6.7 Exponential and logarithmic models (Page 4/16)

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Given the basic exponential growth equation $A = A_{0} e^{k t},$ doubling time can be found by solving for when the original quantity has doubled, that is, by solving $2 A_{0} = A_{0} e^{k t} .$

The formula is derived as follows:

\begin{array}{l} 2 A_{0} = A_{0} e^{k t} \\ 2 = e^{k t} & Divide by A_{0} . \\ \ln 2 = k t & Take the natural logarithm . \\ t = \frac{\ln 2}{k} & Divide by the coefficient of t . \end{array}

Thus the doubling time is

t = \frac{\ln 2}{k}

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 formula is derived as follows:

\begin{array}{l} t = \frac{\ln 2}{k} & The doubling time formula . \\ 2 = \frac{\ln 2}{k} & Use a doubling time of two years . \\ k = \frac{\ln 2}{2} & Multiply by k and divide by 2 . \\ A = A_{0} e^{\frac{\ln 2}{2} t} \begin{matrix}  \end{matrix} & Substitute k into the continuous growth formula . \end{array}

The function is $A = A_{0} e^{\frac{\ln 2}{2} t} .$

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Try It

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.

$f (t) = A_{0} e^{\frac{\ln 2}{3} t}$

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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

T (t) = a e^{k t} + T_{s}

This formula is derived as follows:

\begin{array}{l} T (t) = A b^{c t} + T_{s} \\ T (t) = A e^{\ln (b^{c t})} + T_{s} \begin{matrix}  \end{matrix} & Laws of logarithms . \\ T (t) = A e^{c t \ln b} + T_{s} & Laws of logarithms . \\ T (t) = A e^{k t} + T_{s} & Rename the constant c \ln b, calling it k . \end{array}

A General Note

Newton’s law of cooling

The temperature of an object, $T,$ in surrounding air with temperature $T_{s}$ will behave according to the formula

T (t) = A e^{k t} + T_{s}

where

$t$ is time
$A$ is the difference between the initial temperature of the object and the surroundings
$k$ is a constant, the continuous rate of cooling of the object

How To

Given a set of conditions, apply Newton’s Law of Cooling.

Set $T_{s}$ equal to the y -coordinate of the horizontal asymptote (usually the ambient temperature).
Substitute the given values into the continuous growth formula $T (t) = A e^{k}^{t} + T_{s}$ to find the parameters $A$ and $k .$
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 $165°F,$ and is placed into a $35°F$ refrigerator. After 10 minutes, the cheesecake has cooled to $150°F .$ If we must wait until the cheesecake has cooled to $70°F$ 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^{k t} + 35

We know the initial temperature was 165, so $T (0) = 165 .$

\begin{array}{l} 165 = A e^{k 0} + 35 & Substitute (0, 165) . \\ A = 130 & Solve for A . \end{array}

We were given another data point, $T (10) = 150,$ which we can use to solve for $k .$

\begin{array}{l} 150 = 130 e^{k 10} + 35 & Substitute (10, 150) . \\ 115 = 130 e^{k 10} & Subtract 35 . \\ \frac{115}{130} = e^{10 k} & Divide by 130 . \\ \ln (\frac{115}{130}) = 10 k & Take the natural log of both sides . \\ k = \frac{\ln (\frac{115}{130})}{10} = - 0.0123 & Divide by the coefficient of k . \end{array}

This gives us the equation for the cooling of the cheesecake: $T (t) = 130 e^{- 0.0123 t} + 35 .$

Now we can solve for the time it will take for the temperature to cool to 70 degrees.

\begin{array}{l} 70 = 130 e^{- 0.0123 t} + 35 & Substitute in 70 for T (t) . \\ 35 = 130 e^{- 0.0123 t} & Subtract 35 . \\ \frac{35}{130} = e^{- 0.0123 t} & Divide by 130 . \\ \ln (\frac{35}{130}) = - 0.0123 t & Take the natural log of both sides \\ t = \frac{\ln (\frac{35}{130})}{- 0.0123} \approx 106.68 & Divide by the coefficient of t . \end{array}

It will take about 107 minutes, or one hour and 47 minutes, for the cheesecake to cool to $70°F .$

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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