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How can I find the equation relating $x$ and $y$ directly without an additional parameter $t$, which both are related to initially.

For example, $\frac{\mathrm{d}y}{\mathrm{d}t} = 2t+1$, $\frac{\mathrm{d}x}{\mathrm{d}t} = 2t-1$, and $y = t^2+t$ , $x = t^2-t$

Finding $\frac{\mathrm{d}y}{\mathrm{d}x}$, it is $\frac{2t + 1}{2t - 1}$. However, from here, would it be ever possible to solve to find an equation directly relating $x$ and $y$? From what I've learnt in school, the question usually ends by asking you to find the value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ given a certain $t$ value, so I'd like to find out more about this.

Thank you!

BLAZE
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mathworker123
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2 Answers2

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$t^2=y-t=x+t \implies y-x=2t$

$\cfrac{\mathrm{d}y}{\mathrm{d}x}=\cfrac{2t + 1}{2t - 1}\times\cfrac{2t + 1}{2t + 1}=\cfrac{(2t + 1)^{2}}{4t^2 - 1}=\cfrac{(y-x + 1)^{2}}{4(y-t) - 1}=\cfrac{(y-x + 1)^{2}}{4\left(y-\frac{y-x}{2}\right) - 1}=\color{blue}{\cfrac{(y-x + 1)^{2}}{4\left(\frac{3y+x}{2}\right) - 1}}$

BLAZE
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Putting $x+y=:u$, $\ y-x=:v$ you obtain $$\dot u=4t,\quad\dot v=2\ ,$$ which integrates to $$u(t)=2t^2+u_0,\quad v(t)=2t+v_0\ .\tag{1}$$ Here $u_0$ and $v_0$ denote the values of the variables $u$ and $v$ at time $t=0$. We now can eliminate $t$ from $(1)$and obtain $$u-u_0={1\over2}(v-v_0)^2\ .$$ In terms of $x$ and $y$ this can be rewritten as $$(x-x_0)+(y-y_0)={1\over2}\bigl((x-x_0)-(y-y_0)\bigr)^2\ .$$ This is a parabola through the point $(x_0,y_0)$. You can find $y'(x)$ by solving for $y$, or by means of implicit differentiation.

Christian Blatter
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