Math Help - parametric position function and vectors

1. parametric position function and vectors

I am reviewing for the AP Calc test and I came across this very confusing a challenging problem (confusing to me but maybe not to you). Please explain how to get the answer thanks.

$R=3cos\frac{\pi}{3}ti+2sin\frac{\pi}{3}tj$ is the (position) vector xi + yj from the origin to a moving point P(x,y) at time t.

a. a single equation in x and y for the path of the point is:
(A) $x^2+y^2=13$
(B) $9x^2+4y^2=36$
(C) $2x^2+3y^2=13$ (D) $4x^2+9y^2=1$
(E) $4x^2+9y^2=36$

b. when t= 3, the speed of the particle is?

c. The magnitude of the acceleration when t=3 is?

d. at the point where $t=\frac{1}{2}$ , the slope of the curve along which the particle moves is?

2. Hello, yoman360!

Here's the first part . . .

$R\:=\:\left(3\cos\tfrac{\pi}{3}t\right)i +\left(2\sin\tfrac{\pi}{3}t\right)j$ .is the position vector $xi + yj$

. . from the origin to a moving point $P(x,y)$ at time $t.$

a. A single equation in $x$ and $y$ for the path of the point is:

$(A)\;x^2+y^2\:=\:13\qquad (B)\;9x^2+4y^2\:=\;36 \qquad
$4x^2+9y^2\:=\:1 \qquad (E)\;4x^2+9y^2\:=\:36$
We have: . $\begin{array}{cccccccc}x &=& 3\cos\frac{\pi}{3}t & \Longrightarrow & \dfrac{x}{3} &=& \cos\frac{\pi}{3}t & [1] \\ \\[-3mm]
y &=& 2\sin\frac{\pi}{3}t & \Longrightarrow & \dfrac{y}{2} &=& \sin\frac{\pi}{3}t & [2] \end{array}$

Square [1] and [2]: . $\begin{array}{ccc}\dfrac{x^2}{9} &=& \cos^2\!\frac{\pi}{3}t \\ \\[-3mm] \dfrac{y^2}{4} &=& \sin^2\!\frac{\pi}{3}t \end{array}$

. . . . . . .Add: . . $\frac{x^2}{9} + \frac{y^2}{4} \;=\;\underbrace{\cos^2\!\frac{\pi}{3}t + \sin^2\!\frac{\pi}{3}t}_{\text{This is 1}}$

. . . . . . . . . . . . $\frac{x^2}{9} + \frac{y^2}{4} \;=\;1$

Multiply by 36: . $4x^2 + 9y^2 \;=\;36$ . . . answer {E}

3. Originally Posted by yoman360
$R=3cos\frac{\pi}{3}ti+2sin\frac{\pi}{3}tj$ is the (position) vector xi + yj from the origin to a moving point P(x,y) at time t.

b. when t= 3, the speed of the particle is?

c. The magnitude of the acceleration when t=3 is?

d. at the point where $t=\frac{1}{2}$ , the slope of the curve along which the particle moves is?
$x(t) = 3\cos\left(\frac{\pi t}{3}\right)
$

$y(t) = 2\sin\left(\frac{\pi t}{3}\right)$

(b) speed, $|v(t)| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

(c) magnitude of acceleration, $|a(t)| = \sqrt{\left(\frac{d^2x}{dt^2}\right)^2 + \left(\frac{d^2y}{dt^2}\right)^2}
$

(d) slope ... $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$