Question

Derivadas!!! me ajudem por favor urgente...
Encontre o ponto (x,y) onde a tangente á curva dada seja horizontal, o ângulo da tangente com o eixo dos x seja 60°, que esse ângulo seja - 30°escreva a equação da reta tangente e faça os graficos

1) $y=x^2+2x-3$

Answer 1

$y=x^2+2x-3$


Encontrando a derivada da função (usando as regras básicas de derivação):

$y'=(x^2+2x+3)'\\\\ y'=2x+2$


Agora, vamos encontrar retas tangentes pedidas.

O coeficiente angular é a derivada no ponto, e tem o valor da tangente do ângulo de inclinação

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•   Pontos onde a reta tangente é horizontal:(inclinação 0° ou 180°)

$y'\big|_{x=x_P}=tg\,0^\circ=tg\,180^\circ=0\\\\\\ y'\big|_{x=x_P}=0\\\\\\ 2x_{_P}+2=0\\\\ 2x_{_P}=-2\\\\ x_{_P}=\dfrac{-2}{2}\\\\ x_{_P}=-1$


•   Calculando o valor da função neste ponto:

$y_{_P}=x_{_P}^2+2x_{_P}-3\\\\ y_{_P}=(-1)^2+2\cdot (-1)-3\\\\ y_{_P}=1-2-3\\\\ y_{_P}=1-5\\\\ y_{_P}=-4$


•   Equação da reta tangente (como a reta é horizontal, então):

$y-y_{_P}}=0\\\\y-(-4)=0\\\\ y+4=0$

$\fbox{ y=-4 }$   <———   reta tangente horizontal

no ponto $(-1,\,-4).$

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•   Pontos onde a reta tangente tem inclinação 60°:

$y'\big|_{x=x_P}=tg\,60^\circ\\\\\\ y'\big|_{x=x_P}=\sqrt{3}\\\\\\ 2x_{_P}+2=\sqrt{3}\\\\ 2x_{_P}=\sqrt{3}-2\\\\ x_{_P}=\dfrac{1}{2}\,(\sqrt{3}-2)$


•   Calculando o valor da função nesse ponto:

$y_{_{P}}=x_{_{P}}^2+2x_{_{P}}-3\\\\ y_{_P}=\left[\dfrac{1}{2}\,(\sqrt{3}-2)\right]^2+\diagup\!\!\!\! 2\cdot \left[\dfrac{1}{\diagup\!\!\!\! 2}\,(\sqrt{3}-2)\right]-3\\\\\\ y_{_P}=\dfrac{1}{4}\,(\sqrt{3}-2)^2+(\sqrt{3}-2)-3\\\\\\ y_{_P}=\dfrac{1}{4}\,(7-4\sqrt{3})+(\sqrt{3}-2)-3$

$y_{_P}=\dfrac{1}{4}\,(7-4\sqrt{3})+\dfrac{1}{4}\,(4\sqrt{3}-20)\\\\\\ y_{_P}=\dfrac{7}{4}-\dfrac{20}{4} \\\\\\y_{_P}=-\,\dfrac{13}{4}$


•   Equação da reta tangente:

$y-y_{_P}=y'\big|_{x=x_P}\cdot (x-x_{_P})~\qquad\text{onde }x_{_P}=\dfrac{1}{2}\,(\sqrt{3}-2)\\\\\\ y-\bigg(\!\!-\dfrac{13}{4}\bigg)=\sqrt{3}\cdot \left[x-\dfrac{1}{2}\,(\sqrt{3}-2) \right]\\\\\\ y+\dfrac{13}{4}=\sqrt{3}\,x-\dfrac{\sqrt{3}}{2}\,(\sqrt{3}-2)\\\\\\ y=\sqrt{3}\,x-\dfrac{3}{2}+\sqrt{3}-\dfrac{13}{4}\\\\\\ y=\sqrt{3}\,x+\sqrt{3}-\dfrac{6}{4}-\dfrac{13}{4}$

$\fbox{ y=\sqrt{3}\,x+\sqrt{3}-\dfrac{19}{4} }$   <———    equação da reta tangente

no ponto $\bigg(\dfrac{1}{2}\,(\sqrt{3}-2),\,-\,\dfrac{13}{4}\bigg).$

__________

•   Pontos onde a reta tangente tem inclinação – 30°:

$y'\big|_{x=x_P}=tg(-30^\circ)\\\\\\ y'\big|_{x=x_P}=-\,\dfrac{\sqrt{3}}{3}\\\\\\ 2x_{_P}+2=-\,\dfrac{\sqrt{3}}{3}\\\\\\ 2x_{_P}=-\,\dfrac{\sqrt{3}}{3}-2\\\\\\ x_{_P}=\dfrac{1}{2}\bigg(\!\!-\dfrac{\sqrt{3}}{3}-2\bigg)\\\\\\ x_{_P}=-\,\dfrac{1}{2}\bigg(\dfrac{\sqrt{3}}{3}+\dfrac{6}{3}\bigg)\\\\\\ x_{_P}=-\,\dfrac{1}{6}\,(\sqrt{3}+6)$


•   Calculando o valor da função neste ponto:

$y_{_{P}}=x_{_{P}}^2+2x_{_{P}}-3\\\\ y_{_P}=\left[-\,\dfrac{1}{6}\,(\sqrt{3}+6) \right ]^2+2\left[-\,\dfrac{1}{6}\,(\sqrt{3}+6) \right ]-3\\\\\\ y_{_P}=\dfrac{1}{36}\,(\sqrt{3}+6)^2-\,\dfrac{1}{3}\,(\sqrt{3}+6)-3\\\\\\ y_{_P}=\dfrac{1}{36}\,(39+12\sqrt{3})-\,\dfrac{12}{36}\,(\sqrt{3}+6)-\dfrac{108}{36}$

$y_{_P}=\dfrac{1}{36}\,\big[(39+12\sqrt{3})-12\,(\sqrt{3}+6)-108\big]\\\\\\ y_{_P}=\dfrac{1}{36}\,\big[39+12\sqrt{3}-12\sqrt{3}-72-108\big]\\\\\\ y_{_P}=\dfrac{1}{36}\cdot (-141)\\\\\\ y_{_P}=-\,\dfrac{47}{12}$


•   Equação da reta tangente:

$y-y_{_P}=y'\big|_{x=x_P}\cdot (x-x_{_P})~\qquad\text{onde }x_{_P}=-\,\dfrac{1}{6}\,(\sqrt{3}+6)\\\\\\ y-\left(\!-\,\dfrac{47}{12}\right)=-\,\dfrac{\sqrt{3}}{3}\cdot \left[x-\left(\!-\,\dfrac{1}{6}\,(\sqrt{3}+6)\right) \right ]\\\\\\ y+\dfrac{47}{12}=-\,\dfrac{\sqrt{3}}{3}\cdot \left[x+\dfrac{1}{6}\,(\sqrt{3}+6)\right ]\\\\\\ y=-\,\dfrac{\sqrt{3}}{3}\,x-\dfrac{\sqrt{3}\,(\sqrt{3}+6)}{18}-\dfrac{47}{12}\\\\\\ y=-\,\dfrac{\sqrt{3}}{3}\,x-\dfrac{3+6\sqrt{3}}{18}-\dfrac{47}{12}\\\\\\ y=-\,\dfrac{\sqrt{3}}{3}\,x-\dfrac{6\sqrt{3}}{18}-\dfrac{47}{12}-\dfrac{3}{18}$

$y=-\,\dfrac{\sqrt{3}}{3}\,x-\dfrac{\sqrt{3}}{3}-\dfrac{47}{12}-\dfrac{1}{6}\\\\\\ y=-\,\dfrac{\sqrt{3}}{3}\,x-\dfrac{\sqrt{3}}{3}-\dfrac{47}{12}-\dfrac{2}{12}$

$\fbox{ y=-\,\dfrac{\sqrt{3}}{3}\,x-\dfrac{\sqrt{3}}{3}-\dfrac{49}{12} }$    <———     equação da reta tangente

no ponto $\left(-\,\dfrac{1}{6}\,(\sqrt{3}+6),\;-\,\dfrac{47}{12}\right).$

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A representação do gráfico da função e das retas tangentes estão em anexo.


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Dúvidas? Comente.


Bons estudos! :-)


Tags: reta tangente derivada gráfico função esboço ângulo cálculo diferencial