Question

Seja f uma função diferenciável em P e $\vec{u} \ = \ \frac{\sqrt{3}}{2} \ \vec{i} \ + \ \frac{1}{2} \ \vec{j}$

e  $\vec{v} \ = \ \frac{1}{2} \ \vec{i} \ + \ \frac{\sqrt{3}}{2} \ \vec{j}$
duas direções, tal que

$\frac{\partial f}{\partial \vec{u}} \ (P) \ = \ \sqrt{3}$


$\frac{\partial f}{\partial \vec{v}} \ (P) \ = \ - 2 \sqrt{3}$

Letra a) $\bigtriangledown f \ (P)$

Letra b) $\frac{\partial f}{\partial \vec{w}} \ (P) \ = \ ?$ na direção $\vec{i} \ + \ \vec{j}$

Answer 1

Temos uma função f diferenciável no ponto P e dois vetores direções

     $\mathsf{\overset{\to}{u}=\dfrac{\sqrt{3}}{2}\,\overset{\to}{i}+\dfrac{1}{2}\,\overset{\to}{j}=\Big(\dfrac{\sqrt{3}}{2},\,\dfrac{1}{2}\Big)}$

     $\mathsf{\overset{\to}{v}=\dfrac{1}{2}\,\overset{\to}{i}+\dfrac{\sqrt{3}}{2}\,\overset{\to}{j}=\Big(\dfrac{1}{2},\,\dfrac{\sqrt{3}}{2}\Big)}$


de modo que as derivadas direcionais de f no ponto P nas direções acima são

     $\mathsf{\dfrac{\partial f}{\partial \overset{\to}{u}}(P)=\sqrt{3}}$

     $\mathsf{\dfrac{\partial f}{\partial \overset{\to}{v}}(P)=-2\sqrt{3}}$


Observe que os vetores dados já são unitários, ou seja,

     $\mathsf{\|\overset{\to}{u}\|=\Big\|\Big(\dfrac{\sqrt{3}}{2},\,\dfrac{1}{2}\Big)\Big\|=1}\\\\\\ \mathsf{\|\overset{\to}{v}\|=\Big\|\Big(\dfrac{1}{2},\,\dfrac{\sqrt{3}}{2}\Big)\Big\|=1}$


Isso facilitará os nossos cálculos posteriormente.


a) Calcular o vetor $\mathsf{\nabla f(P).}$

Suponha que o vetor gradiente de f em P seja

     $\mathsf{\nabla f(P)=a\overset{\to}{i}+b\overset{\to}{j}=(a,\,b).}$


Como f é diferenciável em P, temos que

     $\mathsf{\dfrac{\partial f}{\partial \overset{\to}{u}}(P)=\nabla f(P)\cdot \dfrac{\overset{\to}{u}}{\|\overset{\to}{u}\|}}\\\\\\ \mathsf{\sqrt{3}=(a,\,b)\cdot \dfrac{(\frac{\sqrt{3}}{2},\,\frac{1}{2})}{1}}\\\\\\ \mathsf{\sqrt{3}=a\cdot \dfrac{\sqrt{3}}{2}+b\cdot \dfrac{1}{2}}\\\\\\ \mathsf{\sqrt{3}\,a+b=2\sqrt{3}\qquad\quad(i)}$


     $\mathsf{\dfrac{\partial f}{\partial \overset{\to}{v}}(P)=\nabla f(P)\cdot \dfrac{\overset{\to}{v}}{\|\overset{\to}{v}\|}}\\\\\\ \mathsf{-2\sqrt{3}=(a,\,b)\cdot \dfrac{(\frac{1}{2},\,\frac{\sqrt{3}}{2})}{1}}\\\\\\ \mathsf{-2\sqrt{3}=a\cdot \dfrac{1}{2}+b\cdot \dfrac{\sqrt{3}}{2}}\\\\\\ \mathsf{a+\sqrt{3}\,b=-4\sqrt{3}\qquad\quad(ii)}$


Resolva o sistema formado pelas equações (i) e (ii).

