[SPAM] Re: [obm-l] outra tangente a curva

["fabricio82@xxxxxx" <fabricio82@xxxxxx>]

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Tio Cabri st wrote:
> Boa noite, será que alguém entre uma ceia e outra poderia me auxiliar nesta:
>
> Determine uma reta paralela a x+y=1 (daqui extraio que esta reta tem
> coeficiente angular -1) e tangente à curva y^3 +xy+ x^3=0 em um ponto
> (x0,y0), com x0<0 e y0<0.
>
> Reposta: x + y = -1
>
> Obrigado Cabri
>
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>
>   
Derivando implicitamente a equação, você obtém:

3y^2.y' + y + x.y' + 3x^2 = 0

y'.(3y^2 + x) = -(3x^2 + y)

y' = - (3x^2 + y)/(3y^2 + x)

Buscamos o ponto onde a inclinação (derivada) tem valor -1.

y' = -1

ou seja,

- (3x^2 + y)/(3y^2 + x) = -1

3x^2 + y = 3y^2 + x

3x^2 - 3y^2 + y - x = 0

3(x^2 - y^2) - 1(x - y) = 0

3(x + y)(x - y) - 1(x - y) = 0

(x - y).(3x + 3y -1) = 0

Logo

(A) x - y = 0

E a intersecção de x - y = 0 com

y^3 + xy + x^3 = 0 é o ponto (-1/2; -1/2)


ou

(B) 3x + 3y - 1 = 0

impossível, pois como x + y = 1/3, não podemos ter ambas coordenadas 
negativas.



A reta tangente, com inclinação -1 que passa pelo ponto (-1/2; -1/2) é:

x + y = -1


Pio.
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