1. ## parametric equations

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

2. Originally Posted by Rose Wanjohi
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
I'm not sure how you'd prove it but you can use the chain rule to find the gradient:

$\frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}$

3. i have tried that but it won't work how about simplifying LHS and RHS interms of t no no changed my mind any other suggestions?

4. Originally Posted by Rose Wanjohi
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
$x+ y= t^3+ 2t^2+ t$ and $x^2+ xy= x(x+ y)= (t^3+ t^2)(t^3+ 2t^2+ t)= t^6+ 3t^5+ 3t^4+ t^3= (t^2+ t)^3= y^3$

Differentiating both sides of $y^3= x^2+ xy$ to get $3y^2y'= 2x+ y+ xy'$ so $(3y^2- x)y'= 2x+ y$ and $y'= \frac{2x+ y}{3y^2- x}$.

5. Hello, Rose!

Show that function: . $\begin{array}{cccc} {\color{blue}[1]} & x&=&t^3+t^2 \\ {\color{blue}[2]} & y&=&t^2+t \end{array}$ .can be written: . $y^3\:=\:x^2+xy$
$\text{From }{\color{blue}[1]}\text{, we have: }\:x \:=\:t\underbrace{(t^2+t)}_{\text{This is }y} \quad\Rightarrow\quad x \:=\:ty \quad \Rightarrow\quad t \:=\:\frac{x}{y}$

Substitute into [2]: . $y \:=\:\left(\frac{x}{y}\right)^2 + \frac{x}{y} \quad\Rightarrow\quad y \:=\:\frac{x^2}{y^2} + \frac{x}{y}$

Multiply by $y^2\!:\quad y^3 \:=\:x^2 + xy$