calculuscurvesparametricparametrization
Find a set of parametric equations $x(m)$, $y(m)$ that represent the graph of $y= 2x^2-3x+1$ such that the parameter $m$ is the slope $m = \frac{dy}{dx}$ at the point $(x,y)$.
I am guessing I need first convert the polynomial into parametric first? Honestly, I have no clue how to even start this. I've looked for hours on the internet but can't find an explanation that I understand.
If $x=\cos(t)$, then $t=\arccos(x)$, at least when $0\leq t \leq \pi/2$. Thus, your curve can be expressed as $y=e^{\arccos(x)}$, which you can integrate over $[0,1]$.
I certainly agree that a picture like the one below is useful. You could plot both the parametric plot and the function plot using a tool like WolframAlpha. It then becomes apparent that you should subtract 1 from the integral. Of course, you can also ask WolframAlpha for help with the integral, which might be a bit tricky.
We have $$x=\frac{3t}{1+t^3}\tag 1$$ $$y=\frac{3t^2}{1+t^3}\tag 2$$ Dividing (2) by (1), we get $$\frac{y}{x}=t\iff t=\frac{y}{x}$$ Now setting the value of $t$ in (1), we get $$x=\frac{3\frac{y}{x}}{1+\frac{y^3}{x^3}}$$ $$x=\frac{3\frac{y}{x}}{\frac{x^3+y^3}{x^3}}$$ $$x^3+y^3-3xy=0$$ Hence, the ordinary equation in the cartesian coordinates $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x^3+y^3-3xy=0}}$$
Best Answer
Since $m=y'=4x-3$ then $$x=\frac{m+3}{4}$$ Then the graph is $$(x,y)=(x,2x^2-3x+1)=\left(\frac{m+3}4,\frac{m^2-6m+9}8-\frac{3m+9}4+1\right)$$ or $$(x(m),y(m))=\left(\frac{m+3}4,\frac{m^2-12m-1}8\right)$$
Note that it is crucial that you can solve for $x$ in the equation of the derivative, so the polynomial should have a degree at most $2$. Or, at least, the derivative of the polynomial should be an injective function.