Given the basic
exponential growth equation
$\text{\hspace{0.17em}}A={A}_{0}{e}^{kt},$ doubling time can be found by solving for when the original quantity has doubled, that is, by solving
$\text{\hspace{0.17em}}2{A}_{0}={A}_{0}{e}^{kt}.$
Finding a function that describes exponential growth
According to Moore’s Law, the doubling time for the number of transistors that can be put on a computer chip is approximately two years. Give a function that describes this behavior.
The function is
$\text{\hspace{0.17em}}A={A}_{0}{e}^{\frac{\mathrm{ln}2}{2}t}.$
Recent data suggests that, as of 2013, the rate of growth predicted by Moore’s Law no longer holds. Growth has slowed to a doubling time of approximately three years. Find the new function that takes that longer doubling time into account.
$f(t)={A}_{0}{e}^{\frac{\mathrm{ln}2}{3}t}$
Using newton’s law of cooling
Exponential decay can also be applied to temperature. When a hot object is left in surrounding air that is at a lower temperature, the object’s temperature will decrease exponentially, leveling off as it approaches the surrounding air temperature. On a graph of the temperature function, the leveling off will correspond to a horizontal asymptote at the temperature of the surrounding air. Unless the room temperature is zero, this will correspond to a
vertical shift of the generic
exponential decay function. This translation leads to
Newton’s Law of Cooling , the scientific formula for temperature as a function of time as an object’s temperature is equalized with the ambient temperature
The temperature of an object,
$\text{\hspace{0.17em}}T,$ in surrounding air with temperature
$\text{\hspace{0.17em}}{T}_{s}\text{\hspace{0.17em}}$ will behave according to the formula
$$T(t)=A{e}^{kt}+{T}_{s}$$
where
$t\text{\hspace{0.17em}}$ is time
$A\text{\hspace{0.17em}}$ is the difference between the initial temperature of the object and the surroundings
$k\text{\hspace{0.17em}}$ is a constant, the continuous rate of cooling of the object
Given a set of conditions, apply Newton’s Law of Cooling.
Set
$\text{\hspace{0.17em}}{T}_{s}\text{\hspace{0.17em}}$ equal to the
y -coordinate of the horizontal asymptote (usually the ambient temperature).
Substitute the given values into the continuous growth formula
$\text{\hspace{0.17em}}T(t)=A{e}^{k}{}^{t}+{T}_{s}\text{\hspace{0.17em}}$ to find the parameters
$\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}k.$
Substitute in the desired time to find the temperature or the desired temperature to find the time.
Using newton’s law of cooling
A cheesecake is taken out of the oven with an ideal internal temperature of
$\text{\hspace{0.17em}}\text{165\xb0F,}\text{\hspace{0.17em}}$ and is placed into a
$\text{\hspace{0.17em}}\mathrm{35\xb0F}\text{\hspace{0.17em}}$ refrigerator. After 10 minutes, the cheesecake has cooled to
$\text{\hspace{0.17em}}\text{150\xb0F}\text{.}\text{\hspace{0.17em}}$ If we must wait until the cheesecake has cooled to
$\text{\hspace{0.17em}}\text{70\xb0F}\text{\hspace{0.17em}}$ before we eat it, how long will we have to wait?
Because the surrounding air temperature in the refrigerator is 35 degrees, the cheesecake’s temperature will decay exponentially toward 35, following the equation
$$T(t)=A{e}^{kt}+35$$
We know the initial temperature was 165, so
$\text{\hspace{0.17em}}T(0)=165.$
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