x^2 − y)/h)] = lim_(h→0) [((((4xh)/( (√(85)))) + ((4h^2 )/(85)) + ((9h)/( (√(85)))))/h)] = lim_(h→0) [((4x)/( (√(85)))) + ((4h)/(85)) + (9/( (√(85))))] = ((4x)/( (√(85)))) + ((4(0))/(85)) + (9/( (√(85)))) = ((4x)/( (√(85)))) + (9/( (√(85))))](https://www.tinkutara.com/question/Q228205.png)
$$\mathrm{Vector},\:{a}\:=\:\langle\mathrm{2},\:\mathrm{9}\rangle \\ $$$$\mathrm{Unit}\:\mathrm{Vector},\:{u}\:=\:\langle\frac{\mathrm{2}}{\:\sqrt{\mathrm{85}}},\:\frac{\mathrm{9}}{\:\sqrt{\mathrm{85}}}\rangle \\ $$$$ \\ $$$$\:_{{u}} {f}\left({x}_{\mathrm{0}} ,\:{y}_{\mathrm{0}} \right)\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\left({x}\:+\:\frac{\mathrm{2}{h}}{\:\sqrt{\mathrm{85}}}\right)^{\mathrm{2}} \:+\:\left({y}\:+\:\frac{\mathrm{9}{h}}{\:\sqrt{\mathrm{85}}}\right)\:−\:\left({x}^{\mathrm{2}} \:+\:{y}\right)}{{h}}\right] \\ $$$$=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{{x}^{\mathrm{2}} \:+\:\frac{\mathrm{4}{xh}}{\:\sqrt{\mathrm{85}}}\:+\:\frac{\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{85}}\:+\:{y}\:+\:\frac{\mathrm{9}{h}}{\:\sqrt{\mathrm{85}}}\:−\:{x}^{\mathrm{2}} \:−\:{y}}{{h}}\right] \\ $$$$=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\frac{\mathrm{4}{xh}}{\:\sqrt{\mathrm{85}}}\:+\:\frac{\mathrm{4}{h}^{\mathrm{2}} }{\mathrm{85}}\:+\:\frac{\mathrm{9}{h}}{\:\sqrt{\mathrm{85}}}}{{h}}\right] \\ $$$$=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{\mathrm{4}{x}}{\:\sqrt{\mathrm{85}}}\:+\:\frac{\mathrm{4}{h}}{\mathrm{85}}\:+\:\frac{\mathrm{9}}{\:\sqrt{\mathrm{85}}}\right] \\ $$$$=\:\frac{\mathrm{4}{x}}{\:\sqrt{\mathrm{85}}}\:+\:\frac{\mathrm{4}\left(\mathrm{0}\right)}{\mathrm{85}}\:+\:\frac{\mathrm{9}}{\:\sqrt{\mathrm{85}}} \\ $$$$=\:\frac{\mathrm{4}{x}}{\:\sqrt{\mathrm{85}}}\:+\:\frac{\mathrm{9}}{\:\sqrt{\mathrm{85}}} \\ $$
Commented by Kassista last updated on 02/Apr/26
$${thanks}! \\ $$