This note covers exponential models for growth, decay and cooling, and sinusoidal models for repeating patterns such as tides, daylight hours and rotating wheels. You will learn what each parameter means in context and how to use your GDC to solve for unknowns.
Key ideas
Exponential models. A quantity that is multiplied by the same factor in each equal time step follows an exponential model. Two common forms are:
- \(f(x) = ka^x + c\), where \(a \gt 0\) and \(a \neq 1\). If \(a \gt 1\) this is growth. If \(0 \lt a \lt 1\) this is decay. The same model can be written \(f(x) = ka^{-x} + c\), where now \(a \gt 1\) gives decay.
- \(f(x) = ke^{rx} + c\), where \(e \approx 2.718\). If \(r \gt 0\) this is growth and if \(r \lt 0\) it is decay. Here \(a = e^r\) in the first form.
The parameters have meanings. The value \(k\) is the starting amount above the asymptote, because \(f(0) = k + c\). The value \(c\) sets the horizontal asymptote \(y = c\), which the graph approaches but never reaches. The growth factor \(a\) tells you the multiplier per unit of \(x\). A factor of 1.08 means 8% growth per step, and a factor of 0.85 means 15% decay per step.
Many real models have \(c = 0\) (population, investment, depreciation). A cooling object follows a model with \(c \neq 0\), where \(c\) is the surrounding temperature. A medicine in the blood usually has \(c = 0\) as the concentration falls towards zero.
To solve for an unknown exponent, for example \(24000(0.85)^t = 10000\), use the GDC to find where the graph meets a horizontal line, or use the equation solver. You can also use logarithms: take \(\ln\) of both sides and use \(\ln(a^t) = t\ln a\).
Sinusoidal models. A repeating wave pattern is modelled by \(f(x) = a\sin(b(x - c)) + d\) or \(f(x) = a\cos(b(x - c)) + d\).
- The amplitude is \(|a|\). It is half the distance between the maximum and the minimum.
- The period is \(\dfrac{2\pi}{b}\) when \(x\) is measured in radians, or \(\dfrac{360^\circ}{b}\) in degrees. It is the length of one full cycle.
- The principal axis is the horizontal line \(y = d\), halfway between the maximum and the minimum. So \(d = \dfrac{\max + \min}{2}\).
- The phase shift \(c\) moves the graph \(c\) units to the right.
To build a model from information, find the amplitude and the principal axis from the highest and lowest values, find \(b\) from the period, then choose sine or cosine. A cosine curve starts at a maximum at \(x = c\), which is handy when the context begins at a high point. Without a shift, a sine curve starts on the principal axis and rises.
Check that your GDC is in the correct mode. For a model such as \(\cos\left(\dfrac{2\pi}{12.4}t\right)\) the argument is in radians, so set the GDC to radians. Because a sinusoidal equation has many solutions, use the graph to see which ones lie in the interval of the question.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(f(x) = ka^x + c\) | Exponential model. Growth if \(a \gt 1\), decay if \(0 \lt a \lt 1\). Asymptote \(y = c\). |
| \(f(x) = ke^{rx} + c\) | Exponential model with base \(e\). Growth if \(r \gt 0\), decay if \(r \lt 0\). |
| Horizontal asymptote | A line the graph gets arbitrarily close to. For \(ka^x + c\) it is \(y = c\). |
| \(\ln(a^t) = t\ln a\) | Used to solve for an exponent. Also \(\ln e^x = x\). |
| \(f(x) = a\sin(b(x - c)) + d\) | Sinusoidal model with amplitude \(|a|\), period \(\dfrac{2\pi}{b}\), shift \(c\) to the right, principal axis \(y = d\). |
| \(d = \dfrac{\max + \min}{2}\) | Height of the principal axis. |
| \(|a| = \dfrac{\max - \min}{2}\) | Amplitude. |
| Period \(= \dfrac{2\pi}{b}\) | Length of one cycle (radians). Use \(\dfrac{360^\circ}{b}\) in degrees. |
Worked examples
Example 1 A new van is worth €24 000. Its value after \(t\) years is modelled by \(V(t) = 24000(0.85)^t\). Find (a) the value after 3 years, (b) the yearly percentage loss, (c) when the value first falls below €10 000.
(a) \(V(3) = 24000 \times 0.85^3 = 24000 \times 0.614125 = 14739\). The value is about €14 700.
(b) The factor 0.85 means the van keeps 85% of its value each year, so it loses 15% each year.
(c) Solve \(24000(0.85)^t = 10000\). Divide by 24 000: \(0.85^t = \dfrac{10}{24}\).
Take natural logarithms: \(t\ln 0.85 = \ln\dfrac{10}{24}\). So \(t = \dfrac{\ln(10/24)}{\ln 0.85} = 5.387\ldots\)
Check: \(V(5) = 24000 \times 0.85^5 = 10649\), which is still above €10 000, so the value falls below €10 000 after 5.39 years.
