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If a quantity grows exponentially, the time it takes for the quantity to double remains constant. In other words, it takes the same amount of time for a population of bacteria to grow from $100$ to $200$ bacteria as it does to grow from $\mathrm{10,000}$ to $\mathrm{20,000}$ bacteria. This time is called the doubling time. To calculate the doubling time, we want to know when the quantity reaches twice its original size. So we have
If a quantity grows exponentially, the doubling time is the amount of time it takes the quantity to double. It is given by
Assume a population of fish grows exponentially. A pond is stocked initially with $500$ fish. After $6$ months, there are $1000$ fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches $\mathrm{10,000}.$ When will the owner’s friends be allowed to fish?
We know it takes the population of fish $6$ months to double in size. So, if t represents time in months, by the doubling-time formula, we have $6=\left(\text{ln}\phantom{\rule{0.2em}{0ex}}2\right)\text{/}k.$ Then, $k=\left(\text{ln}\phantom{\rule{0.2em}{0ex}}2\right)\text{/}6.$ Thus, the population is given by $y=500{e}^{\left(\left(\text{ln}\phantom{\rule{0.2em}{0ex}}2\right)\text{/}6\right)t}.$ To figure out when the population reaches $\mathrm{10,000}$ fish, we must solve the following equation:
The owner’s friends have to wait $25.93$ months (a little more than $2$ years) to fish in the pond.
Suppose it takes $9$ months for the fish population in [link] to reach $1000$ fish. Under these circumstances, how long do the owner’s friends have to wait?
$38.90$ months
Exponential functions can also be used to model populations that shrink (from disease, for example), or chemical compounds that break down over time. We say that such systems exhibit exponential decay, rather than exponential growth. The model is nearly the same, except there is a negative sign in the exponent. Thus, for some positive constant $k,$ we have $y={y}_{0}{e}^{\text{\u2212}kt}.$
As with exponential growth, there is a differential equation associated with exponential decay. We have
Systems that exhibit exponential decay behave according to the model
where ${y}_{0}$ represents the initial state of the system and $k>0$ is a constant, called the decay constant .
The following figure shows a graph of a representative exponential decay function.
Let’s look at a physical application of exponential decay. Newton’s law of cooling says that an object cools at a rate proportional to the difference between the temperature of the object and the temperature of the surroundings. In other words, if $T$ represents the temperature of the object and ${T}_{a}$ represents the ambient temperature in a room, then
Note that this is not quite the right model for exponential decay. We want the derivative to be proportional to the function, and this expression has the additional ${T}_{a}$ term. Fortunately, we can make a change of variables that resolves this issue. Let $y(t)=T(t)-{T}_{a}.$ Then ${y}^{\prime}(t)={T}^{\prime}(t)-0={T}^{\prime}(t),$ and our equation becomes
From our previous work, we know this relationship between y and its derivative leads to exponential decay. Thus,
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