Question

Determine as equações das retas tangente e normal à curva y = x² − 1 em x = 2.

Answer 1

✅ Após desenvolver os cálculos, concluímos que as equações reduzidas das retas tangente e normal à referida curva pelo respectivo ponto de tangência são respectivamente:

        $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf t: y = 4x - 5\:\:\:}}\end{gathered} }$

    $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf n : y = -\frac{1}{4}x + \frac{14}{4}\:\:\:}}\end{gathered} }$

Sejam os dados:

                $\Large\begin{cases} y = x^{2} - 1\\x = 2\end{cases}$

Observe que:

                 $\Large\displaystyle\begin{gathered} y = f(x)\end{gathered}$

Para resolver esta questão, devemos:

• Determinar as coordenadas do ponto de tangência "T" entre as curvas. Para isso, fazemos:

             $\Large\displaystyle\begin{gathered} T = (X_{T},\,Y_{T})\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = \left[x,\,f(x)\right]\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = \left[2,\,2^{2} - 1\right]\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = \left[2,\,4 - 1\right]\end{gathered}$

                  $\Large\displaystyle\begin{gathered} = \left[2,\,3\right]\end{gathered}$

           $\Large\displaystyle\begin{gathered} \therefore\:\:\:T(2, 3)\end{gathered}$

• Calcular a declividade - coefiiciente angular - da reta tangente "t". Para isso, devemos calcular a derivada primeira da função no ponto "T". Então, fazemos:

                 $\Large\displaystyle\begin{gathered} m_{t} = \frac{\partial}{\partial x}\codt f(x)\end{gathered}$

                         $\Large\displaystyle\begin{gathered} = \frac{\partial}{\partial x}\codt (x^{2} - 3x)\end{gathered}$

                         $\Large\displaystyle\begin{gathered} = 2\cdot2^{2 - 1} - 0\end{gathered}$

                         $\Large\displaystyle\begin{gathered} = 2\cdot2\end{gathered}$

                         $\Large\displaystyle\begin{gathered} = 4\end{gathered}$

              $\Large\displaystyle\begin{gathered} \therefore\:\:\:m_{t} = 4\end{gathered}$

• Montar a equação da reta tangente "t". Para isso, devemos utilizar a fórmula do "ponto/declividade", ou seja:

        $\Large\displaystyle\begin{gathered} \bf(I)\end{gathered}$        $\Large\displaystyle\begin{gathered} y - y_{T} = m_{t}\cdot(x - x_{T})\end{gathered}$

        Substituindo os valores das incógnitas na equação "I", temos:

                       $\Large\displaystyle\begin{gathered} y - 3 = 4\cdot(x - 2)\end{gathered}$

                       $\Large\displaystyle\begin{gathered} y - 3 = 4x - 8\end{gathered}$

                                $\Large\displaystyle\begin{gathered} y = 4x - 8 + 3\end{gathered}$

                                $\Large\displaystyle\begin{gathered} y = 4x - 5\end{gathered}$

• Calcular a reta normal:

                     $\Large\displaystyle\text{ \begin{gathered} y - y_{T} = -\frac{1}{m_{t}}\cdot(x - x_{T})\end{gathered} }$  

                        $\Large\displaystyle\begin{gathered} y - 3 = -\frac{1}{4}\cdot(x - 2)\end{gathered}$    

                         $\Large\displaystyle\begin{gathered} y - 3 = \frac{-x}{4} + \frac{2}{4}\end{gathered}$

                                  $\Large\displaystyle\begin{gathered} y = \frac{-x}{4} + \frac{2}{4} + 3\end{gathered}$  

                                  $\Large\displaystyle\begin{gathered} y = \frac{-x + 2 + 12}{4}\end{gathered}$

                                  $\Large\displaystyle\begin{gathered} y = \frac{-x + 14}{4}\end{gathered}$

                                  $\Large\displaystyle\begin{gathered} y = -\frac{1}{4}x + \frac{14}{4}\end{gathered}$

✅ Portanto, as equações das retas tangente e normal são:

              $\Large\displaystyle\begin{gathered} t: y = 4x - 5\end{gathered}$

              $\Large\displaystyle\begin{gathered} n : y = -\frac{1}{4}x + \frac{14}{4}\end{gathered}$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$