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IB Mathematics: Applications and Interpretation HL · AHL 6: Complex Numbers · AHL 6.10

Adding sinusoids Problem set B Worked solutions

Skills

  1. Write \(7\cos(5t + 1.1) + 8\cos(5t + 4)\) in the form \(A\cos(5t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

    • \(7\cos(5t + 1.1) + 8\cos(5t + 4) = \mathrm{Re}\left(7e^{(5t + 1.1)i} + 8e^{(5t + 4)i}\right)\)
    • \(= \mathrm{Re}\left(e^{5ti}(7e^{1.1i} + 8e^{4i})\right)\)
    • From the calculator: \(7e^{1.1i} + 8e^{4i} = -2.05 + 0.184i = 2.06e^{3.05i}\)
    • \(= \mathrm{Re}\left(2.06e^{(5t + 3.05)i}\right) = 2.06\cos(5t + 3.05)\)

  2. Write \(8\sin(2t - 0.6) + 5\sin(2t + 1.5)\) in the form \(A\sin(2t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

    • \(8\sin(2t - 0.6) + 5\sin(2t + 1.5) = \mathrm{Im}\left(8e^{(2t - 0.6)i} + 5e^{(2t + 1.5)i}\right)\)
    • \(= \mathrm{Im}\left(e^{2ti}(8e^{-0.6i} + 5e^{1.5i})\right)\)
    • From the calculator: \(8e^{-0.6i} + 5e^{1.5i} = 6.96 + 0.470i = 6.97e^{0.0675i}\)
    • \(= \mathrm{Im}\left(6.97e^{(2t + 0.0675)i}\right) = 6.97\sin(2t + 0.0675)\)

  3. Use complex exponentials.

    1. (a) Show that \(9\sin(t + 1.2) + 4\sin(t + 2.6) = \mathrm{Im}\left(9e^{(t + 1.2)i} + 4e^{(t + 2.6)i}\right)\).

      • \(e^{\theta i} = \cos\theta + i\sin\theta \text{, so } \mathrm{Im}\left(Ae^{\theta i}\right) = A\sin\theta\)
      • \(\mathrm{Im}(z_1 + z_2) = \mathrm{Im}(z_1) + \mathrm{Im}(z_2) = 9\sin(t + 1.2) + 4\sin(t + 2.6)\)

    2. (b) Hence write \(9\sin(t + 1.2) + 4\sin(t + 2.6)\) in the form \(A\sin(t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(9\sin(t + 1.2) + 4\sin(t + 2.6) = \mathrm{Im}\left(9e^{(t + 1.2)i} + 4e^{(t + 2.6)i}\right)\)
      • \(= \mathrm{Im}\left(e^{ti}(9e^{1.2i} + 4e^{2.6i})\right)\)
      • From the calculator: \(9e^{1.2i} + 4e^{2.6i} = -0.166 + 10.5i = 10.5e^{1.59i}\)
      • \(= \mathrm{Im}\left(10.5e^{(t + 1.59)i}\right) = 10.5\sin(t + 1.59)\)

  4. Write each expression in the form \(A\cos(x + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

    1. (a) \(9\cos x + 40\sin x\)

      • \(9\cos x + 40\sin x = \mathrm{Re}\left((9 - 40i)e^{xi}\right)\)
      • From the calculator: \(9 - 40i = 41e^{-1.35i}\)
      • \(= 41\cos(x - 1.35)\)

    2. (b) \(6\cos x - 2\sin x\)

      • \(6\cos x - 2\sin x = \mathrm{Re}\left((6 + 2i)e^{xi}\right)\)
      • From the calculator: \(6 + 2i = 6.32e^{0.322i}\)
      • \(= 6.32\cos(x + 0.322)\)

  5. The graph shows \(y = 6\cos x + 2\cos(x - 1.5)\) for \(0 \le x \le 6.5\). P is the maximum, Q is the minimum and R is the right-hand x-intercept.

    123456−9−6−3369xyPQR

    1. (a) Write \(y\) in the form \(A\cos(x + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(6\cos(x) + 2\cos(x - 1.5) = \mathrm{Re}\left(6e^{xi} + 2e^{(x - 1.5)i}\right)\)
      • \(= \mathrm{Re}\left(e^{xi}(6 + 2e^{-1.5i})\right)\)
      • From the calculator: \(6 + 2e^{-1.5i} = 6.14 - 1.99i = 6.46e^{-0.314i}\)
      • \(= \mathrm{Re}\left(6.46e^{(x - 0.314)i}\right) = 6.46\cos(x - 0.314)\)

    2. (b) Hence find the coordinates of P, Q and R.

      • \(P: x - 0.314 = 0 \Rightarrow x = 0.314,\ y = 6.46\)
      • \(Q: x - 0.314 = \pi \Rightarrow x = 3.46,\ y = -6.46\)
      • \(R: x - 0.314 = \tfrac{3\pi}{2} \Rightarrow x = 5.03,\ y = 0\)

  6. Two generators feed one lamp. At time \(t\) milliseconds the voltages, in volts, are \(V_1 = 14\cos(60t + 0.5)\) and \(V_2 = 9\cos(60t - 1)\).

    1. (a) Write the total voltage \(V(t) = V_1 + V_2\) in the form \(A\cos(60t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(14\cos(60t + 0.5) + 9\cos(60t - 1) = \mathrm{Re}\left(14e^{(60t + 0.5)i} + 9e^{(60t - 1)i}\right)\)
      • \(= \mathrm{Re}\left(e^{60ti}(14e^{0.5i} + 9e^{-1i})\right)\)
      • From the calculator: \(14e^{0.5i} + 9e^{-1i} = 17.1 - 0.861i = 17.2e^{-0.0502i}\)
      • \(= \mathrm{Re}\left(17.2e^{(60t - 0.0502)i}\right) = 17.2\cos(60t - 0.0502)\)

    2. (b) Find the greatest voltage.

      • \(\text{greatest } V = A = 17.2\) volts

In context

  1. A paddle boarder is on a lake. At time \(t\) seconds she is lifted by waves from a motor boat, \(H_1(t) = 0.6\sin(3t + 1)\text{ m}\), and from a jet ski, \(H_2(t) = 0.4\sin(3t - 0.5)\text{ m}\).

    1. (a) Write the total height \(H(t) = H_1 + H_2\) in the form \(A\sin(3t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(0.6\sin(3t + 1) + 0.4\sin(3t - 0.5) = \mathrm{Im}\left(0.6e^{(3t + 1)i} + 0.4e^{(3t - 0.5)i}\right)\)
      • \(= \mathrm{Im}\left(e^{3ti}(0.6e^{1i} + 0.4e^{-0.5i})\right)\)
      • From the calculator: \(0.6e^{1i} + 0.4e^{-0.5i} = 0.675 + 0.313i = 0.744e^{0.434i}\)
      • \(= \mathrm{Im}\left(0.744e^{(3t + 0.434)i}\right) = 0.744\sin(3t + 0.434)\)

    2. (b) Find the times in the first 6 seconds when \(H(t) = 0\).

      • \(0.744\sin(3t + 0.434) = 0 \Rightarrow 3t + 0.434 = k\pi\)
      • From the calculator, solving on \(0 \le t \le 6\): \(t = 0.902, 1.95, 3.00, 4.04, 5.09\)

    3. (c) Find the first time at which the total height is greatest.

      • \(3t + 0.434 = \tfrac{\pi}{2} \text{ (smallest positive } t)\)
      • \(t = 0.379\)

    4. (d) At that time, what percentage of the wave is due to the motor boat?

      • \(\dfrac{0.6\sin(3t + 1)}{H(t)} \text{ at } t = 0.379: \dfrac{0.506}{0.744} \times 100 = 68.1\%\)

  2. A footbridge deck sways from its mean position because of wind and walkers. At time \(t\) seconds, in centimetres, \(D_1 = 2.4\cos(0.8t)\) and \(D_2 = 1.1\cos(0.8t + 1.3)\).

    1. (a) Write \(D(t) = D_1 + D_2\) in the form \(A\cos(0.8t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(2.4\cos(0.8t) + 1.1\cos(0.8t + 1.3) = \mathrm{Re}\left(2.4e^{0.8ti} + 1.1e^{(0.8t + 1.3)i}\right)\)
      • \(= \mathrm{Re}\left(e^{0.8ti}(2.4 + 1.1e^{1.3i})\right)\)
      • From the calculator: \(2.4 + 1.1e^{1.3i} = 2.69 + 1.06i = 2.90e^{0.375i}\)
      • \(= \mathrm{Re}\left(2.90e^{(0.8t + 0.375)i}\right) = 2.90\cos(0.8t + 0.375)\)

    2. (b) Find the greatest value of \(D\).

      • \(\text{greatest } D = A = 2.90\)

    3. (c) Find the first time \(t \ge 0\) at which \(D\) is greatest.

      • \(0.8t + 0.375 = 0 \Rightarrow t = 7.39\)

Stretch

  1. Three lamps shine on a sensor. At time \(t\) seconds the light intensities, in lux, are \(S_1 = 4\cos(3t + 0.5)\), \(S_2 = 6\cos(3t - 1)\) and \(S_3 = 3\cos(3t + 2.5)\).

    1. (a) Write \(S(t) = S_1 + S_2 + S_3\) in the form \(A\cos(3t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(4\cos(3t + 0.5) + 6\cos(3t - 1) + 3\cos(3t + 2.5) = \mathrm{Re}\left(4e^{(3t + 0.5)i} + 6e^{(3t - 1)i} + 3e^{(3t + 2.5)i}\right) = \mathrm{Re}\left(e^{3ti}(4e^{0.5i} + 6e^{-1i} + 3e^{2.5i})\right)\)
      • From the calculator: \(4e^{0.5i} + 6e^{-1i} + 3e^{2.5i} = 4.35 - 1.34i = 4.55e^{-0.298i}\)
      • \(= 4.55\cos(3t - 0.298)\)

    2. (b) State the greatest value of \(S\).

      • \(\text{greatest } S = A = 4.55\)

  2. Two seismic waves add at a sensor. At time \(t\) seconds the displacements, in micrometres, are \(S_1 = 9\sin(5t)\) and \(S_2 = 4\sin(5t + 2.4)\).

    1. (a) Write \(R(t) = S_1 + S_2\) in the form \(A\sin(5t + B)\), where \(A > 0\) and \(-\pi < B \le \pi\).

      • \(9\sin(5t) + 4\sin(5t + 2.4) = \mathrm{Im}\left(9e^{5ti} + 4e^{(5t + 2.4)i}\right)\)
      • \(= \mathrm{Im}\left(e^{5ti}(9 + 4e^{2.4i})\right)\)
      • From the calculator: \(9 + 4e^{2.4i} = 6.05 + 2.70i = 6.63e^{0.420i}\)
      • \(= \mathrm{Im}\left(6.63e^{(5t + 0.420)i}\right) = 6.63\sin(5t + 0.420)\)

    2. (b) At \(t = 0.05\), what percentage of the signal is due to \(S_1\)?

      • \(S_1(0.05) = 2.23,\ R(0.05) = 4.11\)
      • \(\dfrac{2.23}{4.11} \times 100 = 54.1\%\)

jmaths.xyz · AI HL · AHL 6.10 Adding sinusoidal functions with complex exponentials