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Thread: Parametric equations

  1. #1
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    Parametric equations

    The curve C has parametric equations

    $\displaystyle x=cos^3\theta$

    $\displaystyle y=sin^3\theta$

    $\displaystyle 0<=\theta.=\frac{\pi}{2}$

    a) find the arc length of C.

    b) The curve C is rotated through 4 right angles about the x-axis. find the area of the curve surface generated.


    OK so I've learned everything now for FP3 it's just a case of putting it into practice. Also my notes are slightly hard to understand in some places lol.

    part A i would think differentiate x, then differentiate y, then form dy/dx

    getting

    $\displaystyle \frac{dy}{dx}=\frac{-3cos^2\theta sin\theta}{-3sin^2\theta cos\theta}$

    thats what i did initally but looking at my notes it looks more like differentiation from first principles ... and pythagoras ... so i'm a bit stumped
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Use the arclength form for parametric equations:

    s = integral {[ (dx/dt)^2 + (dy/dt)^2]^1/2 dt}

    integrated from 0 to pi/2
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by djmccabie View Post
    The curve C has parametric equations

    $\displaystyle x=cos^3\theta$

    $\displaystyle y=sin^3\theta$

    $\displaystyle 0<=\theta.=\frac{\pi}{2}$

    a) find the arc length of C.

    b) The curve C is rotated through 4 right angles about the x-axis. find the area of the curve surface generated.


    OK so I've learned everything now for FP3 it's just a case of putting it into practice. Also my notes are slightly hard to understand in some places lol.

    part A i would think differentiate x, then differentiate y, then form dy/dx

    getting

    $\displaystyle \frac{dy}{dx}=\frac{3cos^2\theta}{-3sin^2\theta}$

    thats what i did initally but looking at my notes it looks more like differentiation from first principles ... and pythagoras ... so i'm a bit stumped

    When you have parametric equaitons the arc length is given by the fomula

    $\displaystyle L=\int_{t_0}^{t_1}\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}$

    $\displaystyle \int_{0}^{\pi/2}\sqrt{(3cos^2(\theta)(-sin(\theta)))^2+(3\sin^2(\theta))\cos(\theta))^2}d \theta$

    $\displaystyle 9\int\sqrt{\cos^4(\theta)\sin^2(\theta)+\sin^4(\th eta)\cos^2(\theta)}d\theta$

    $\displaystyle 9\int\sqrt{\cos^4(\theta)\sin^2(\theta)+\sin^2(\th eta)(1-\cos^2(\theta))\cos^2(\theta)}d\theta$

    $\displaystyle 9\int \sqrt{\sin^2(\theta)\cos^2(\theta)}d\theta=9\int |\sin(\theta)\cos(\theta)|d\theta$

    You can drop the absolute value becuase they are both posative on $\displaystyle \theta=[0,\frac{\pi}{2}]$
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  4. #4
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    Hello, djmccabie!

    The curve $\displaystyle C$ has parametric equations: .$\displaystyle \begin{array}{ccc}x&=&\cos^3\!\theta \\ y&=&\sin^3\!\theta\end{array} \quad 0 \leq \theta \leq \tfrac{\pi}{2}$

    a) Find the arc length of $\displaystyle C.$
    There is an Arc Length formula for parametric functions: .$\displaystyle L \;=\;\int^b_a\sqrt{\left(\frac{dx}{d\theta}\right) ^2 + \left(\frac{dy}{d\theta}\right)^2}\,d\theta $

    We have: .$\displaystyle \begin{array}{ccccccc}\dfrac{dx}{d\theta} &=& \text{-}3\cos^2\!\theta\sin\theta & \Rightarrow & \left(\dfrac{dx}{d\theta}\right)^2 &=& 9\sin^2\!\theta\cos^4\!\theta \\ \\[-4mm]
    \dfrac{dy}{d\theta} &=& 3\sin^2\!\theta\cos\theta & \Rightarrow & \left(\dfrac{dy}{d\theta}\right)^2 &=& 9\sin^4\!\theta\cos^2\!\theta \end{array}$

    Then: .$\displaystyle \left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 \;=\;9\sin^2\!\cos^4\!\theta + 9\sin^4\!\theta\cos^2\!\theta$

    . . . . . . . . . . $\displaystyle =\;9\sin^2\!\theta\cos^2\!\theta\underbrace{(\cos^ 2\!\theta + \sin^2\!\theta)}_{\text{This is 1}} \;=\;9\sin^2\!\theta\cos^2\!\theta $

    . . Hence: .$\displaystyle \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \;=\;\sqrt{9\sin^2\!\theta\cos^2\!\theta} \;=\;3\sin\theta\cos\theta $


    Therefore: .$\displaystyle L \;=\;3\int^{\frac{\pi}{2}}_0 \sin\theta\cos\theta\,d\theta \quad\hdots $ Got it?

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  5. #5
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    thanks with help i now understand my notes
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