Question

Considere a curva cuja equação (2 - x)y² = x³. Obtenha a equação da reta tangente ao gráfico da curva em (1,1).

Answer 1

$\displaystyle (2-\text x)\text y^2=\text x^3$

Reta tangente no ponto da curva (1,1).

$\text y = \text {mx}+\text n$

o coeficiente angular da reta é a derivada da curva, ou seja :

$\displaystyle \text {m}=\text y ' \\\\ \ [\ (2-\text x)\text y^2 \ ] '=(\text x^3)' \\\\ \underline{\text{Vamos derivar implicitamente e depois s{\'o} arrumar}} : \\\\ \ [ \ 2\text y^2-\text x\text y^2\ ]'=3\text x^2 \\\\ 4\text y.\text y' -(\text x.2.\text y.\text y'+\text y^2) =3\text x^2 \\\\ 4\text y.\text y '-2\text{xy}.\text y'-\text y^2=3\text x^2 \\\\ \text y'(4\text y-2\text{xy})=3\text x^2+\text y^2 \\\\ \text y'=\frac{3\text x^2+\text y^2}{2\text y(2-\text x)}$

Substituindo o ponto (1,1) :

$\displaystyle \text y' = \frac{3.1^2+1^2}{2.1(2-1)} \\\\\\ \text y' = \frac{3+1}{2} \\\\ \text y'= 2 \\\\ \text{Ent{\~a}o }: \\\\ \boxed{\text m =2}$

Equação da reta tangente :

$\displaystyle \text y = 2\text x+\text n \\\\ \text{Substituindo o ponto (1,1)}: \\\\ 1 = 2+\text n \\\\ \text n = 1-2 \\\\ \text n =-1 \\\\ \text{Da{\'i}}: \\\\ \text y = 2\text x-1 \\\\ \underline{\text{Portanto a equa{\c c}{\~a}o da reta tangente ao gr{\'a}fico da curva em (1,1) {\'e}}}: \\\\\\ \huge\boxed{2\text x-\text y-1=0\ }\checkmark$