2008-01-10

Today's nonsense
- Several is three or more; a couple is two. According to Webster, a few is a small number of units or individuals -- should be three.
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Derivatives of Parametric Equations

1) A particle follows the following path:
x = 4 cos (2T)
y = 3 sin (T)
In what quadrant and in which direction is the particle moving when T = 12?

So, when T = 12, x = 1.697 and y = -1.610
We had to evaluate to find x and y at that time, and then we know what quadrant, right? Yes. So that's the 4th quadrant.

dx/dt if positive, then it's moving to the right. Then find dy/dt and figure out if it is moving up (positive) or down (negative).
dx/dt = -8sin(2T)
dy/dt = 3cos(T)
So when T = 12, what is the value of dx/dt and what is the value of dy/dt?
dx/dT @ T = 12 = -8sin(24) = 7.245
dy/dT @ T = 12 = 3cos(12) = 2.532
So it's moving up and to the right.
FYI - if you were to have graphed that graph, it's going to look like a parabola. And so, where was this thing at T = 12? It was somewhere over here. And it was moving in that direction. Where is it at T =13, 14, 15? Somewhere it bounces back. So did I draw it in reverse? Not necessarily, but only for the time period involving time period T = 12, but once you hit that edge of the parabola, it's going to go back and do it in 'reverse' again. How would you represent directionality? You would have arrows going both ways on the graph.

x = 8 cos^{^(2)}(T)
y = 6(sin(T))^{^(2)}
They are already squared. So it is not an ellipse. Without a calculator, sketch the graph. Which direction is the particle at t=60?
(x/8) + (y/6) = 1. So it's linear, but it's not necessarily a line, so it could be a ray or a line segment.
x cannot be greater than 8, because cos^2 can't be greater than one. And y cannot be greater than 6, nor less than 0.
This is going to be a line segment (directionality: show arrows going in both directions) and it's just going to be bouncing back and forth for increasing T.

To figure out the direction of the particle at T=60, take the derivatives.
dx/dt = -16costsint
dy/dt = 12sintcost
So, when T = 60, dx/dt = -4.645 and dy/dt = 3.484
And then what about dy/dx? Solve for x/8 + y/6 = 1 for dy/dx. It turns out that dy/dx = -3/4. To get this you could have done (dy/dt) / (dx/dt) to get your dy/dx.

Parametric Form of the Derivative

If a smooth curve, c, is given by the equations x = f(T) and y=g(T), then the slope of c @ (x,y) is
(dy/dx) = (dy/dT) / (dx/dT) , where dx/dt =! 0