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Question as in the title. Also, if a $2$-dimensional parametric equation does not have a Cartesian form, how would one go about proving this?

Two particular parametric equations I believe do not have a Cartesian form. (Or at least, I was unable to find the Cartesian form.)

First set of equations: (The square roots are difficult to eliminate.)

$x = \sqrt{60t} - t$

$y = (\sqrt{60t} - t)\left(\frac{60}{\sqrt{60t}}-1\right)$

Second set of equations: (The term $f(t)$ is difficult to eliminate.)

Let $f$ be some differentiable relation and let $r \in \mathbb{R}$. Define

$x = t - r\sin(\arctan(f'(t)))$

$y = f(t) + r\cos(\arctan(f'(t)))$

One interesting thing to note about the second set of parametric equations is that the set of points $(x,y)$ is an exact duplicate of $f(t)$, but $r$ units away along the normal through $(t,f(t))$.

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  • $\begingroup$ You can google for the specific terms "conversion parametric implicit", there are a bunch of research paper about this subject, all quite long, so it seems difficult to summarize the general problem in simple terms.In particular this one looks accessible graphics.stanford.edu/courses/cs348a-17-winter/Handouts/… $\endgroup$ – zwim Nov 17 '17 at 5:09
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I believe that you can always make one variable in an equation of two variables the "subject" of the equation. Maybe not in terms of elementary functions in very specific cases, but I'm pretty sure it's always possible. (And then substituting what you found for $t$ into the other equation is trivial.)

So, for instance, with your first set:

$$\sqrt{60t} - t = x$$ $$\sqrt{60t} = x + t$$ $$60t = (t+x)^2$$ $$60 t-t^2-2 t x-x^2 = 0$$ Solve this quadratic in $t$ to get $$t = 30 - x \pm \sqrt{900-60x}$$

Plugging in is the easy part. Replace $t$ with $30 - x \pm \sqrt{900-60x}$ to get

$$y = \sqrt{60} \, \Bigg(-1 + \frac{\sqrt{60}}{\sqrt{30 \pm \sqrt{900 - 60 x} - x}} \Bigg) \sqrt{30 \pm \sqrt{900 - 60 x} - x}$$ which looks like

y(x)

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Let x=f(t) and y=g(t). Let F(f(t))=t, in other words, F is the inverse of f, supposing f is invertible. Then y = g(F(x)) will give you y in terms of x.

Additionally, principles of differential equations can be used.

dy/dx = g'(t)/f'(t). If that is continuous and it's derivative with respect to t divided by g'(t) is also continuous, then there exists a unique solution of y in terms of x. Mind the points where the functions are not defined, i.e. avoid division by 0.

http://faculty.sfasu.edu/judsontw/ode/html/firstlook06.html

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