Calc1131W9T2 - Parameterisation

Tuesday, 1 May 2018

8:30 PM

Machine generated alternative text: Curves are usually specified in terms of a single equation. For example the equation of a circle of radius 1 is given by Every point [x, y] on the curve satisfies this relation. Alternatively, points on the curve can be specified in terms of a parameter u: [cos u, sin u] 0 < u 2m. Another possibility is the rational parametrisation: a(t) which allows you to find points on the unit circle with infinite precision by substituting t values. For example [-3/5, 4/5] [4/5, -3/5] For a general parametrised curve the slope of the tangent at some point when t c is given by dy dt And the slope of the normal line at some point when t c is the negative reciprocal of the slope of the tangent line, namely dc Hence, using the rational parametrisation of the unit circle above -6/25 -8/25 • The slope of the tangent line to the curve at [ -3/5, • The slope of the normal line to the curve at [ -3/5, 4/5] is — 4/5] is dc dt -4/3

 

Machine generated alternative text: Given the rational parametrisation of the unit circle from the previous question 1+t2 Move the t slider in the GeoGebra app below to t -2 a(t) 1.5 -1.5 a .4) 1.5 0.5 1.5 -2 t = -3.4 0.5 The equation of the tangent (displayed above in blue) at a(—3) is (-4/3) * Y The equation of the normal (displayed above in red) at a(—3) is Y 1+t2 2 (x + 4/5) - 3/5 (x + 4/5) - 3/5

 

Machine generated alternative text: Consider the curve with parametric equation The equation of the: • tangent to the curve at the point a(l) is y • normal to the curve at the point a(l) is y 3t+3,3t2 a(t) 8/3 * (x-6) + 10 -3/8 * (x-6) + 10 ,tek. By eliminating the parameter t , we find that the Cartesian equation of the curve is: Y (xA2 - + 9)/3 + 2/3 * (X-3) The curve (in black), tangent (in blue) and normal (in red) are displayed below. -4 20 18 16 12 10 8 6 4 2 -2 2 4 6 8 x 10 14 16

 

Machine generated alternative text: A projectile follows the trajectory 49 p 20 t, 200t 10 where t is in seconds and distances are in meters. Here the and y coordinates represent the horizontal and vertical displacements. Then i) The projectile's height at t = 4 is (to the nearest meter) ii) Its maximum height is (to the nearest meter) 2041 722 iii) The slope of its trajectory at t iv) The slope of its trajectory at t 4 is (as a fraction) 14 is (as a fraction) 201/25 157/50

 

Machine generated alternative text: 3 Let's return to the Folium of Descartes x + y 3xy . This curve has rational parametrisation 3t2 3t a(t) 1+t3 1+t3 Set t —3 in the GeoGebra app below. x3-3xy+Y3 _4 -2 3 2 -2 [9/26, -27/26] 2 3 4 -3) Find the point The equation of the tangent to the curve at the point a(—3) The equation of the normal to the curve at the point a(—3) Note: the Maple syntax for the point [1, 2] is [1, 2] . is -87/55 * (x-9 / 26) - 27/26 is 55/87 * (x-9 / 26) - 27/26

 

 

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