Finding the equation using Implict differentiation

**Paymemoney** · April 22nd 2010, 05:18 PM

Hi
I need help on the following question;

Find the equation to the normal to the curve $x^3 + y^3 -2xy + 4x = 8$ at the point (0,2)

I have found the eqaution but i don't understand why the answer is x=0.

$3x^2 + 3y^2 \frac{dy}{dx} - (2x \frac{dy}{dx} + 2y) +4 = 0$

$\frac{dy}{dx} 3y^2 \frac{dy}{dx} +2y -3x^2$

$\frac{dy}{dx} = \frac{-4-2y-3x^2}{3y^2-2x}$

to find the equation i sub the values (0,2)

which gets: $\frac{-2}{3}$

so equation i get is $y=\frac{3}{2}x + 2$

P.S

**skeeter** · April 22nd 2010, 05:58 PM

Originally Posted by Paymemoney

Hi
I need help on the following question;

Find the equation to the normal to the curve $x^3 + y^3 -2xy + 4x = 8$ at the point (0,2)

I have found the eqaution but i don't understand why the answer is x=0.

$3x^2 + 3y^2 \frac{dy}{dx} - (2x \frac{dy}{dx} + 2y) +4 = 0$

$\frac{dy}{dx} 3y^2 \frac{dy}{dx} +2y -3x^2$

$\frac{dy}{dx} = \frac{-4-2y-3x^2}{3y^2-2x}$

to find the equation i sub the values (0,2)

which gets: $\frac{-2}{3}$

so equation i get is $y=\frac{3}{2}x + 2$

P.S

sign error ...

$\frac{dy}{dx} = \frac{-4+2y-3x^2}{3y^2-2x}$

have a look see at the graph ...

**Paymemoney** · April 22nd 2010, 11:15 PM

that means when i sub (0,2) into the equation $\frac{dy}{dx} = 0$ , i don't understand how the book's answer is x=0,

**skeeter** · April 23rd 2010, 04:43 AM

Originally Posted by Paymemoney

that means when i sub (0,2) into the equation $\frac{dy}{dx} = 0$ , i don't understand how the book's answer is x=0,

the tangent line at (0,2) is y = 2

the line perpendicular to a horizontal line is a vertical line ... the vertical line passing through the point (0,2) is x = 0 ... the y-axis.

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