1. ## implicit differentiation help

A curve is defined implicitly by the equation

$x^3 + y^3 = 6xy$

a)Using Implicit Differentiation
b)Determine the location of each point on the curve whose tangent line has a slope of -1.

what i got for a) is

$y' = \frac{6y-3x^2}{3y^2-6x}$

i just want get clarified if thats correct or not,

and also, how do i get the answer to b? i don't know where to start.

can anyone help?
thanks!

Now set y ' = -1

6y - 3x^2 =-3y^2 +6x

3x^2 + 6x + 3y^2 - 6y = 0

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

Complete the squares

3(x+1)^2 + 3(y-1)^2 = 6

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

so at all pts on this circle the tangent line to the curve defined by

has a slope of -1

3. Originally Posted by break
A curve is defined implicitly by the equation

$x^3 + y^3 = 6xy$

a)Using Implicit Differentiation
b)Determine the location of each point on the curve whose tangent line has a slope of -1.

what i got for a) is

$y' = \frac{6y-3x^2}{3y^2-6x}$

i just want get clarified if thats correct or not,

and also, how do i get the answer to b? i don't know where to start.

can anyone help?
thanks!
$x^3 + y^3 = 6xy$

$\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(6xy)$

$3x^2 + \frac{d}{dx}(y^3) = 6\left[x\frac{d}{dx}(y) + y\frac{d}{dx}(x)\right]$

$3x^2 + \frac{d}{dy}(y^3)\,\frac{dy}{dx} = 6x\,\frac{dy}{dx} + 6y$

$3x^2 + 3y^2\,\frac{dy}{dx} = 6x\,\frac{dy}{dx} + 6y$

$3y^2\,\frac{dy}{dx} - 6x\,\frac{dy}{dx} = 6y - 3x^2$

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

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

You need to determine where the tangent line has slope $= -1$.

So $-1 = \frac{2y - x^2}{y^2 - 2x}$

$2x - y^2 = 2y - x^2$

$y^2 + 2y = x^2 + 2x$

$y^2 + 2y + 1^2 = x^2 + 2x + 1^2$

$(y + 1)^2 = (x + 1)^2$

$y + 1 = |x + 1|$

$y = |x + 1| - 1$

4. i'm sort of lost, am i not supposed to find points?

points as in the example below,

ex:
(3,6)
(-3,-6)

5. A function is still a set of points...

If you have $y = |x + 1| - 1$, it really means

$y = x + 1 - 1$ if $x + 1 \geq 0$, i.e. $x \geq -1$

$y = x$ if $x \geq -1$.

It also means

$y = -(x + 1) - 1$ if $x + 1 < 0$, i.e. $x < -1$

$y = -x - 2$ if $x < -1$.

So all the points which satisfy these two linear equations will be your solution.