Unit 7 Notes: Exponential Models via Differential Equations (AP Calculus AB)

Exponential model

A model where the rate of change of a quantity is proportional to the quantity itself, typically written as dy/dt = ky.

Proportional rate of change

A rate statement meaning there exists a constant k such that dy/dt = k·y (the derivative is always a constant multiple of the current amount).

Differential equation

An equation involving a function and its derivatives (e.g., dy/dt = ky) that describes how a quantity changes.

Proportionality constant (k)

The constant in dy/dt = ky; it equals the constant relative growth rate (per unit time) and determines growth (k>0) or decay (k<0).

Exponential growth

Behavior when k>0 in dy/dt = ky; the quantity increases and the slope becomes steeper as the quantity gets larger.

Exponential decay

Behavior when k

Instantaneous relative growth rate

(1/y)(dy/dt); in an exponential model it is constant and equals k (an “instantaneous percent rate” as a decimal).

Constant percent change (continuous)

Another description of exponential behavior: the relative growth rate stays constant over time, so (1/y)(dy/dt)=k.

Separation of variables

A method for solving certain differential equations by rearranging so all y-terms are on one side and all t-terms are on the other (e.g., (1/y)dy = k dt).

Natural logarithm (ln)

The log base e; used when integrating 1/y (giving ln|y|) and when solving exponential equations for k or t.

Constant of integration (C)

The arbitrary constant that appears after integrating; for dy/dt = ky it leads to the solution form y = Ce^{kt}.

General solution to dy/dt = ky

y(t) = Ce^{kt}, including the special case y=0 when C=0.

Initial condition

A given value of the function at a specific time (e.g., y(0)=y0) used to determine the constant C in the exponential solution.

Finding C from y(t0)=y0

If y=Ce^{kt} and y(t0)=y0, then C = y0·e^{-k t0} (so C equals y(0) only when t0=0).

Using two points to find k (ratio method)

From y(t)=Ce^{kt}, using y(t1)=y1 and y(t2)=y2 gives k = (1/(t2−t1))·ln(y2/y1).

Finding k from instantaneous rate data

If dy/dt = ky and at some time you know both y and dy/dt, then k = (dy/dt)/y (with units “per time”).

Doubling time (Td)

The time it takes an exponential growth model to multiply by 2; Td = ln(2)/k (for k>0).

Half-life (T1/2)

The time it takes an exponential decay model to multiply by 1/2; T1/2 = −ln(2)/k (positive when k<0).

Time-to-reach formula

For y(t)=Ce^{kt}, the time to reach y* is t = (1/k)·ln(y*/C) (requires logarithms to solve).

Equilibrium (limiting) value (L)

The long-term value that a quantity approaches in models like dy/dt = k(y−L); solutions level off at y=L.

Exponential approach to equilibrium

A model of the form dy/dt = k(y−L) where the rate is proportional to the distance from equilibrium; typically k<0 gives a stable approach toward L.

Shifted exponential solution

The solution to dy/dt = k(y−L): y(t)=L+Ce^{kt}, meaning the exponential behavior occurs in the difference y−L.

Newton’s Law of Cooling/Heating

A temperature model stating dT/dt is proportional to the difference between object temperature and ambient temperature: dT/dt = k(T−Ts), usually with k<0.

Ambient (surrounding) temperature (Ts)

The constant surrounding temperature in Newton’s Law of Cooling/Heating; the temperature difference T−Ts decays toward 0.

Concavity of y=Ce^{kt} (for y>0)

Since d²y/dt² = k²y, exponential solutions with y>0 are concave up even during decay (the slope is negative but becomes less negative).