For a set of topic=3/30parametric equations/topic:

(x=f(t), y=g(t) )

The derivative can be found by differentiating each equation with respect to (t), and then dividing.

(\dfrac{dy}{dx} = \dfrac{dy}{dt} ÷ \dfrac{dx}{dt} = \dfrac{dy}{\cancel{dt}} × \dfrac{\cancel{dt}}{dx} )

(\implies \boxed{\dfrac{dy}{dx} = \dfrac{g'(t)}{f'(t)}} )

[b]uEquations of tangents and normals[/u]/b

Once the gradient function (\dfrac{dy}{dx} ) is found, the equation of the tangent or normal at any specific point on the curve can be found in the topic=7/76usual way/topic.

[b]uTurning points[/u]/b

Turning points can be found in the usual way, by solving (\dfrac{dy}{dx} = 0 ).

The solution to the equation (\dfrac{dy}{dx} = 0 ) will be in terms of the parameter (e.g. (t)).

Substituting (t) in the original parametric equation and solving will provide the (x)- and (y)-coordinate of the turning point(s).

[b]uParametric differentiation[/u]/b

(x=f(t), y=g(t) )

(\dfrac{dy}{dx} = \dfrac{g'(t)}{f'(t)} )