Question

Determine o gradiente da função f(x, y, z)= x³ -xy² - z, no ponto P(1, 1, 0).

Answer 1

O vetor gradiente da função $\large\displaystyle\begin{gathered} \tt f(x,y,z)=x^3-xy^2-z\end{gathered}$ no ponto $\large\displaystyle\begin{gathered} \tt P(1,1,0)\end{gathered}$ é igual a:

$\Large\displaystyle\begin{gathered} \tt \overset{\rightarrow}{\nabla} f(1,1,0)=\left(2 , -2 ,-1\right)\end{gathered}$

Desejamos calcular o vetor gradiente da função $\large\displaystyle\begin{gathered} \tt f(x,y,z)=x^3-xy^2-z\end{gathered}$ no ponto $\large\displaystyle\begin{gathered} \tt P(1,1,0)\end{gathered}$. Para isso, temos que o gradiente de uma função é dado da seguinte forma:

$\Large\displaystyle\begin{gathered} \tt \overset{\rightarrow}{\nabla} f(x,y,z)=\left(\frac{\partial f}{\partial x} \ ,\ \frac{\partial f}{\partial y} \ ,\ \frac{\partial f}{\partial z} \right)\end{gathered}$

Com isso, temos que o gradiente da função $\large\displaystyle\begin{gathered} \tt f(x,y,z)=x^3-xy^2-z\end{gathered}$ é dado da seguinte forma:  $\Large\displaystyle\begin{gathered} \tt \overset{\rightarrow}{\nabla} f=\left(\frac{\partial \left(x^3-xy^2-z\right)}{\partial x} , \frac{\partial \left(x^3-xy^2-z\right)}{\partial y} , \frac{\partial \left(x^3-xy^2-z\right)}{\partial z} \right)\end{gathered}$Calculando as derivadas parciais, temos que:

$\Large\displaystyle\begin{gathered} \tt \frac{\partial}{\partial x}\left(x^3-xy^2-z\right)= 3x^2-y^2-0\end{gathered}$

$\Large\displaystyle\begin{gathered} \tt \frac{\partial}{\partial y}\left(x^3-xy^2-z\right)= 0-2xy-0\end{gathered}$

$\Large\displaystyle\begin{gathered} \tt \frac{\partial}{\partial z}\left(x^3-xy^2-z\right)= 0-0-1\end{gathered}$

Feito isso, temos que o gradiente da função $\large\displaystyle\begin{gathered} \tt f(x,y,z)=x^3-xy^2-z\end{gathered}$ é igual a:

$\Large\displaystyle\begin{gathered} \tt \overset{\rightarrow}{\nabla} f(x,y,z)=\left(3x^2-y^2 , -2xy ,-1\right)\end{gathered}$

No ponto $\large\displaystyle\begin{gathered} \tt P(1,1,0)\end{gathered}$, temos que:

$\Large\displaystyle\begin{gathered} \tt \overset{\rightarrow}{\nabla} f(1,1,0)=\left(3(1)^2-(1)^2 , -2(1)(1) ,-1\right)\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered}\therefore \green{\underline{\boxed{ \tt \overset{\rightarrow}{\nabla} f(1,1,0)=\left(2 , -2 ,-1\right)}}}\ \ (\checkmark).\end{gathered} }$

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