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Thread: Parametric equations

  1. #1
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    Parametric equations

    1) Let c be given parametrically by x= cost and y=sint with t[0,pi]. Rotate this curve about the x axis. Set up integral from surface area.


    2)The curve c is defined by the parametric equation x=t^3 + 3t and y= t^2 e^-t. determine any values of t where the derivative is horizontal.


    3) Consider the curve C given parametrically by x= tant and y= sect with t[-pi/2, pi/2]. eliminate the parameter to get a familiar equation.

    Do as many as you can please. I really need these
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  2. #2
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    Quote Originally Posted by tiga killa View Post
    1) Let c be given parametrically by x= cost and y=sint with t[0,pi]. Rotate this curve about the x axis. Set up integral from surface area.

    $\displaystyle x = \cos t$

    $\displaystyle y =\sin t$

    $\displaystyle x^2+y^2=\sin^2 t+\cos^2 t=1$

    This is a circle with centre (0, 0) and radius 1 unit. With the limits, [0, pi] . See graph.

    $\displaystyle x = \cos 0=1, \;y = \sin 0 =0$ so, point (1,0)

    $\displaystyle x = \cos \pi=-1, \;y = \sin \pi =0$ so, point (-1,0)

    $\displaystyle \frac{dx}{dt}=- \sin t $

    $\displaystyle \Rightarrow dx = - \sin t \;dt$

    $\displaystyle \frac{dy}{dt}=-\cos t$

    $\displaystyle \frac{dy}{dx}=\frac{\cos t}{-\sin t}$

    $\displaystyle
    1+\left(\frac{dy}{dx}\right)^2= 1+\left(\frac{\cos t}{-\sin t}\right)^2
    $

    $\displaystyle 1+\left(\frac{dy}{dx}\right)^2=\frac{1}{\sin^2 t}$

    Surface Area

    $\displaystyle S = 2\pi \int\limits_0^\pi {y\sqrt {1 + \left( {\frac{{dy}}
    {{dx}}} \right)^2 } } .dx \hfill \\$

    $\displaystyle = 2\pi \int\limits_0^\pi {\sin t\sqrt {\left( {\frac{1}
    {{\sin ^2 t}}} \right)} } .\left( { - \sin t{\text{ }}dt} \right) \hfill \\$

    $\displaystyle = - 2\pi \int\limits_0^\pi {\sin t{\text{ }}dt} \hfill \\$

    $\displaystyle = 4\pi \hfill \\ $
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  3. #3
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    Quote Originally Posted by tiga killa View Post
    1)

    3) Consider the curve C given parametrically by x= tant and y= sect with t[-pi/2, pi/2]. eliminate the parameter to get a familiar equation.
    $\displaystyle x^2 = \tan ^2 t \hfill \\$

    $\displaystyle y^2 = \sec ^2 t \hfill \\$

    $\displaystyle y^2 - x^2 = \sec ^2 t - \tan ^2 t = 1 \hfill \\$

    $\displaystyle y^2 - x^2 = 1 \hfill \\ $

    Now , you recognize, what type of curve it is? Hyperbola?
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  4. #4
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    Hello, tiga killa!

    2) The curve $\displaystyle C$ is defined by: .$\displaystyle \begin{array}{ccc}x&=&t^3 + 3t \\ \\[-4mm] y&=& t^2\, e^{-t}\end{array}$

    Determine any values of $\displaystyle t$ where the derivative is horizontal.
    The derivative is: .$\displaystyle \frac{dy}{dx} \:=\:\frac{\frac{dy}{dt}}{\frac{dx}{dt}} $

    The tangent is horizontal when $\displaystyle \frac{dy}{dx} \,=\,0$
    . . That is, when $\displaystyle \frac{dy}{dt} = 0$

    We have: .$\displaystyle \frac{dy}{dt} \;=\;-t^2e^{-t} + 2te^{-t} \:=\:0 \quad\Rightarrow\quad -t\cdot e^{-t}\cdot(t - 2) \:=\:0$

    . . Hence: .$\displaystyle \begin{array}{ccccccc}-t &=& 0 & \Rightarrow & t\:=\:0 \\
    e^{-t} &=& 0 & \Rightarrow & \text{no solution} \\
    t-2 &=& 0 & \Rightarrow & t \:=\: 2 \end{array}$


    Therefore, horizontal tangents when $\displaystyle t \;=\;0,\:2$

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  5. #5
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    Quote Originally Posted by tiga killa View Post
    2)The curve c is defined by the parametric equation x=t^3 + 3t and y= t^2 e^-t. determine any values of t where the derivative is horizontal.
    $\displaystyle {\text{For, derivative to be horizontal, slope = 0, }}\frac{{dy}}
    {{dx}} = 0 \hfill \\
    $

    finish it.
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