The approach model Problem set A One question at a time

In the model \(D=a+bk^{-t}\) with \(a=15\), the starting value is \(D(0)=3\). Find b.

• \(D(0)=a+b=3\)
• \(b=3-15\)
• b = −12

A model is \(D=9+bk^{-t}\) with \(k>1\), \(D(0)=0\) and \(D(1)=7.2\).

1. (a) Find b.

   • \(D(0)=9+b=0\)
   • b = −9

2. (b) Find k.

   • \(9-9k^{-1}=7.2\), so \(\dfrac{9}{k}=1.8\)
   • k = 5

A model is \(D=20-20\times1.4^{-t}\), with D in km and t in hours. Find the distance still to travel after 5 hours.

• \(D(5)=20-20\times1.4^{-5}=16.28\)
• Remaining = 20 − 16.28
• 3.72 km

The graph shows \(D=a+bk^{-t}\) with \(k>1\). The dashed line is \(D=10\). Explain why a is the total distance travelled, and write its value.

• 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 = 10

A hiker walks towards a mountain hut 6 km from the start of the trail. Her distance from the start after t hours is D km, where \(D=a+bk^{-t}\) and \(k>1\). She starts at the trailhead, so \(D(0)=0\), and after 1 hour \(D(1)=3.6\).

1. (a) State the value of a.

   • a is the total distance: 6

2. (b) Find b.

   • \(D(0)=6+b=0\)
   • b = −6

3. (c) Find k.

   • \(6-6k^{-1}=3.6\)
   • \(6\div k=2.4\), k = 2.5

4. (d) Find the distance still to go after 3 hours.

   • \(D(3)=6-6\times2.5^{-3}\)
   • Remaining = \(6\times2.5^{-3}\) = 0.384 km

5. (e) Find the time, to the nearest hour, when D = 5.9.

   • \(6-6\times2.5^{-t}=5.9\)
   • Solve for t: t = 4.47
   • 4 hours

A submersible sinks towards a wreck. Its depth below the surface after t hours is D m, where \(D=12+bk^{-t}\) with \(k>1\). It starts at the surface, and after 2 hours its depth is 10.08 m.

1. (a) Find k.

   • \(D(0)=0\) gives \(b=−12\)
   • \(D(2)=12-12k^{-2}=10.08\), so \(k^{2}=6.25\)
   • k = 2.5 (k > 1)

2. (b) Find the time, to the nearest hour, when D = 11.5.

   • \(12-12\times2.5^{-t}=11.5\)
   • Solve for t: t = 3.47
   • 3 hours