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

Slope fields and solution curves Problem set A Answers

Skills

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

In context

    1. (a) P: falls at first (gradient −1.5), then rises, bending towards the line through Q. Q: the straight line y = x − 2 (gradient 1 all along it). R: flat at first (gradient 0), then rises towards 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) P = 0 and P = 8
    2. (b) 6 and −10
    3. (c) It falls: the gradient is negative, so P goes down towards 8 but does not reach it
    1. (a) A decreasing curve that flattens towards T = 20
    2. (b) About 31 °C (accept 30 to 32)
    3. (c) 20 °C
    1. (a) \(N=0\) and \(N=15\)
    2. (b) \(N=15\) is stable; \(N=0\) is unstable
    3. (c) \(N=7.5\); 1.5 thousand per hour

Stretch

    1. (a) P: 4, Q: −4, R: 2
    2. (b) Q
    1. (a) At P = 0 and P = 6 the gradient is 0 for every t
    2. (b) An S-shaped curve; about 2.7 years (accept 2.4 to 3)
    3. (c) It falls and flattens towards P = 6. P = 6 is a solution curve, and solution curves do not cross

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