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Math Help - urgent help....

  1. #1
    Junior Member
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    urgent help....

    Write the equation of the tangent line to the parametric curve x=sint y=sect when t=pi/4


    SET UP the integral to calculate the arc length of the curve if t∈[0,pi/3]
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  2. #2
    Senior Member Pinkk's Avatar
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    \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=  \frac{sec(t)tan(t)}{cos(t)}=\frac{sin(t)}{cos^{3}(  t)}

    \frac{dy}{dx}(\frac{\pi}{4})=\frac{\frac{\sqrt{2}}  {2}}{(\frac{\sqrt{2}}{2})^{3}}=(\frac{\sqrt{2}}{2}  )^{-2}=2

    ^That is the slope of your tangent.

    x(\frac{\pi}{4})=\frac{\sqrt{2}}{2},y(\frac{\pi}{4  })=\sqrt{2}

    The equation of your tangent line in the form y-b=m(x-a):

    y-\sqrt{2}=2(x-\frac{\sqrt{2}}{2})

    Arc length:

    \int_0^{\frac{\pi}{3}}\sqrt{[x'(t)]^{2}+[y'(t)]^{2}}\,\,\,dt

    \int_0^{\frac{\pi}{3}}\sqrt{cos^{2}(t)+sec^{2}(t)t  an^{2}(t)}\,\,\,dt
    Last edited by Pinkk; March 23rd 2009 at 02:07 PM.
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  3. #3
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    Hello, tiga killa!

    Write the equation of the tangent line to the parametric curve
    . . x\:=\:\sin t,\;\;y\:=\:\sec t \:\text{ when }t\,=\,\tfrac{\pi}{4}

    When t = \tfrac{\pi}{4} \!:\;\begin{Bmatrix}x \:=\:\sin\frac{\pi}{4} \:=\:\frac{\sqrt{2}}{2} \\ y \:=\:\sec\frac{\pi}{4} \:=\:\sqrt{2} \end{Bmatrix}\quad \Rightarrow \text{ The point is: }\:\left(\tfrac{\sqrt{2}}{2},\:\sqrt{2}\right)


    The derivatives are: . \begin{Bmatrix}\dfrac{dy}{dt} \:=\:\sec t\tan t \\ \dfrac{dx}{dt} \:=\:\cos t \end{Bmatrix} \quad\Rightarrow\quad \frac{dy}{dx} \:=\:\frac{\sec t\tan t}{\cos t} \:=\:\sec^2\!t\tan t

    When t = \tfrac{\pi}{4}\!:\;\;\frac{dy}{dx} \;=\;\sec^2\tfrac{\pi}{4}\tan\tfrac{\pi}{4} \;=\;(\sqrt{2})^2(1) \;=\;2


    Now write the equation of the line through \left(\tfrac{\sqrt{2}}{2},2\right) with slope 2.




    SET UP the integral to calculate the arc length of the curve if t \in \left[0,\,\tfrac{\pi}{3}\right]

    Formula: . L \;=\;\int^b_a\sqrt{\left(\tfrac{dx}{dt}\right)^2 + \left(\tfrac{dy}{dt}\right)^2}\,dt

    Therefore: . L \;=\;\int^{\frac{\pi}{3}}_0\sqrt{\cos^2\!t + \sec^2\!t\tan^2\!t}\:dt

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