$Explain parameter elimination for complex curves$

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$Explain parameter elimination for complex curves$

I'm reading Stewart's Calculus and in "Calculus with Parametric Curves" he shows the formula which comes from the Chain rule:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
But what I don't understand is that for some complex curves for example:
$x = 2\sin\left(1+3t\right), y = 2t^{3}$
If I try to retreive y(x) I will get
$y\ =\ 2\left(\frac{\arcsin\left(\frac{x}{2}\right)-1}{3}\right)^{3}$
Which will represent only small part of the parametric curve(graph).
And only for some small range of t $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. For example when t = 1, $y(x)$ is not defined.
So the question is: Am I missing something, or $\frac{dy}{dx}$ can't always represent $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$?

Calculus

Babaduras

106

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We have:

$y\ =\ 2\left(\frac{\arcsin\left(\frac{x}{2}\right)-1}{3}\right)^{3}$

So,

$\frac{dy}{dx} = 6\times \frac{\frac{1}{2}}{3\times \sqrt{1-\frac{x^{2}}{4}}}(\frac{arcsin(\frac{x}{2})-1}{3})^{2}$

Which simplifies to:

$\frac{dy}{dx} = \frac{1}{\sqrt{1-\frac{x^{2}}{4}}}(\frac{arcsin(\frac{x}{2})-1}{3})^{2}$

$t=\frac{{arcsin(\frac{x}{2})-1}}{3}$

$sin(1+3t)=\frac{x}{2}$

$cos^{2}(1+3t)=1-\frac{x^{2}}{4}$

$cos(1+3t)=\sqrt{1-\frac{x^{2}}{4}}$

So,

$\frac{dy}{dx} ==\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{t^{2}}{cos(1+3t)}=\frac{({\frac{{\arcsin\left(\frac{x}{2}\right)-1}}{3}})^{2}}{\sqrt{1-\frac{x^{2}}{4}}}$

t cannot be a constant and it varies.

x is a function of the variable t.

y is also a function of the variable t.

y is also a function of x.

dy/dx can be calculated for various t. But, when calculating dy/dt, t is not constant and is a variable.

Kav10

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Babaduras

0

That’s not exactly what I’ve asked. As I’ve said if we take a look into graph of parametric equation it has complex shape and many negative derivatives(negative angles of tangent lines). But y(x) is just a small part of the graph of parametric equation, and it doesn’t have those negative derivatives(angles of tangent lines), since it grows from left to right. That’s why I said that dy/dx can’t even have similar values to dy/dt/dx/dt for some of the t values( for example 1).
Babaduras

0

I will tip you additional 5$. Just give me some better explanation please.
Kav10

0

I see you accepted. Do you still need aditional help? I explained above that dy/dx always equals (dy/dt)/(dx/dt). Not sure what you mean by when t = 1, y(x) is not defined.
- Babaduras
  
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$Explain parameter elimination for complex curves$

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