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IB Mathematics: Applications and Interpretation HL · AHL 9: Differential Equations · AHL 9.4

Slope fields and solution curves Problem set B Answers

Skills

1. (a) Gradients for x = 0, 1, 2. y = −1: 1, 0, −1; y = 0: 0, −1, −2; y = 1: 1, 0, −1; y = 2: 4, 3, 2
2. (b) (0, 0), (1, −1), (1, 1)
1. (a) Gradients for x = −2, −1, 0, 1, 2. y = 2: −2, 0, 2, 4, 6; y = 1: −3, −1, 1, 3, 5; y = 0: −4, −2, 0, 2, 4; y = −1: −5, −3, −1, 1, 3; y = −2: −6, −4, −2, 0, 2
2. (b) (−1, 2), (0, 0), (1, −2)
1. (a) Below y = 4 the segments rise, on y = 4 they are flat, above y = 4 they fall
2. (b) The curve falls, gets flatter, and gets closer to y = 4 without reaching or crossing it
3. (c) y = 4

1. (a) P: falls slightly at first (gradient −0.5), then rises, bending upwards away from the line through Q. Q: the straight line y = −x − 2 (gradient −1 all along it). R: falls at first (gradient −0.5), flattens, then rises, staying above the line through Q
2. (b) At each point there is only one gradient, so only one direction for a curve. Two curves that met would both have to go the same way, so they would be the same curve
1. (a) V = 0 and V = 12
2. (b) 10.8 and −18
3. (c) It falls: the gradient is negative, so V goes down towards 12 but does not reach it
1. (a) An increasing curve that flattens towards c = 25
2. (b) About 19 g/L (accept 18 to 20)
3. (c) 25 g/L
1. (a) \(R=0\) and \(R=8\)
2. (b) \(R=8\) is stable; \(R=0\) is unstable
3. (c) \(R=4\); 0.5 hundred per year

1. (a) P: 2, Q: 4, R: −2
2. (b) R
1. (a) At V = 0 and V = 4 the gradient is 0 for every t
2. (b) An S-shaped curve; about 2.4 months (accept 2.2 to 2.7)
3. (c) It falls and flattens towards V = 4. V = 4 is a solution curve, and solution curves do not cross

jmaths.xyz · AI HL · AHL 9.4 Slope fields and solution curves