     $\left\{\begin{array}{rcrcrc} \mathsf{\sqrt{3}\,a}&\!\!\!+\!\!\!&\mathsf{b}&\!\!\!=\!\!\!&\mathsf{2\sqrt{3}}&\qquad\mathsf{(i)}\\ \mathsf{a}&\!\!\!+\!\!\!&\mathsf{\sqrt{3}\,b}&\!\!\!=\!\!\!&\mathsf{-4\sqrt{3}}&\qquad\mathsf{(ii)} \end{array}\right.$


Multiplique a equação (i) por −√3, e depois some com a equação (ii):

     $\mathsf{-\sqrt{3}\cdot (\sqrt{3}\,a+b)+a+\sqrt{3}\,b=-\sqrt{3}\cdot (2\sqrt{3})-4\sqrt{3}}\\\\ \mathsf{-3a-\sqrt{3}\,b+a+\sqrt{3}\,b=-6-4\sqrt{3}}\\\\ \mathsf{-2a=-6-4\sqrt{3}}\\\\ \mathsf{a=\dfrac{-6-4\sqrt{3}}{-2}}\\\\\\ \mathsf{a=3+2\sqrt{3}}$


Substitua na equação (i) para encontrar o valor de b:

     $\mathsf{\sqrt{3}\cdot (3+2\sqrt{3})+b=2\sqrt{3}}\\\\ \mathsf{3\sqrt{3}+6+b=2\sqrt{3}}\\\\ \mathsf{b=2\sqrt{3}-3\sqrt{3}-6}\\\\ \mathsf{b=-6-\sqrt{3}}$


Logo, o vetor gradiente de f em P é

     $\mathsf{\nabla f(P)=(3+2\sqrt{3})\overset{\to}{i}+(-6-\sqrt{3})\overset{\to}{j}=(3+2\sqrt{3},\,-6-\sqrt{3}).}$


b) Calcular a derivada direcional de f em P na direção do vetor

     $\mathsf{\overset{\to}{w}=\overset{\to}{i}+\overset{\to}{j}=(1,\,1).}$


Como f é diferenciável em P, podemos novamente usar o cálculo do produto escalar:

     $\mathsf{\dfrac{\partial f}{\partial \overset{\to}{w}}(P)=\nabla f(P)\cdot \dfrac{\overset{\to}{w}}{\|\overset{\to}{w}\|}}\\\\\\ \mathsf{\dfrac{\partial f}{\partial \overset{\to}{w}}(P)=(3+2\sqrt{3},\,-6-\sqrt{3})\cdot \dfrac{(1,\,1)}{\|(1,\,1)\|}}\\\\\\ \mathsf{\dfrac{\partial f}{\partial \overset{\to}{w}}(P)=(3+2\sqrt{3},\,-6-\sqrt{3})\cdot \dfrac{(1,\,1)}{\sqrt{1^2+1^2}}}\\\\\\ \mathsf{\dfrac{\partial f}{\partial \overset{\to}{w}}(P)=\dfrac{(3+2\sqrt{3},\,-6-\sqrt{3})\cdot (1,\,1)}{\sqrt{2}}}\\\\\\ \mathsf{\dfrac{\partial f}{\partial \overset{\to}{w}}(P)=\dfrac{(3+2\sqrt{3})\cdot 1+(-6-\sqrt{3})\cdot 1}{\sqrt{2}}}\\\\\\ \mathsf{\dfrac{\partial f}{\partial \overset{\to}{w}}(P)=\dfrac{3+2\sqrt{3}-6-\sqrt{3}}{\sqrt{2}}}\\\\\\ \mathsf{\dfrac{\partial f}{\partial \overset{\to}{w}}(P)=\dfrac{-3+\sqrt{3}}{\sqrt{2}}}$


Dúvidas? Comente.


Bons estudos! :-)