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Thread: parametric position function and vectors

  1. #1
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    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.

    $\displaystyle R=3cos\frac{\pi}{3}ti+2sin\frac{\pi}{3}tj$ 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) $\displaystyle x^2+y^2=13$
    (B) $\displaystyle 9x^2+4y^2=36$
    (C) $\displaystyle 2x^2+3y^2=13$ (D) $\displaystyle 4x^2+9y^2=1$
    (E) $\displaystyle 4x^2+9y^2=36$

    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 $\displaystyle t=\frac{1}{2}$ , the slope of the curve along which the particle moves is?
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  2. #2
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    Hello, yoman360!

    Here's the first part . . .


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

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

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

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


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


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

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


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

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  3. #3
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    Quote Originally Posted by yoman360 View Post
    $\displaystyle R=3cos\frac{\pi}{3}ti+2sin\frac{\pi}{3}tj$ 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 $\displaystyle t=\frac{1}{2}$ , the slope of the curve along which the particle moves is?
    $\displaystyle x(t) = 3\cos\left(\frac{\pi t}{3}\right)
    $

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

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

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

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