Exponential Models with Differential Equations

3. Visual Connection

5. Concept Development

An exponential differential-equation model begins with the statement that a quantity's rate of change is proportional to its current amount:

The proportionality constant \(k\) is the continuous relative growth rate. Separating variables shows why the solution is exponential:

If \(Q_0=0\), the equilibrium solution is \(Q=0\). Otherwise the solution keeps the sign of \(Q_0\) and never crosses zero.

Interpreting the parameters

The absolute rate \(Q'=kQ\) is generally not constant. In a growth model, a larger amount produces a larger absolute increase during the same length of time.

Equivalent exponential forms

The base-\(e\) form connects directly to the differential equation. The base-\(b\) form can make a per-period multiplier easier to interpret. If the amount is multiplied by \(b\) each time unit, the matching continuous constant is \(k=\ln b\).

Finding the constant from two measurements

If \(Q(t_1)=Q_1>0\) and \(Q(t_2)=Q_2>0\), divide the two model equations:

A convenient point-centered model is then

Positive \(k\) should agree with increasing data and negative \(k\) with decreasing data.

Doubling time and half-life

For growth, set \(Q(D)=2Q_0\); for decay, set \(Q(H)=Q_0/2\):

These times do not depend on the starting amount. Equivalent forms are

Continuous rate versus period-by-period percent

The statement “the continuous relative rate is \(8\%\) per year” means \(k=0.08\), so one-year multiplication is \(e^{0.08}\approx1.0833\). The statement “the amount increases by \(8\%\) each year” gives annual multiplier \(1.08\); a continuous model passing through the annual data uses \(k=\ln(1.08)\). These are close, but not identical.

Exponential approach to an equilibrium

Sometimes the difference from a fixed level \(L\), rather than the quantity itself, changes proportionally. If

then \(Z=Q-L\) satisfies \(Z'=-rZ\). Therefore

This shifted exponential form models processes such as an object's temperature approaching a constant ambient temperature.

When is the model appropriate?

Check whether ratios over equal time intervals are approximately constant, or equivalently whether \(\ln Q\) is approximately linear in time for positive data. Exponential growth assumes unlimited proportional growth, so resource limits may eventually make a logistic model more appropriate. A continuous model can also return noninteger values for a population; interpret those values in context.

6. Detailed Worked Example and Error Check

Example 1: Derive and verify the model

Suppose \(P'=0.18P\) and \(P(0)=250\). The exponential solution is

Differentiation gives \(P'=45e^{0.18t}=0.18P\), and \(P(0)=250\). If \(t\) is measured in years, \(0.18\) has units \(\text{year}^{-1}\).

Example 2: Determine a model from two data points

A culture has \(500\) cells initially and \(800\) cells after \(6\) hours. Assuming exponential growth,

Thus

The time to reach \(1000\) cells satisfies \(2=e^{kt}\), so

Example 3: Doubling-time form

A population starts at \(300\) and doubles every \(6\) hours:

After \(15\) hours, \(P(15)=300\cdot2^{2.5}=1200\sqrt2\).

Example 4: Radioactive half-life

A sample begins at \(120\) grams and has half-life \(8\) years. Then

After \(20\) years, \(A(20)=120(1/2)^{2.5}=15\sqrt2\approx21.21\) grams.

Example 5: Continuous versus annual growth

An account increasing by exactly \(12\%\) at the end of each year has annual multiplier \(1.12\). The continuous exponential curve matching those annual balances is

so its differential equation is \(B'=(\ln1.12)B\), not \(B'=0.12B\). The latter would produce annual factor \(e^{0.12}\).

Example 6: Temperature approaching room level

An object begins at \(90^\circ\mathrm C\) in a \(20^\circ\mathrm C\) room, and its temperature difference from the room halves every \(10\) minutes. Then

The corresponding differential equation is

The temperature decreases while remaining above \(20^\circ\mathrm C\).

Example 7: Recover the differential equation

Given \(Q(t)=75e^{-0.04t}\), differentiate:

The model starts at \(75\), decays continuously at relative rate \(0.04\) per time unit, and has half-life \(\ln2/0.04\approx17.33\) time units.

Example 8: Recognize exponential data

At times \(0,2,4\), a quantity has values \(100,150,225\). Each two-unit interval multiplies the amount by \(1.5\), so

The model satisfies \(Q'=(\ln1.5/2)Q\). Constant differences would indicate a linear model; constant ratios indicate an exponential model.

Common modeling errors

• Using \(Q'=k\) when the rate is proportional to the amount.
• Forgetting the initial multiplier \(Q_0\).
• Giving \(k\) the wrong sign in a decay context.
• Writing \(k=0.08\) for an \(8\%\) period multiplier without checking whether the rate is continuous.
• Using \(Q_2-Q_1\) instead of the ratio \(Q_2/Q_1\) to determine \(k\).
• Applying exponential growth indefinitely when the context has a limiting capacity.