11.7 – Parametric Equations Precalculus 11.7 – Parametric Equations

Find the equation of the given graph using a parametric equation The easiest way to write a parametric equation of a line is to set the interval to be 0 < t < 1 When t = 0, the graph needs to be at (1, 3) and when t = 1, the graph needs to be at (4, -2) So far we have x(t) = 1 + a*t y(t) = 3 + b*t; 0 < t < 1 because this way when t = 0, x = 1 and y = 3 What “a” would make x = 4 when t = 1? 4 = 1 + a*1 3 = a What “b” would make y = -2 when t = 1? -2 = 3 + b*1 -5 = b Thus the equation is x(t) = 1 + 3t y(t) = 3 – 5t; 0 < t < 1 Notice that <3, -5> is the vector from the first point to the second point!

Find the equation of the given graph using a parametric equation Try this one on your own Check by graphing on your calculator when you have your equation Possible Answer: x(t) = 4t y(t) = -2 + 5t 0 < t < 1

Find the equation of the given graph using a parametric equation This is the graph of part of a circle Notice that the center is at (-2, 1) and the radius is 5 Thus we have the equation x(t) = 5 cos(t) – 2 y(t) = 5 sin(t) + 1 In order to end up with just part of a circle we set the interval to match up with the angles (in radians) 0 < t < 3π/2 Thus the equation is x(t) = 5 cos(t) – 2 y(t) = 5 sin(t) + 1 0 < t < 3π/2

Find the equation of the given graph using a parametric equation Try this one on your own Check by graphing on your calculator when you have your equation Possible Answer: x(t) = 2 cos(t) y(t) = 6 sin(t) 0 < t < 2π