Find dy/dx given x = 3at/(1+t³), y = (3at²)/(1+t³)

Find $\dfrac{dy}{dx}$ given $x = \dfrac{3at}{1+t^{3}}$, $y = \dfrac{3at^{2}}{1+t^{3}}$.

Step 1: Differentiate $x$

Quotient rule with $u = 3at$, $u' = 3a$, $v = 1+t^{3}$, $v' = 3t^{2}$: \[x'_{t} = \frac{3a\bigl(1+t^{3}\bigr) - 3at\cdot 3t^{2}}{\bigl(1+t^{3}\bigr)^{2}} = \frac{3a\bigl(1+t^{3}-3t^{3}\bigr)}{\bigl(1+t^{3}\bigr)^{2}} = \frac{3a\bigl(1-2t^{3}\bigr)}{\bigl(1+t^{3}\bigr)^{2}}.\]

Step 2: Differentiate $y$

Quotient rule with $u = 3at^{2}$, $u' = 6at$: \[y'_{t} = \frac{6at\bigl(1+t^{3}\bigr) - 3at^{2}\cdot 3t^{2}}{\bigl(1+t^{3}\bigr)^{2}} = \frac{3at\bigl(2+2t^{3}-3t^{3}\bigr)}{\bigl(1+t^{3}\bigr)^{2}} = \frac{3at\bigl(2-t^{3}\bigr)}{\bigl(1+t^{3}\bigr)^{2}}.\]

Step 3: Form the ratio

The common factor $\dfrac{3a}{\bigl(1+t^{3}\bigr)^{2}}$ cancels entirely: $\frac{dy}{dx} = \frac{t\bigl(2-t^{3}\bigr)}{1-2t^{3}}$.

Step 4: Identifying the curve

Dividing the two coordinates gives $\dfrac{y}{x} = t$, so $y = tx$. Substituting back shows the curve satisfies

\[ x^{3}+y^{3} = 3axy, \]

the folium of Descartes, with the parameter $t$ being the slope of the ray from the origin.

Step 5: Verification

\[t = 1:\quad (x,y) = \left(\tfrac{3a}{2},\tfrac{3a}{2}\right),\quad \frac{dy}{dx} = \frac{1\cdot 1}{1-2} = -1 \;\;\checkmark\]

\[\text{Implicitly: } 3x^{2}+3y^{2}y' = 3a\bigl(y+xy'\bigr) \;\Longrightarrow\; y' = \frac{ay-x^{2}}{y^{2}-ax}; \text{ at } x=y=\tfrac{3a}{2}:\ y' = -1 \;\;\checkmark\]

\[t = 0:\quad (x,y) = (0,0),\quad \frac{dy}{dx} = 0 \text{ — the horizontal branch through the node} \;\;\checkmark\]