# Horizontal Tangent - Implicit Differentiation

• Dec 7th 2010, 01:21 PM
rawkstar
Horizontal Tangent - Implicit Differentiation
a curve in the plane is given by the following equation
x^4+2x^2y+3y^2=6

A) Find an expression for dy/dx using implicit differentiation
B) Find the four points on the curve where the tangent line is horizontal

Ok I know how to solve for the derivative. I got
dy/dx = (4x^3-4xy)/(2x^2+6y)

However Im not quite sure how to solve when the tangent is horizontal. I that the tangent is horizontal when the derivative equals 0. However I don't know how solve for this when it is an implicit derivative.
• Dec 7th 2010, 01:35 PM
skeeter
Quote:

Originally Posted by rawkstar
a curve in the plane is given by the following equation
x^4+2x^2y+3y^2=6

A) Find an expression for dy/dx using implicit differentiation
B) Find the four points on the curve where the tangent line is horizontal

Ok I know how to solve for the derivative. I got
dy/dx = (4x^3-4xy)/(2x^2+6y)

However Im not quite sure how to solve when the tangent is horizontal. I that the tangent is horizontal when the derivative equals 0. However I don't know how solve for this when it is an implicit derivative.

I am assuming your derivative is correct ...

$\displaystyle \frac{dy}{dx} = 0$ if $\displaystyle 4x^3 - 4xy = 0$

$\displaystyle 4x(x^2-y) = 0$

$\displaystyle x = 0$ , $\displaystyle y = x^2$

sub $\displaystyle x = 0$ and find the corresponding y-value(s)

sub $\displaystyle y = x^2$ and find the corresponding x-values
• Dec 7th 2010, 01:42 PM
rawkstar
sub it into the original equation?
• Dec 7th 2010, 01:51 PM
skeeter
Quote:

Originally Posted by rawkstar
sub it into the original equation?

yes ... you are trying to find points on the curve.
• Dec 7th 2010, 01:54 PM
rawkstar
ok so i subbed those into the orignial equation and got two x values and two y values
do i then plug those into the original to get the corresponding opposite value
• Dec 7th 2010, 02:01 PM
skeeter
Quote:

Originally Posted by rawkstar
ok so i subbed those into the orignial equation and got two x values and two y values
do i then plug those into the original to get the corresponding opposite value

when you sub in x = 0 , you get two values for y ... the two points will be $\displaystyle (0,\sqrt{2})$ and $\displaystyle (0,-\sqrt{2})$

when you sub in $\displaystyle x^2$ for $\displaystyle y$, you get two values for y ... square those to get the corresponding x-values.