What kinds of situation are modelled by exponential growth and decay?

Growth : populations of animals or humans, and investments under compound interest. Decay : levels of radioactivity, the amount of a drug in the bloodstream, and depreciation in value.

Why does an exponential model usually apply only for a limited time?

Because unlimited growth or decay is not realistic: a population cannot increase for ever, and an old car does not become worth nothing at all. The model describes a phase of the situation rather than the whole of it.

ASMathsEdexcelPureFlashcardsExponentials & LogarithmsModelling with Exponentials & Logarithms

Modelling with Exponentials & Logarithms (Edexcel AS Maths: Pure): Flashcards

Exam code: 8MA0

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FrontExponential Growth & Decay
Exponential growth is modelled by $y = A e^{______}$ and exponential decay by $y = A e^{______}$ , where $t$ is time and $k > 0$ .

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Cards in this collection (11)

Exponential growth is modelled by $y = A e^{______}$ and exponential decay by $y = A e^{______}$ , where $t$ is time and $k > 0$ .
Exponential growth is modelled by $y = A e^{k t}$ and exponential decay by $y = A e^{- k t}$ , where $t$ is time and $k > 0$ .
Because $k$ is taken positive, the sign in the power alone tells you which of the two it is.
In $y = A e^{k t}$ , what does $A$ represent?
The initial value, that is the value of $y$ when $t = 0$ .
Substituting $t = 0$ gives $e^{0} = 1$ , so $y = A$ .
Why is $t$ used rather than $x$ in these models?
Because most of the situations being modelled are time-dependent.
The mathematics is identical; only the label changes, to match what the variable actually stands for.
Why can any exponential $a^{x}$ be rewritten as $e^{k x}$ ?
Because a suitable $k$ can always be found, namely $k = \ln a$ , since $a = e^{\ln a}$ .
Working in base $e$ throughout makes the calculus much simpler, which is why models are written that way.
True or False?
The graph of $y = \ln x$ is the reflection of $y = e^{x}$ in the line $y = x$ .
True.
They are inverse functions, and the graph of any inverse is that reflection.
So $y = \ln x$ passes through $(1, 0)$ where $y = e^{x}$ passes through $(0, 1)$ .
What does $k$ control in $y = A e^{k t}$ ?
The rate of the growth or decay: a larger $k$ means a faster change.
So $A$ fixes where the model starts and $k$ fixes how quickly it moves.
What kinds of situation are modelled by exponential growth and decay?
Growth: populations of animals or humans, and investments under compound interest.
Decay: levels of radioactivity, the amount of a drug in the bloodstream, and depreciation in value.
Why does an exponential model usually apply only for a limited time?
Because unlimited growth or decay is not realistic: a population cannot increase for ever, and an old car does not become worth nothing at all.
The model describes a phase of the situation rather than the whole of it.
If $y = A e^{k t}$ then:
$\frac{d y}{d t} =______$
$\frac{d y}{d t} = A k e^{k t}$
That is $k$ times $y$ itself, so the rate of change is proportional to the quantity, which is the defining feature of exponential behaviour.
True or False?
An exponential decay model eventually reaches zero.
False.
The curve approaches zero but never gets there, because $e^{- k t}$ stays positive for every $t$ .
In a real context that is one of the places the model stops being realistic.
How would you find how long a sample takes to halve, given $N = 2000 e^{- k t}$ ?
Set $N = 1000$ and solve $1000 = 2000 e^{- k t}$ for $t$ .
Dividing gives $e^{- k t} = 0.5$ , and taking natural logarithms brings $t$ down out of the power.

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