calculus

Notice that the curve given by the parametric equations x=25−t^2 y=t^3−16t is symmetric about the
x-axis. (If t gives us the point (x,y),then −t will give (x,−y)).
At which x value is the tangent to this curve horizontal? x = ?
At which t value is the tangent to this curve vertical? t = ?
The curve makes a loop which lies along the x-axis. What is the total area inside the loop? Area = ?

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  1. for the horizontal tangent, you want dy/dx = 0

    dy/dt = 3t^2-16
    dx/dt = -2t

    dy/dx = (3t^2-16)/(-2t)
    dy/dx = 0 when t=±4/√3
    That is, where x = 59/3

    for the vertical tangent, you want dx/dy = 0, or t=0

    The loop is formed where y=0. That is, t(16t^2-1) = 0
    Or, t=0 or ±4

    So, the area inside the loop where -4 <= t <= 4 is

    ∫[-4,4] y(t) dx/dt dt
    = ∫[-4,4] (3t^2-16)(-2t) dt
    = 256

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