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Math Help - parametric equations

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by Rose Wanjohi View Post
    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}
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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by Rose Wanjohi View Post
    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}.
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  5. #5
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    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

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