# Finding the equation using Implict differentiation

• Apr 22nd 2010, 05:18 PM
Paymemoney
Finding the equation using Implict differentiation
Hi
I need help on the following question;

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

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

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

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

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

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

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

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

P.S
• Apr 22nd 2010, 05:58 PM
skeeter
Quote:

Originally Posted by Paymemoney
Hi
I need help on the following question;

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

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

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

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

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

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

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

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

P.S

sign error ...

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

have a look see at the graph ...
• Apr 22nd 2010, 11:15 PM
Paymemoney
that means when i sub (0,2) into the equation $\displaystyle \frac{dy}{dx} = 0$, i don't understand how the book's answer is x=0,
• Apr 23rd 2010, 04:43 AM
skeeter
Quote:

Originally Posted by Paymemoney
that means when i sub (0,2) into the equation $\displaystyle \frac{dy}{dx} = 0$, 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.