Answer: (a) €14 700; (b) 15% per year; (c) after 5.39 years.
Example 2 A mug of tea is at 90 °C in a room at 22 °C. Its temperature after \(t\) minutes is \(T(t) = 22 + 68e^{-0.07t}\). Find (a) the temperature after 15 minutes, (b) the time to cool to 40 °C, (c) the temperature the tea approaches.
Check: \(T(0) = 22 + 68 = 90\). This matches the starting temperature.
(a) \(T(15) = 22 + 68e^{-1.05} = 22 + 68 \times 0.34993 = 45.8\ldots\) °C.
(b) Solve \(22 + 68e^{-0.07t} = 40\). Then \(68e^{-0.07t} = 18\) and \(e^{-0.07t} = \dfrac{18}{68}\).
Take \(\ln\): \(-0.07t = \ln\dfrac{18}{68}\), so \(t = \dfrac{\ln(18/68)}{-0.07} = 18.98\ldots\) minutes. A GDC graph intersection with \(y = 40\) gives the same value.
(c) As \(t\) grows, \(e^{-0.07t} \to 0\), so \(T \to 22\). The horizontal asymptote is \(T = 22\), the room temperature.
Answer: (a) 45.8 °C; (b) 19.0 minutes; (c) 22 °C.
Example 3 In a harbour, high tide has depth 7.4 m and low tide has depth 1.8 m. The time between successive high tides is 12.4 hours. High tide occurs at \(t = 0\). (a) Find a model of the form \(d(t) = a\cos(bt) + k\). (b) Find the depth at \(t = 3\). (c) Boats need at least 5 m of water. For how long in each cycle can they enter?
(a) Amplitude \(a = \dfrac{7.4 - 1.8}{2} = 2.8\). Principal axis \(k = \dfrac{7.4 + 1.8}{2} = 4.6\). Period 12.4, so \(b = \dfrac{2\pi}{12.4} = 0.5067\ldots\)
A cosine curve starts at its maximum, which suits high tide at \(t = 0\). The model is \(d(t) = 2.8\cos(0.5067t) + 4.6\).
(b) With the GDC in radian mode, \(d(3) = 2.8\cos(1.520) + 4.6 = 4.74\ldots\) m.
(c) Solve \(2.8\cos(0.5067t) + 4.6 = 5\), so \(\cos(0.5067t) = \dfrac{0.4}{2.8}\).
The first positive solution is \(t = 2.817\ldots\) hours. By symmetry about \(t = 0\), the depth is at least 5 m from \(t = -2.817\) to \(t = 2.817\).
The time is \(2 \times 2.817 = 5.634\) hours.
Answer: (a) \(d(t) = 2.8\cos(0.507t) + 4.6\); (b) 4.74 m; (c) 5.63 hours per cycle.
Common errors
- Leaving the GDC in degree mode when the model uses radians, which gives wrong values for every sine or cosine.
- Reading the amplitude as the full height from minimum to maximum. The amplitude is half of it.
- Writing the period as \(b\) instead of \(\dfrac{2\pi}{b}\). The value \(b\) is the number of cycles in \(2\pi\) units, not the length of a cycle.
- Treating a 15% decrease as a factor of 0.15. The factor is \(1 - 0.15 = 0.85\).
- Forgetting that \(c\) in \(ka^x + c\) is the asymptote, so the starting value is \(k + c\), not \(k\).
Exam tip: After building a model, test it with one data point from the question, such as the starting value. If it fails, find the error before you answer the later parts.
Check yourself
- For \(f(x) = 3(0.5)^x + 2\), state the horizontal asymptote and the value of \(f(0)\).
Answer
The asymptote is \(y = 2\). \(f(0) = 3(1) + 2 = 5\). - A town has population \(P(t) = 1200e^{0.03t}\) after \(t\) years. How long does it take to double?
Answer
Solve \(e^{0.03t} = 2\). Then \(0.03t = \ln 2\), so \(t = \dfrac{0.6931}{0.03} = 23.1\) years. - For \(y = 3\sin(4x) + 7\) with \(x\) in radians, state the amplitude, period, and the minimum and maximum values.
Answer
Amplitude 3. Period \(\dfrac{2\pi}{4} = \dfrac{\pi}{2} \approx 1.57\). The minimum is \(7 - 3 = 4\) and the maximum is \(7 + 3 = 10\). - The height in metres of a point on a Ferris wheel is \(h(t) = 15 - 12\cos\left(\dfrac{\pi t}{20}\right)\), with \(t\) in seconds. Find the period and the first time the height is 21 m.
Answer
Period \(= \dfrac{2\pi}{\pi/20} = 40\) s. Solve \(15 - 12\cos\left(\dfrac{\pi t}{20}\right) = 21\), so \(\cos\left(\dfrac{\pi t}{20}\right) = -0.5\). The first positive solution has \(\dfrac{\pi t}{20} = \dfrac{2\pi}{3}\), so \(t = \dfrac{40}{3} = 13.3\) s.