Elliptical Motion: Proving w/ Displacement & Acceleration Vectors

[danago]

Given the following displacement vector (and thus, also, acceleration vector):

$$\begin{array}{l} \overrightarrow r (t) = \left( {\begin{array}{*{20}c} {1 + 3\cos (\frac{{\pi t}}{2})} \\ {2 + 4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) \\ \overrightarrow a (t) = \left( {\begin{array}{*{20}c} {\frac{{ - 3\pi ^2 }}{4}\cos (\frac{{\pi t}}{2})} \\ { - \pi ^2 \sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) \\ \end{array}$$

I need to show that the body is undergoing eliptical motion. This is how i proceeded:
$$\begin{array}{l} \overrightarrow r (t) = \left( {\begin{array}{*{20}c} {1 + 3\cos (\frac{{\pi t}}{2})} \\ {2 + 4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} {3\cos (\frac{{\pi t}}{2})} \\ {4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) + \left( {\begin{array}{*{20}c} 1 \\ 2 \\ \end{array}} \right) \\ \therefore\left( {\begin{array}{*{20}c} {3\cos (\frac{{\pi t}}{2})} \\ {4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) = \overrightarrow r (t) - \left( {\begin{array}{*{20}c} 1 \\ 2 \\ \end{array}} \right) \\ \\ \overrightarrow a (t) = \left( {\begin{array}{*{20}c} {\frac{{ - 3\pi ^2 }}{4}\cos (\frac{{\pi t}}{2})} \\ { - \pi ^2 \sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) = \frac{{ - \pi ^2 }}{4}\left( {\begin{array}{*{20}c} {3\cos (\frac{{\pi t}}{2})} \\ {4\sin (\frac{{\pi t}}{2})} \\ \end{array}} \right) = \frac{{ - \pi ^2 }}{4}\left[ {\overrightarrow r (t) - \left( {\begin{array}{*{20}c} 1 \\ 2 \\ \end{array}} \right)} \right] \\ \end{array}$$

Is that how i should do it? By showing that the acceleration is proportional to the displacement vector?

Thanks in advance,
Dan.

 

[danago]

I just thought about using the cartesian equation of motion.

$$\begin{array}{l} x = 1 + 3\cos (\frac{{\pi t}}{2}) \\ \therefore\left( {\frac{{x - 1}}{3}} \right)^2 = \cos ^2 (\frac{{\pi t}}{2}) \\ \\ y = 2 + 4\sin (\frac{{\pi t}}{2}) \\ \therefore\left( {\frac{{y - 2}}{4}} \right)^2 = \sin ^2 (\frac{{\pi t}}{2}) \\ \\ \therefore\left( {\frac{{x - 1}}{3}} \right)^2 + \left( {\frac{{y - 2}}{4}} \right)^2 = 1 \\ \end{array}$$

Which is the cartesian equation of an ellipse. Is this method also mathematically valid?

 

[HallsofIvy]

Yes, that's the obvious way to go!

In fact, saying that "the acceleration is proporional to the displacement vector" would be saying that the acceleration vector points toward the center. That's true for circles but NOT for ellipses.

The planets move in ellipses and their acceleration is, of course, toward the sun- which is at a focus, not the center.

 

[danago]

Yes, that's the obvious way to go!

In fact, saying that "the acceleration is proporional to the displacement vector" would be saying that the acceleration vector points toward the center. That's true for circles but NOT for ellipses.

The planets move in ellipses and their acceleration is, of course, toward the sun- which is at a focus, not the center.

Thanks for the reply

In regards to my first post, how does it show that the acceleration is to the center? Wouldnt the $$- \left( {\begin{array}{*{20}c} 1 \\ 2 \\ \end{array}} \right)$$ part mean that the acceleration isn't towards the centre? Or were you referring to the comment i made about it being proportional to the displacement vector?

 

[HallsofIvy]

You are right. that $$- \left( {\begin{array}{*{20}c} 1 \\ 2 \\\end{array}} \right)$$ means it is not pointing toward the center.

You said "Is that how i should do it? By showing that the acceleration is proportional to the displacement vector?

For a vector to be proportional to another vector, i.e. a constant times the vector, they would have to be parallel or anti-parallel. I thought you were saying that you had shown that the acceleration vector was anti-parallel to the displacement vector- pointing toward the center.