Parametric to Polar Conversion

[Zach Knight]

Homework Statement

Convert the two equations x=x(t) and y=y(t) to a polar equation of the form r=r($$\theta$$)

Homework Equations

x=r*cos($$\theta$$)
y=r*sin($$\theta$$)
r$$^{2}$$=x$$^{2}$$+y$$^2$$

The Attempt at a Solution

Perhaps I'm over-thinking this, but in order to eliminate the parameter t, I solved one of the two parametric equations, say x(t), for t, giving an equation t=t(x). I then substituted this equation into y(t), giving y=y(t(x)). Afterward, I used the identities listed above to convert y to r. Is this valid? Is there a way to do this without having to invert one of the functions?

 

[Mark44]

Homework Statement

Convert the two equations x=x(t) and y=y(t) to a polar equation of the form r=r($$\theta$$)

Homework Equations

x=r*cos($$\theta$$)
y=r*sin($$\theta$$)
r$$^{2}$$=x$$^{2}$$+y$$^2$$

The Attempt at a Solution

Perhaps I'm over-thinking this, but in order to eliminate the parameter t, I solved one of the two parametric equations, say x(t), for t, giving an equation t=t(x). I then substituted this equation into y(t), giving y=y(t(x)). Afterward, I used the identities listed above to convert y to r. Is this valid? Is there a way to do this without having to invert one of the functions?

It doesn't seem valid to me. For one thing, when you solve for t in terms of x, you are finding the inverse of the original function, which may or may not exist. For example, if x = f(t) = t² + 3, f is not one-to-one, so doesn't have an inverse.

The usual approach is to eliminate the parameter t, and then replace x and y using the identities you show.

For example, if x = t and y = t², x² - y = t² - t² = 0,
so r²cos²(theta) - rsin(theta) = 0.

This is equivalent to rcos²(theta) - sin(theta) = 0, or r = sin(theta)/cos²(theta), so here we have r as a function of theta. Eliminating one factor of r is legitimate in this case since there is at least one value of theta for which sin(theta)/cos²(theta) = 0 (namely theta = 0, and others), so we haven't lost any solutions by getting rid of the factor of r.

There is one conversion formula that you didn't show, that is sometimes useful: theta = tan^-1(y/x).

 

[HallsofIvy]

But where are your parametric equations you want to convert to polar coordinates? $x= rcos(\theta)$ and $y= r sin(\theta)$ are NOT parametric equations- they are true for all points in the plane.

 

[Zach Knight]

The thing is, I don't have the parametric equations; I'm trying to find them via a differential equation. I'm trying to formulate
$$\frac{d^2\vec{r}}{dt^2}=\frac{-MG}{|r|^3}\vec{r}$$
in terms of $$\theta$$ because the only way I could find to solve the above equation was to assume $$|r|$$ was a constant. I think I have it now though. I rewrote the problem in terms of the unit vector
$$\hat{r}=cos(\theta)\hat{i}+sin(\theta)\hat{j}$$
and got
$$(\frac{d^2r}{dt^2}-r(\frac{d\theta}{dt})^2)\hat{r}+(2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2})\frac{d\hat{r}}{dt} = \frac{-MG}{r^2}\hat{r}$$,
a differential equation I think I can actually solve.
Sorry if any of the math or notation is wrong; I haven't had a calculus class yet, so my education isn't very formalized yet.