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IB Mathematics: Applications and Interpretation SL · 12.5: Functions · 12.5.5

The approach model Problem set B One question at a time

Try each question on paper first. Open the working a line at a time, then say whether you had it.

Skills

  1. In the model \(D=a+bk^{-t}\) with \(a=18\), the starting value is \(D(0)=4\). Find b.

    • \(D(0)=a+b=4\)
    • \(b=4-18\)
    • b = −14

  2. A model is \(D=14+bk^{-t}\) with \(k>1\), \(D(0)=0\) and \(D(1)=12.6\).

    1. (a) Find b.

      • \(D(0)=14+b=0\)
      • b = −14

    2. (b) Find k.

      • \(14-14k^{-1}=12.6\), so \(\dfrac{14}{k}=1.4\)
      • k = 10

  3. A model is \(D=30-30\times1.3^{-t}\), with D in km and t in hours. Find the distance still to travel after 6 hours.

    • \(D(6)=30-30\times1.3^{-6}=23.78\)
    • Remaining = 30 − 23.78
    • 6.22 km

  4. The graph shows \(D=a+bk^{-t}\) with \(k>1\). The dashed line is \(D=15\). Explain why a is the total distance travelled, and write its value.

    24681051015t (hours)D (km)D
    • As t increases, \(k^{t}\) grows, so \(k^{-t}\to0\)
    • Then \(D\to a\): the curve levels off at the dashed line and never passes it
    • So a is the total distance: a = 15

In context

  1. A skateboarder rolls towards a barrier 9 m from the start of a ramp. Her distance from the start after t hours is D m, where \(D=a+bk^{-t}\) and \(k>1\). She starts at the top, so \(D(0)=0\), and after 1 hour \(D(1)=6.3\).

    1. (a) State the value of a.

      • a is the total distance: 9

    2. (b) Find b.

      • \(D(0)=9+b=0\)
      • b = −9

    3. (c) Find k.

      • \(9-9k^{-1}=6.3\)
      • \(9\div k=2.7\), \(k=\dfrac{10}{3}=3.33\) to 3 s.f.

    4. (d) Find the distance still to go after 4 hours.

      • \(D(4)=9-9\times\left(\dfrac{10}{3}\right)^{-4}\)
      • Remaining = \(9\times\left(\dfrac{10}{3}\right)^{-4}\) = 0.0729 m

    5. (e) Find the time, to the nearest hour, when D = 8.8.

      • \(9-9\times\left(\dfrac{10}{3}\right)^{-t}=8.8\)
      • Solve for t: t = 3.16
      • 3 hours

Stretch

  1. A weather balloon rises towards a cloud layer. Its height above the launch point after t hours is D m, where \(D=15+bk^{-t}\) with \(k>1\). It starts at the launch point, and after 2 hours its height is 11.25 m.

    1. (a) Find k.

      • \(D(0)=0\) gives \(b=−15\)
      • \(D(2)=15-15k^{-2}=11.25\), so \(k^{2}=4\)
      • k = 2 (k > 1)

    2. (b) Find the time, to the nearest hour, when D = 14.5.

      • \(15-15\times2^{-t}=14.5\)
      • Solve for t: t = 4.91
      • 5 hours

jmaths.xyz · AI SL · 12.5.5 Review: the approach model and graph transformations