AP Calculus AB and BC glossary

Exponential model

An exponential model sets a quantity's rate of change proportional to the amount present, so its solution is the starting value times e to the kt. Two measurements pin it down: the natural log of their ratio divided by the time between them gives k, and either one then gives the starting value.

\frac{dy}{dt} = ky, \quad y(0) = y_{0} \quad \Longrightarrow \quad y = y_{0}e^{kt}

Separating variables turns the equation into $\int \frac{dy}{y} = \int k\,dt$ , so $\ln\lvert y \rvert = kt + C$ , and exponentiating gives $y = y_{0}e^{kt}$ . The initial condition lands as $y_{0}$ , the amount at $t = 0$ , because $e^{0} = 1$ .

The solved model carries two unknowns, so it takes two facts to fit. Given $y = y_{1}$ at $t = t_{1}$ and $y = y_{2}$ at $t = t_{2}$ , dividing one equation by the other cancels $y_{0}$ , and a logarithm then isolates $k$ ; substituting back recovers $y_{0}$ . A colony measured at 200 when $t = 0$ and 260 when $t = 5$ has $k = \frac{\ln 1.3}{5} \approx 0.0525$ per unit time.

k = \frac{\ln\left(\frac{y_{2}}{y_{1}}\right)}{t_{2} - t_{1}}, \qquad y_{0} = y_{1}e^{-kt_{1}}

That $k$ is the relative rate of change, $k = \frac{1}{y}\cdot\frac{dy}{dt}$ , which is why it is quoted as a percent per unit of time rather than as an amount per unit of time. Keep it unrounded until the last step: $k$ sits in an exponent, so an error in $k$ is stretched by the elapsed time before it reaches the answer.

The mistake

Reading $t$ off the calendar. The $y_{0}$ in the solution is the amount at $t = 0$ , so once you call 2010 time zero, a 2015 reading is $t = 5$ , not $t = 2015$ . Using the raw year divides $\ln\left(\frac{y_{2}}{y_{1}}\right)$ by 2015 instead of by 5.

Appears in: Unit 7: Differential Equations