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.

Define logarithmic axis .

An axis on which the logarithm of a quantity is plotted, rather than the quantity itself. A graph with at least one such axis is called a log graph.

A LevelMathsEdexcelPureFlashcardsExponentials & LogarithmsModelling with Exponentials & Logarithms

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

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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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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.
Why plot $\ln y$ against $t$ instead of $y$ against $t$ ?
Because it turns an exponential curve into a straight line, which is far easier to fit and to read values from.
It also compresses a wide range of values onto a manageable scale.
Define logarithmic axis.
An axis on which the logarithm of a quantity is plotted, rather than the quantity itself.
A graph with at least one such axis is called a log graph.
Taking $\ln$ of both sides of $y = A e^{k t}$ gives:
$\ln y =______+______$
$\ln y = k t + \ln A$
That is $y = m x + c$ with $\ln y$ as the vertical variable, so plotting it against $t$ gives a straight line.
On a graph of $\ln y$ against $t$ for $y = A e^{k t}$ , what do the gradient and intercept give you?
The gradient is $k$ , and the vertical intercept is $\ln A$ .
So $A$ is found by raising $e$ to the power of that intercept, not by reading it off directly.
How do you straighten a model of the form $y = A b^{k x}$ ?
Take logarithms of both sides, which brings the power down: $\ln y = k x \ln b + \ln A$ .
Plotting $\ln y$ against $x$ then gives a straight line, this time of gradient $k \ln b$ .
True or False?
Plotting $\ln y$ against $\ln t$ would also straighten $y = A e^{k t}$ .
False.
An exponential model straightens when $\ln y$ is plotted against $t$ itself.
Plotting $\ln y$ against $\ln t$ straightens a different kind of model, one in which $t$ is raised to a power.

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