# Thread: Stationary Point of a parametric function.

1. ## Stationary Point of a parametric function.

A curve is defined by the equation $\displaystyle x^2y+y^3 = 2x+1$

a)Workings

$\displaystyle \frac{d}{dx}(x^2y)+\frac{d}{dx}(y^3) = \frac{d}{dx}(2x+1)$

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

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

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

$\displaystyle \frac{dy}{dx}= \frac{2-2(1)(2)}{(2)^2+3(1)^2}$

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

$\displaystyle \frac{dy}{dx}= -\frac{2}{7}$

b) Show that the x-cordinate of any stationary point of this curve satisfies the equation

$\displaystyle \frac{1}{x^3}=x+1$

b) Workings

$\displaystyle \frac{dy}{dx}=0$

$\displaystyle \frac{2-2xy}{x^2+3y^2}=0$

now what? :S

2. so you have 2 - 2xy = 0

from which y =1/x

Plug this into the original equation:

3. isnt $\displaystyle 2-2xy = 3y^2+x^2$

4. From

What is 0 * (x^2 +3*y^2) ?

5. Gosh you make me feel so stupid :{

6. We all have our moments -- I've spent hours on pblms because of simple mistakes

Don't forget your interpretation of the problem and mechanics were excellent.