# 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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