Parametric Equations Parametric vs Cartesian Equations Sketching Parametric

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Parametric Equations

Parametric Equations

Parametric vs Cartesian Equations

Parametric vs Cartesian Equations

Sketching Parametric Curves Let’s just try a few values for the parameter and see

Sketching Parametric Curves Let’s just try a few values for the parameter and see what coordinates we get. . . t -3 -2 -1 0 1 2 3 x 9 4 1 0 1 4 9 y -6 -4 -2 0 2 4 6 ? 6 4 2 O -2 -4 -6 2 ? 4 6 8

Solving Problems involving Parametric Equations Q ?

Solving Problems involving Parametric Equations Q ?

Solving Problems involving Parametric Equations Q ? ?

Solving Problems involving Parametric Equations Q ? ?

Test Your Understanding Jan 2013 a? b?

Test Your Understanding Jan 2013 a? b?

Exercise for practice 7 1 a c e ? ? ? 2 a c

Exercise for practice 7 1 a c e ? ? ? 2 a c e ? 8 ? ? ? ? 9 5 a b ? ? ? 10 ?

Solving Problems involving Parametric Equations

Solving Problems involving Parametric Equations

Solving Problems involving Parametric Equations

Solving Problems involving Parametric Equations

Solving Problems involving Parametric Equations

Solving Problems involving Parametric Equations

Converting from Parametric Cartesian We can use substitution or well known identities to convert

Converting from Parametric Cartesian We can use substitution or well known identities to convert parametric equations into Cartesian ones. This involves eliminating the parametric variable and forming a single equation. Remember, sin²Ө + cos² = 1 ? X=2 sinӨ x/2 = sinӨ (X/2)²=sin²Ө So (x/2)²=1 -cos²Ө=1 -(x/2)² y=cosӨ y²=cos²Ө Sub? in for cos²Ө y²=1 -(x/2)²

Converting from Parametric Cartesian We can use substitution or well known identities to convert

Converting from Parametric Cartesian We can use substitution or well known identities to convert parametric equations into Cartesian ones. This involves eliminating the parametric variable and forming a single equation. ? y=sinӨ y²=sin²Ө ?

Test Your Understanding Jan 2011 ?

Test Your Understanding Jan 2011 ?

Exercise for practice 3 a b ? ? 4 5 ? ? ? C

Exercise for practice 3 a b ? ? 4 5 ? ? ? C 4 Jan 2008 Q 7 ?

Parametric equations - applied Sales of Ice cream in fonction of sales of Sunscreen

Parametric equations - applied Sales of Ice cream in fonction of sales of Sunscreen y = (t - 21)2 y = x 2 x = t - 21 What can you conclude from the graph? What is the root cause? How can you find the equation y=f(t) from the parametric equations? Which information are you losing?

Parametric equations – A real life application Working out path of objects Watch from

Parametric equations – A real life application Working out path of objects Watch from 2: 27 to 3: 07 Why is this path not possible?

Parametric equations – A real life application Working out path of objects No acceleration

Parametric equations – A real life application Working out path of objects No acceleration along x 5 m/s Which equations will describe Wile E Coyote path? x(t) = 20 + 5 t 100 m No initial speed along y Acceleration = gravity = 9. 8 m/s 2 20 m 1) How will you draw precisely the real path of Wile E Coyote?

Parametric equations – A real life application x(t) = 20 + 5 t 1)

Parametric equations – A real life application x(t) = 20 + 5 t 1) How will you draw precisely the real path of Wile E Coyote? Height (m) t = 0 100 t(s) x(m) y(m) 0 1 50 2 3 4 … … 0 10 20 Distance (m)

Parametric equations – A real life application x(t) = 20 + 5 t 1)

Parametric equations – A real life application x(t) = 20 + 5 t 1) How will you draw precisely the real path of Wile E Coyote? Height (m) t = 0 100 t(s) x(m) y(m) 0 20 100 1 25 95. 1 2 30 80. 4 3 35 55. 9 4 40 21. 6 … … 0 50 10 20 Distance (m)

Parametric equations – A real life application x= 20 + 5 t 2) How

Parametric equations – A real life application x= 20 + 5 t 2) How can you work out the cartesian equation of the path? Rewrite t in terms of x, and then substitute it into the y equation… 3) Which information are you losing?