1. ## implicit differentiation help

A curve is defined implicitly by the equation

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

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

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

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

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

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

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

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!
$\displaystyle x^3 + y^3 = 6xy$

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

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

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

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

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

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

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

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

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

$\displaystyle 2x - y^2 = 2y - x^2$

$\displaystyle y^2 + 2y = x^2 + 2x$

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

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

$\displaystyle y + 1 = |x + 1|$

$\displaystyle y = |x + 1| - 1$

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 $\displaystyle y = |x + 1| - 1$, it really means

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

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

It also means

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

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

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