Given x = at/(1+t³), y = (a (4t³+1))/(6 (1+t³)²), find y'

Given $x = \dfrac{at}{1+t^{3}}$, $y = \dfrac{a\bigl(4t^{3}+1\bigr)}{6\bigl(1+t^{3}\bigr)^{2}}$, find $y'$ and determine the algebraic equation relating $x$ and $y'$.

Step 1: Differentiate $x$

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

Step 2: Differentiate $y$

With $N = a\bigl(4t^{3}+1\bigr)$, $N' = 12at^{2}$, $D = 6\bigl(1+t^{3}\bigr)^{2}$, $D' = 36t^{2}\bigl(1+t^{3}\bigr)$: \[y'_{t} = \frac{12at^{2}\cdot 6\bigl(1+t^{3}\bigr)^{2} - a\bigl(4t^{3}+1\bigr)\cdot 36t^{2}\bigl(1+t^{3}\bigr)}{36\bigl(1+t^{3}\bigr)^{4}} = \frac{12at^{2}\bigl(1+t^{3}\bigr)\Bigl[6\bigl(1+t^{3}\bigr)-3\bigl(4t^{3}+1\bigr)\Bigr]}{36\bigl(1+t^{3}\bigr)^{4}}.\]

The bracket simplifies: $6+6t^{3}-12t^{3}-3 = 3\bigl(1-2t^{3}\bigr)$, so

\[ y'_{t} = \frac{at^{2}\bigl(1-2t^{3}\bigr)}{\bigl(1+t^{3}\bigr)^{3}}. \]

Step 3: Form the ratio

The factors $a$ and $\bigl(1-2t^{3}\bigr)$ cancel entirely: \[y' = \frac{y'_{t}}{x'_{t}} = \frac{at^{2}\bigl(1-2t^{3}\bigr)}{\bigl(1+t^{3}\bigr)^{3}}\cdot\frac{\bigl(1+t^{3}\bigr)^{2}}{a\bigl(1-2t^{3}\bigr)} = \frac{t^{2}}{1+t^{3}}.\]

Step 4: Eliminate the parameter

Write $p = x$ and $q = y'$. Dividing one by the other gives the parameter back: \[\frac{q}{p} = \frac{t^{2}}{1+t^{3}}\cdot\frac{1+t^{3}}{at} = \frac{t}{a} \qquad\Longrightarrow\qquad t = \frac{aq}{p}.\]

Clearing the denominator in $p = \dfrac{at}{1+t^{3}}$ gives $p\,t^{3} - a\,t + p = 0$. Substituting $t = \dfrac{aq}{p}$: \[p\cdot\frac{a^{3}q^{3}}{p^{3}} - a\cdot\frac{aq}{p} + p = 0 \;\Longrightarrow\; \frac{a^{3}q^{3}}{p^{2}} - \frac{a^{2}q}{p} + p = 0.\]

Multiplying through by $p^{2}$: $x^{3} + a^{3}y'^{3} = a^{2}x\,y'$.

Step 5: Verification

\[a = 2,\ t = 1:\quad x = 1,\ y' = \tfrac12; \qquad 1 + 8\cdot\tfrac18 = 2 \quad\text{and}\quad 4\cdot 1\cdot\tfrac12 = 2 \;\;\checkmark\]

\[a = 3,\ t = \tfrac12:\quad x = \tfrac{4}{3},\ y' = \tfrac29; \qquad \text{both sides} = \tfrac{8}{3} \;\;\checkmark\]

\[a = 5,\ t = 2:\quad \text{both sides} \approx 12.3457 \;\;\checkmark\]

Step 6: Recognising the curve

Substituting $Y = a\,y'$ turns the relation into

\[ x^{3}+Y^{3} = a\,x\,Y, \]

the folium of Descartes in the variables $\bigl(x,\,ay'\bigr)$ — the same curve met in Problem 196, now traced by the point whose ordinate is $a$ times the slope.