Question

a temperatura num ponto (x,y,z) é dada por T (x,y,z) = e - x^2 - 2y^2 - 3z^2. Identifique a superfície de R³ cujos pontos possuem temperatura igual à do ponto (-1,-1,1)

Answer 1

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$\large\begin{array}{l} \textsf{A temperatura em um em um ponto }\mathsf{(x,\,y,\,z)}\textsf{ \'e dada pela lei}\\\\ \mathsf{T(x,\,y,\,z)=e^{-x^2-2y^2-3z^2}} \end{array}$


$\large\begin{array}{l} \textsf{Logo, no ponto }\mathsf{(-1,\,-1,\,1),}\textsf{ o valor da temperatura \'e}\\\\ \mathsf{T(-1,\,-1,\,1)=e^{-1^2-2\,\cdot\,1^2-3\,\cdot\,1^2}}\\\\ \mathsf{T(-1,\,-1,\,1)=e^{-1-2-3}}\\\\ \mathsf{T(-1,\,-1,\,1)=e^{-6}}\qquad\quad\checkmark \end{array}$


$\large\begin{array}{l} \textsf{A equa\c{c}\~ao da superf\'icie que cont\'em os pontos do }\mathsf{\mathbb{R}^3}\textsf{ com}\\\textsf{temperatura igual \`a do ponto }\mathsf{(-1,\,-1,\,1)}\textsf{ \'e dada por}\\\\ \mathsf{T(x,\,y,\,z)=T(-1,\,-1,\,1)}\\\\ \mathsf{e^{-x^2-2y^2-3z^2}=e^{-6}} \end{array}$


$\large\begin{array}{l} \mathsf{-x^2-2y^2-3z^2=-6}\\\\ \mathsf{x^2+2y^2+3z^2=6}\\\\ \mathsf{\dfrac{x^2+2y^2+3z^2}{6}=\dfrac{6}{6}}\\\\ \mathsf{\dfrac{x^2}{6}+\dfrac{2y^2}{6}+\dfrac{3z^2}{6}=1} \end{array}$


$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\dfrac{x^2}{6}+\dfrac{y^2}{3}+\dfrac{z^2}{2}=1} \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a reposta.}\\\\\\ \textsf{que representa um elipsoide no espa\c{c}o }\mathsf{\mathbb{R}^3.} \end{array}$


$\large\textsf{Bons estudos! :-)}$