Plane motion w/ parametric equations

**aervid** · May 12th 2009, 10:05 PM

Can anyone help me specifically with isolating t and getting the cartesian form of equations? Such as in 1/2sint and 1/2cost.

Also, can some explain the basics of plane motion?

Any help would be greatly appreciated

**HallsofIvy** · May 13th 2009, 06:13 AM

Originally Posted by aervid

Can anyone help me specifically with isolating t and getting the cartesian form of equations? Such as in 1/2sint and 1/2cost.

Also, can some explain the basics of plane motion?

Any help would be greatly appreciated

The basic idea is to eliminate t from the equations x= f(t), y= g(t), however you can. Here, assuming you mean x= (1/2)sin t (as opposed to 1/(2sint)) and y= (1/2)cos t, then $x^2+ y^2= (1/4)sin^2 t+ (1/4) cos^2(t)$ $= (1/4)(sin^2 t+ cos^2 t)= 1/4$ because $sin^2(t)+ cos^2(t)= 1$ for all t. You should be able to recognize $x^2+ y^= 1/4$ as a circle with center at (0,0) and radius 1/2.

You question about the "basics of plane motion" is very general. If x= f(t) and y= g(t), then the "position vector" is $\vec{r}(t)= f(t)\vec{i}+ g(t)\vec{j}$ . Assuming t is "time", then the "velocity vector" is $\vec{v}(t)= \frac{d\vec{r}}{dt}= \frac{df}{dt}\vec{i}+ \frac{dg}{dt}\vec{j}$ and the acceleration vector is $\vec{a}(t)= \frac{d\vec{v}}{dt}= \frac{d^2f}{dt^2}\vec{i}+ \frac{d^2g}{dt^2}\vec{j}$ .

If you are given some force function, $\vec{F}$ then you also need to use the equation $\vec{F}= m\vec{a}$ .

**aervid** · May 13th 2009, 05:53 PM

Oh ok so its basically the same as linear motion involving S(t), V(t), and A(t). And then to get the direction of the V(t) and A(t) we divide j by i right?

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