Unit 4 Topic 4.2

Parametric Functions Modeling Planar Motion

see every function move

3 min readfreeNot practiced

Beyond the exam. The AP® Precalculus exam does not test this unit. Many classes teach it, and it prepares you for calculus.

A parametric function is a natural way to describe something moving around a flat surface: a puck on ice, a robot on a floor, a ball in flight seen from the side. The parameter is time, and the two coordinates tell you where the object is at each moment. This page shows how to read the motion, one direction at a time.

Position at time tt

In planar motion an object's position at time tt is the point (x(t),y(t))(x(t), y(t)). The function x(t)x(t) tracks the horizontal position and y(t)y(t) tracks the vertical position. The two can be described, graphed and analyzed separately, then put back together.

On an interval of timeThe particle moves
x(t)x(t) increasingright
x(t)x(t) decreasingleft
y(t)y(t) increasingup
y(t)y(t) decreasingdown

Combine one row from each pair. If x(t)x(t) is decreasing and y(t)y(t) is increasing, the particle moves left and up.

Turning points

Where x(t)x(t) changes from increasing to decreasing, the particle stops going right and starts going left: that point is a relative maximum of x(t)x(t), the particle's farthest right position nearby. In the same way a maximum of y(t)y(t) is a highest point and a minimum is a lowest point.

A turning point of one coordinate is not a stop. At a maximum of x(t)x(t), if y(t)y(t) is still changing, the particle is moving straight up or down for that moment. It is at rest only when neither coordinate is changing.

Worked example 1. A particle moves with x(t)=t3−6t2+9tx(t) = t^3 - 6t^2 + 9t and y(t)=4t−t2y(t) = 4t - t^2 for 0≤t≤40 \le t \le 4. Describe its motion.

Horizontal. From a table or graph, x(0)=0x(0) = 0, x(1)=4x(1) = 4, x(3)=0x(3) = 0, x(4)=4x(4) = 4, and x(t)x(t) increases on (0,1)(0, 1), decreases on (1,3)(1, 3) and increases on (3,4)(3, 4). So the particle moves right, then left, then right.

Vertical. y(t)=−(t−2)2+4y(t) = -(t - 2)^2 + 4 increases on (0,2)(0, 2) and decreases on (2,4)(2, 4). So it moves up, then down.

Together. On (0,1)(0, 1): right and up. On (1,2)(1, 2): left and up. On (2,3)(2, 3): left and down. On (3,4)(3, 4): right and down. The highest point is at t=2t = 2: (x(2),y(2))(x(2), y(2))=(8−24+18,4)= (8 - 24 + 18, 4)=(2,4)= (2, 4).

The path of a particle with x(t) = t cubed minus 6t squared plus 9t and y(t) = 4t minus t squared for t from 0 to 4. It starts at (0, 0) at t = 0, moves right and up to (4, 3) at t = 1, turns and moves left through (2, 4) at t = 2, the highest point, to (0, 3) at t = 3, then turns again and moves right and down to (4, 0) at t = 4, crossing its earlier path near (2, 1).
Figure 1. The path of Worked example 1. The particle turns around in xx at (4,3)(4, 3) and (0,3)(0, 3) but keeps moving vertically there. LevlPrep original diagram.

Motion with a calculator: a thrown ball

Seen from the side, a ball thrown from a height of 1.5 meters can be modeled with x(t)=15tx(t) = 15t meters across and y(t)y(t)=−4.9t2+12t+1.5= -4.9t^2 + 12t + 1.5 meters up, tt seconds after the throw. The horizontal position grows steadily; the height is a parabola in tt.

Worked example 2. When does the ball land, and how far away?

It lands when y(t)=0y(t) = 0. Graph y=−4.9x2+12x+1.5y = -4.9x^2 + 12x + 1.5 and find its zeros: t≈−0.119t \approx -0.119 and t≈2.568t \approx 2.568. A time before the throw makes no sense, so the ball lands at t≈2.568t \approx 2.568 seconds. Store that value, then compute xx from the stored value: xx≈15(2.568177...)\approx 15(2.568177...)≈38.523\approx 38.523 meters.

The model describes the ball only for 0≤t≤2.5680 \le t \le 2.568. After that the formula gives negative heights, which a ball on the ground cannot have.

Two graphs and one path

Three pictures go with planar motion: the graph of x(t)x(t) against tt, the graph of y(t)y(t) against tt, and the path in the xyxy-plane. The first two tell you when each coordinate rises or falls, so they answer "which way is it moving?". The path answers "where does it go?". Read the direction questions from the graphs against tt, then check your answer on the path.

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