Tangent and concavity of parametric equations

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  1. Tangent and Concavity of Parametric Equations Overview
  1. Find   dydx  \;\frac{dy}{dx}\; and   d2ydx2\;\frac{d^2y}{dx^2}
    1. x=tt2x=t-t^2 , y=3+ty=3+t
    2. x=et x=e^t , y=ety=e^{-t}
  2. Questions Regarding to Tangents and Concavity
    Find the tangent to the cycloid x=r(θsinθ)x=r(\theta - \sin \theta), y=r(1cosθ)y=r(1-\cos \theta) when θ=π4\theta = \frac{\pi}{4} and rr > 00. Determine the concavity for all values of θ\theta. (Do not eliminate the parameter)
    1. Find the point of the parametric curve x=t2+1x=t^2+1 and y=t3+t2y=t^3+t^2, in which the tangent is horizontal.
      1. Find the tangent to the curve x=3costx=3 \cos t, y=4costy=4 \cos t by:
        1. Eliminating the parameter.
        2. Without eliminating the parameter.
      Topic Notes
      In this lesson, we will focus on finding the tangent and concavity of parametric equations. Just like how we can take derivatives of Cartesian equations, we can also do it for parametric equations. First, we will learn to take the derivatives of parametric equations. Then we will look at an application which involves finding the tangents and concavity of a cycloid. After, we will look at special cases of finding a point with a horizontal tangent. Lastly, we will compare the difference of finding tangents by eliminating and without eliminating the parameter.
      We can find the tangent (or derivative) without having to eliminate the parameter tt by using the equation:
      dydx=dydtdxdt   \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \; where   dxdt0\;\frac{dx}{dt} \neq0
      The horizontal tangent occurs when   dydt=0  \;\frac{dy}{dt} =0\; given that   dxdt0\;\frac{dx}{dt} \neq0.
      The vertical tangent occurs when   dxdt=0  \;\frac{dx}{dt} =0\; given that   dydt0\;\frac{dy}{dt} \neq0.

      To find the concavity (or second derivative), we use the following equation: