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Geometry Parametric equations
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Starter: challenge
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Parametric equations: Cartesian equations
KUS objectives BAT convert between parametric and Cartesian equations of a function Starter: previous page Geogebra: parametric eqns
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WB5 Parametric eqns β cartesian eqn example I
Find the Cartesian equations of the following curves a) π₯=π‘β1 π¦= 2π‘ 2 +3π‘ π‘ββ b) π₯= π‘ 2 π¦= 2π‘ 2 +3π‘ π‘β₯0
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WB6 find the Cartesian equations of these (eliminate the parameters)
answers π₯=4π‘ , π¦=3βπ‘ π₯=π‘β1 ,π¦= π‘ 2 +1 π₯=2 π‘ 2 , π¦=4(π‘β1) π₯=π‘+2 , π¦= 1 π‘ π₯= π‘ 2 β1 , π¦= π‘ 2 +1 π₯= π‘ 2 β1 , π¦= π‘ 4 +1 i) ii) iii) iv) v) vi)
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WB 7 Draw the curve given by the Parametric Equations: x= 1 π‘+1 π¦= π‘ 2 for β3β€π‘β€3
-3 -2 -1 1 2 3 x = 1/(t+1) -1/2 -1 ο₯ 1 1/3 y = t2 9 4 1 1 4 9 Hmmmmβ¦ is this enough to sketch this graph? equation of the curve? π₯= 1 π‘+1 π¦= π‘ 2 π¦= 1 π₯ β1 2 π₯(π‘+1)=1 π‘+1= 1 π₯ π¦= 1 π₯ β π₯ π₯ 2 π‘= 1 π₯ β1 π¦= 1βπ₯ π₯ 2 π¦= (1βπ₯) π₯ 2 2
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WB8 Trigonometric Parametric eqns β cartesian eqn
Find the Cartesian equation of the following curve π₯= π ππ π‘ π¦= cos π‘ βπβ€π‘β€ π
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A curve has Parametric equations: π₯=π πππ‘+2 π¦=πππ π‘β3
WB 9 A curve has Parametric equations: π₯=π πππ‘ π¦=πππ π‘β3 a) Find the Cartesian equation of the curve b) Sketch the curve A Cartesian equation is just an equation of a line where the variables used are x and y only π₯=π πππ‘+2 π¦=πππ π‘β3 π₯β2=π πππ‘ π¦+3=πππ π‘ How can we link sin t and cos t in an equation? π π π 2 π‘+ππ π 2 π‘β‘1 (π₯β2 ) 2 +(π¦+3 ) 2 = 1 The equation is that of a circle ο Think about where the centre will be, and its radius 5 (π₯β2 ) 2 +(π¦+3 ) 2 =1 Centre = (2, -3) Radius = 1 -5 5 -5
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Another way of writing this (by squaring the whole of each side)
WB 10 A curve has Parametric equations: π₯=π πππ‘ π¦= sin 2π‘ a) Find the Cartesian equation of the curve b) Sketch the curve π₯=π πππ‘ π¦=π ππ2π‘ double angle formula π₯ 2 =π π π 2 π‘ π¦=2π πππ‘πππ π‘ Replace sint with x π¦=2π₯πππ π‘ π π π 2 π‘+ππ π 2 π‘β‘1 b) Geogebra: parametric eqns ππ π 2 π‘=1βπ π π 2 π‘ Replace sin2t with x2 ππ π 2 π‘=1β π₯ 2 πππ π‘= 1β π₯ 2 π¦=2π₯πππ π‘ We can now replace cos t π¦=2π₯ 1β π₯ 2 Another way of writing this (by squaring the whole of each side) π¦ 2 =4 π₯ 2 (1β π₯ 2 )
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WB11 find the Cartesian equations of these (eliminate the parameters)
answers π₯=3 sin π‘ , π¦=2 cos π‘ π₯= sec π‘ ,π¦=5 tan π‘ π₯=1+ cos π‘ , π¦=1β2 sin π‘ ππ£) π₯= cos π‘ + sin π‘ , π¦=2 cos π‘ + sin π‘ π£) π₯= cos π‘+ π 4 , π¦= 2 sin π‘ π£π) π₯=2 cos π‘ β1 , π¦=3+2 sin π‘ i) ii) iii) iv) v) vi)
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Parametric equations βSummary
Parametric equations are written as: A Cartesian equation would be There are three main types of question in the exam Sketch a graph from parametric equations Eliminate (t) to find the Cartesian equation Differentiating to find gradients, tangents and normals Make a table of values β for t, x and y and plot points (x, y) βZoom inβ on any interesting points β work out more values e.g. to check asymptotes are correct Write one thing you have learned Write one thing you need to improve
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