# Math Help - vector equations

1. ## vector equations

each curve represents the motion of a particle moving in R2

x(t) = (2cos t, 3sin t) , 0 <= t <= 2Pi

x(t) = (3 sec t, 2 tan t) , (-1/3)pi <= t <= (1/3)pi

can someone sketch both these curves showing the direction of motion for me???

and find the velocity and speed of each of the particles with a step by step explanation

the speed of the first one i got as:

sqrt ( 4(sin t)^2 + 6 (cos t)^2 )
is there any way to further simplify this or should i leave it like this???

i also have to locate the points (x,y) where the velocity is horizontal, vertical or 0
i think this means when the derivative or x-component is 0, when derivative or y-component is 0, and when both are zero?
am i correct?
and can you list those points for both equations as well so i can verify them

2. Hello, razorfever!

I'll get you started on this one.

But they really shouldn't use $x$ that way . . .

Each curve represents the motion of a particle moving in R².

. . $f(t) \:= \:(2\cos t,\:3\sin t),\quad0 \leq t \leq 2\pi$
We have the parametric equations: . $\begin{array}{cccc} x\:=\: 2\cos t & {\color{blue}(1)}\\ y\:=\:3\sin t & {\color{blue}(2)}\end{array}$

We can eliminate the parameter . . .

We have: . $\begin{array}{ccccccccc}{\color{blue}(1)} & x \:=\: 2\cos t & \Rightarrow & \dfrac{x}{2} \:=\: \cos t & {\color{blue}(3)}\\ \\[-3mm]
{\color{blue}(2)} & y \:=\: 3\sin t & \Rightarrow & \dfrac{y}{3} \:=\: \sin t & {\color{blue}(4)}\end{array}$

. . $\begin{array}{cccccc}\text{Square {\color{blue}(3)}:} & \dfrac{x^2}{4} \:=\:\cos^2\!t \\ \\[-3mm]
\text{Square {\color{blue}(4)}:} & \dfrac{y^2}{9} \:=\:\sin^2\!t \end{array}$

Add: . $\frac{x^2}{4} + \frac{y^2}{9} \;=\;\underbrace{\cos^2\!t + \sin^2\!t}_{\text{This is 1}}$

And we have the ellipse: . $\frac{x^2}{4} + \frac{y^2}{9} \:=\:1$

Calcuate the points $(x,y)$ for various values of $t.$

. . . $\begin{array}{|c||c|c|}
t & x & y \\ \hline\hline
0 & 2 & 0 \\ \frac{\pi}{2} & 0 & 3 \\ \pi & \text{-}2 & 0 \\ \frac{3\pi}{2} & 0 & \text{-}3 \\ 2\pi & 2 & 0 \end{array}$

We see that the ellipse starts at (2,0) and is generated in the CCW direction.