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10 cards from this deck
They model relationships using a third parameter, often time.
As functions of a third variable ttt: x=f(t)x = f(t)x=f(t), y=g(t)y = g(t)y=g(t).
The horizontal distance travelled in projectile motion.
y(t)=v0sin(θ)⋅t−0.5gt2y(t) = v_0 \sin(\theta) \cdot t - 0.5gt^2y(t)=v0sin(θ)⋅t−0.5gt2.
The resulting curve is a parabola.
Set dy/dt=0dy/dt = 0dy/dt=0 and solve for ttt, then substitute in y(t)y(t)y(t).
x(t)=rcos(ωt)x(t) = r \cos(\omega t)x(t)=rcos(ωt) and y(t)=rsin(ωt)y(t) = r \sin(\omega t)y(t)=rsin(ωt).
x(t)=acos(t)x(t) = a \cos(t)x(t)=acos(t) and y(t)=bsin(t)y(t) = b \sin(t)y(t)=bsin(t).
Typically, ttt represents time in parametric equations.
They can represent complex curves difficult for Cartesian.
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