show that equation x=t^3+t^2, y=t^2+t can also be written as y^3=x^2+xy also, find the gradient from this equation
$\displaystyle x+ y= t^3+ 2t^2+ t$ and $\displaystyle x^2+ xy= x(x+ y)= (t^3+ t^2)(t^3+ 2t^2+ t)= t^6+ 3t^5+ 3t^4+ t^3= (t^2+ t)^3= y^3$
Differentiating both sides of $\displaystyle y^3= x^2+ xy$ to get $\displaystyle 3y^2y'= 2x+ y+ xy'$ so $\displaystyle (3y^2- x)y'= 2x+ y$ and $\displaystyle y'= \frac{2x+ y}{3y^2- x}$.
Hello, Rose!
$\displaystyle \text{From }{\color{blue}[1]}\text{, we have: }\:x \:=\:t\underbrace{(t^2+t)}_{\text{This is }y} \quad\Rightarrow\quad x \:=\:ty \quad \Rightarrow\quad t \:=\:\frac{x}{y}$Show that function: .$\displaystyle \begin{array}{cccc} {\color{blue}[1]} & x&=&t^3+t^2 \\ {\color{blue}[2]} & y&=&t^2+t \end{array}$ .can be written: .$\displaystyle y^3\:=\:x^2+xy$
Substitute into [2]: .$\displaystyle y \:=\:\left(\frac{x}{y}\right)^2 + \frac{x}{y} \quad\Rightarrow\quad y \:=\:\frac{x^2}{y^2} + \frac{x}{y}$
Multiply by $\displaystyle y^2\!:\quad y^3 \:=\:x^2 + xy$