Jo is moving linearly in the xy-plane at a constant speed. She starts from the point (3,1) and moves along the line y = −3x + 10 at a speed of 2 units per second, heading toward the y-axis.
(a) Write parametric equations for Jo’s location t seconds after starting.
In other words, express her horizontal location, x, and vertical location, y, as functions of time t.
She starts at (3,1) and moves to (x,y) while heading left and up toward the y-axis.
The horizontal distance traveled moving left is 3-x. The vertical distance traveled moving up is y-1.
Since distance/time = speed = 2 units/sec., we can write
(3-x)/t = 2 => 3-x = 2t => x = 3 - 2t.
(y-1)/t = 2 => y-1 = 2t => y = 2t + 1.
(b) Express Jo’s distance from the point (5, 8) as a function of t.
At time t, her location is (x,y). The distance between (x,y) and (5,8) is
√((x-5)² + (y-8)²)
= √((3-2t-5)² + (2t+1-8)²)
= √((-2t-2)² + (2t-7)²)
= √((4t²+8t+4) + (4t²-28t+49))
NOTE: Since we are told that she starts at (3,1), it follows that at her start time t=0, we must have x=3 and y=1. Therefore, the Haji Koeu's answer cannot be correct.