Results 1 to 5 of 5

Math Help - Parametric equations

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    31

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Aug 2008
    Posts
    530
    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.

    x = \cos t

     y =\sin t

    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.

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

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

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

    \Rightarrow dx = - \sin t \;dt

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

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

     <br />
1+\left(\frac{dy}{dx}\right)^2= 1+\left(\frac{\cos t}{-\sin t}\right)^2<br />

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

    Surface Area

    S = 2\pi \int\limits_0^\pi  {y\sqrt {1 + \left( {\frac{{dy}}<br />
{{dx}}} \right)^2 } } .dx \hfill \\

     = 2\pi \int\limits_0^\pi  {\sin t\sqrt {\left( {\frac{1}<br />
{{\sin ^2 t}}} \right)} } .\left( { - \sin t{\text{ }}dt} \right) \hfill \\

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

    = 4\pi  \hfill \\
    Attached Thumbnails Attached Thumbnails Parametric equations-g10.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Aug 2008
    Posts
    530
    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.
     x^2  = \tan ^2 t \hfill \\

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

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

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

    Now , you recognize, what type of curve it is? Hyperbola?
    Attached Thumbnails Attached Thumbnails Parametric equations-graph32.jpg  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,660
    Thanks
    600
    Hello, tiga killa!

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

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

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

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

    . . Hence: . \begin{array}{ccccccc}-t &=& 0 & \Rightarrow & t\:=\:0 \\<br />
e^{-t} &=& 0 & \Rightarrow & \text{no solution} \\<br />
t-2 &=& 0 & \Rightarrow & t \:=\: 2 \end{array}


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

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Aug 2008
    Posts
    530
    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.
    {\text{For, derivative to be horizontal, slope  =  0, }}\frac{{dy}}<br />
{{dx}} = 0 \hfill \\ <br />

    finish it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. parametric equations, equations of plane
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 10th 2009, 02:58 AM
  2. Parametric equations to rectangular equations.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 5th 2009, 10:39 PM
  3. Replies: 3
    Last Post: December 2nd 2008, 10:54 AM
  4. Parametric equations
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 12th 2008, 07:09 AM
  5. Replies: 1
    Last Post: September 1st 2007, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum