Algebra 101 - Unit 13

Lesson 81: Applications of Nonlinear Systems

Nonlinear systems are used to model a variety of real-world problems including geometry, physics, economics, and engineering. Solutions often represent intersection points of curves such as parabolas, circles, and ellipses.

Steps for Solving Word Problems

1. Identify the variables and relationships in the problem.
2. Write a nonlinear equation for one relationship and a linear or nonlinear equation for the other.
3. Solve algebraically (substitution or elimination) or graphically.
4. Interpret the solution in the context of the problem.
5. Check the solution in both equations.

Examples

Example 1 — Projectile Motion:

A ball is thrown and its height is given by y = −x² + 6x. A platform is located at y = 5. At what horizontal distance does the ball hit the platform?

Solve: −x² + 6x = 5 → x² − 6x + 5 = 0 → (x − 5)(x − 1) = 0 → x = 1, 5 The ball hits the platform at x = 1 and x = 5 meters.

Example 2 — Geometry (Circle and Line):

Find the points of intersection between the circle x² + y² = 16 and the line y = x + 2.

Substitute y = x + 2 → x² + (x + 2)² = 16 → x² + x² + 4x + 4 = 16 → 2x² + 4x − 12 = 0 → x² + 2x − 6 = 0
x = [−2 ± √(4 + 24)]/2 = [−2 ± √28]/2 = [−2 ± 2√7]/2 = −1 ± √7
y = x + 2 → y = 1 ± √7 → Solutions: (−1 + √7, 1 + √7), (−1 − √7, 1 − √7)

Example 3 — Economics (Profit Optimization):

A company’s profit is modeled by P(x) = −2x² + 20x − 30. Break-even occurs when P(x) = 0. Solve for x:

−2x² + 20x − 30 = 0 → 2x² − 20x + 30 = 0 → x² − 10x + 15 = 0
x = [10 ± √(100 − 60)]/2 = [10 ± √40]/2 = [10 ± 2√10]/2 = 5 ± √10

Practice Problems

1. A ball is thrown: y = −x² + 8x. A platform is at y = 6. Find x-coordinates where it hits.
2. Find intersection points: x² + y² = 25, y = 2x − 1
3. A farmer has a circular field x² + y² = 36 and a straight path y = x + 2. Where does the path intersect the field?
4. Profit model: P(x) = −x² + 12x − 20. Find the break-even points.
5. Intersection of parabola and line: y = x² − 4x + 3, y = x + 1