Question

A curve has equation
3 3
x xy y    3 0
a) Find
dy
dx
in terms of
x
and
y.
b) Find any point where
y x 
.
c) There is one point apart from the origin at which the slope of this curve is 0. Locate this
point and determine whether it is a local maximum or local minimum.

0cb96eabcd2d6b179d8f1c720345dc6b.pdf

Answer 1

Answer:

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

(b) $(0,0), (\frac{3}{2},\frac{3}{2})$

(c) The point is $(\sqrt[3]{2}, \sqrt[3]{4})$, it is a local minimum

Step-by-step explanation:

Given equation of the curve

$x^3-3xy+y^3=0$

Differentiation above w.r.t. x

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

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

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

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

If y=x then

$x^3-3x^2+x^3=0$

$\implies 2x^3-3x^2=0$

$\implies x^2(2x-3)=0$

$\implies x=0,\frac{3}{2}$

The point where y=x  is $(0,0), (\frac{3}{2},\frac{3}{2})$

If the slope of the curve is zero then

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

$\implies \frac{x^2-y}{x+y^2}=0$

$\implies x^2-y=0$

$\implies x^2=y$

Putting the value of y in the original equation of the curve

$x^3-3x\times x^2+(x^2)^3=0$

$\implies x^3-3x^3+x^6=0$

$\implies -2x^3+x^6=0$

$\implies x^3(-2+x^3)=0$

$\implies x=0, \sqrt[3]{2}$

Therefore the other point is $(\sqrt[3]{2}, \sqrt[3]{4})$

if we taken any point left of the above point say$(\sqrt[3]{2}-h, \sqrt[3]{4}-h)$ , the slope is

$\frac{dy}{dx}=\frac{(\sqrt[3]{2}-h)^2-\sqrt[3]{4}-h}{\sqrt[3]{2}-h+(\sqrt[3]{4}-h)^2}$

$\frac{dy}{dx}=\frac{\sqrt[3]{4}-2h\sqrt[3]{2}-\sqrt[3]{4}-h}{\sqrt[3]{2}-h+\sqrt[3]{16}-2h\sqrt[3]{4}}$   (ignoring the higher power of h)

The above value is negative

if we take a point just right of the point  $(\sqrt[3]{2}, \sqrt[3]{4})$ and replace h by -h, the value will be positive

Thus the slope changes from negative to positive, therefore there is a local minima at the point.

Hope this helps.