Question

Gradiente da função f(x,y) = x e^y + cos(xy) no ponto (2,0) é um vetor cujo módulo vale :

a) 2,2
b) 3,0
c) 1,0
d) 1,5
e) 3,6

Answer 1

$\mathtt{f(x,\,y)=x\,e^y+cos(xy)}$


•   Derivando em relação a $\mathtt{x}:$

$\mathtt{\dfrac{\partial f}{\partial x}(x,\,y)=\dfrac{\partial}{\partial x}\big(x\,e^y+cos(xy)\big)}\\\\\\ \mathtt{\dfrac{\partial f}{\partial x}(x,\,y)=e^y-sen(xy)\cdot \dfrac{\partial}{\partial x}(xy)}\\\\\\ \mathtt{\dfrac{\partial f}{\partial x}(x,\,y)=e^y-sen(xy)\cdot y}\\\\\\ \mathtt{\dfrac{\partial f}{\partial x}(x,\,y)=e^y-y\,sen(xy)}$

 
Avaliando no ponto $\mathtt{(2,\,0)},$ temos

$\mathtt{\dfrac{\partial f}{\partial x}(2,\,0)=e^0-0\,sen(2\cdot 0)}\\\\\\ \mathtt{\dfrac{\partial f}{\partial x}(2,\,0)=1-0}\\\\\\ \boxed{\begin{array}{c}\mathtt{\dfrac{\partial f}{\partial x}(2,\,0)=1} \end{array}}$

_________

•   Derivando em relação a $\mathtt{y}:$

$\mathtt{\dfrac{\partial f}{\partial y}(x,\,y)=\dfrac{\partial}{\partial y}\big(x\,e^y+cos(xy)\big)}\\\\\\ \mathtt{\dfrac{\partial f}{\partial y}(x,\,y)=x\,e^y-sen(xy)\cdot \dfrac{\partial}{\partial y}(xy)}\\\\\\ \mathtt{\dfrac{\partial f}{\partial y}(x,\,y)=x\,e^y-sen(xy)\cdot x}\\\\\\ \mathtt{\dfrac{\partial f}{\partial y}(x,\,y)=x\,e^y-x\,sen(xy)}$


Avaliando no ponto $\mathtt{(2,\,0)},$ temos

$\mathtt{\dfrac{\partial f}{\partial y}(2,\,0)=2\cdot e^0-2\,sen(2\cdot 0)}\\\\\\ \mathtt{\dfrac{\partial f}{\partial y}(2,\,0)=2e^0-2\,sen\,0}\\\\\\ \boxed{\begin{array}{c}\mathtt{\dfrac{\partial f}{\partial y}(2,\,0)=2}\end{array}}$

__________

O vetor gradiente de $\mathtt{f}$ no ponto dado é

$\mathtt{\nabla f(2,\,0)=\dfrac{\partial f}{\partial x}(2,\,0)\overrightarrow{\mathtt{i}}+\dfrac{\partial f}{\partial y}(2,\,0)\overrightarrow{\mathtt{j}}}\\\\\\ \mathtt{\nabla f(2,\,0)=1\!\overrightarrow{\mathtt{i}}+2\!\overrightarrow{\mathtt{j}}}$


Calculando o módulo do vetor gradiente no ponto $\mathtt{(2,\,0)}:$

$\mathtt{\|\nabla f(2,\,0)\|=\left\|1\!\overrightarrow{\mathtt{i}}+2\!\overrightarrow{\mathtt{j}}\right\|}\\\\\\ \mathtt{\|\nabla f(2,\,0)\|=\sqrt{1^2+2^2}}\\\\\\ \mathtt{\|\nabla f(2,\,0)\|=\sqrt{1+4}}$

$\mathtt{\|\nabla f(2,\,0)\|=\sqrt{5}}\approx \boxed{\begin{array}{c}\mathtt{2,\!2} \end{array}}$    <———     esta é a resposta.


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Bons estudos! :-)