Question

Convertendo a coordenada polar (4, 3π/4), em coordenadas cartesianas, obtém-se a equação representada na alternativa:

Answer 1

✅ Após resolver os cálculos, concluímos que o ponto "P" em coordenadas cartesianas é:

   $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf P_{c} = (-2\sqrt{2},\:2\sqrt{2})\:\:\:}}\end{gathered} }$

Seja o ponto "P" em coordenadas polares:

            $\Large\displaystyle\begin{gathered} P_{p} = \Bigg(4, \frac{3\pi}{4} \Bigg)\end{gathered}$

Sabendo que:

              $\Large\displaystyle\begin{gathered} \frac{3\pi}{4}rad = 135^{\circ} \end{gathered}$

Convertendo as coordenadas polares em cartesianas:

     $\Large\displaystyle\begin{gathered} P_{c} = (\tau\cdot cos\:\theta, \:\tau\cdot sen\:\theta)\end{gathered}$

            $\Large\displaystyle\text{ \begin{gathered} = \Bigg(4\cdot\frac{(-\sqrt{2})}{2},\:4\cdot\frac{\sqrt{2}}{2} \Bigg)\end{gathered} }$

            $\Large\displaystyle\text{ \begin{gathered} = \Bigg(-\frac{4\sqrt{2}}{2} ,\:\frac{4\sqrt{2}}{2} \Bigg)\end{gathered} }$

            $\Large\displaystyle\begin{gathered} = (-2\sqrt{2}, \:2\sqrt{2})\end{gathered}$

✅ Portanto, as coordenadas cartesianas é:

      $\Large\displaystyle\begin{gathered}P_{c} = (-2\sqrt{2}, \:2\sqrt{2})\end{gathered}$  

Saiba mais:

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