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

0.0(0)
Studied by 0 people
0%Unit 7 Mastery
0%Exam Mastery
Build your Mastery score
multiple choiceAP Practice
Supplemental Materials
call kaiCall Kai
Card Sorting

1/24

Last updated 3:08 PM on 3/12/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai

No analytics yet

Send a link to your students to track their progress

25 Terms

1
New cards

Exponential model

A model where the rate of change of a quantity is proportional to the quantity itself, typically written as dydt=ky\frac{dy}{dt} = ky.

2
New cards

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).

3
New cards

Differential equation

An equation involving a function and its derivatives (e.g., dydt=ky\frac{dy}{dt} = ky) that describes how a quantity changes.

4
New cards

Proportionality constant (k)

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

5
New cards

Exponential growth

Behavior when k>0k > 0 in dydt=ky\frac{dy}{dt} = ky; the quantity increases and the slope becomes steeper as the quantity gets larger.

6
New cards

Exponential decay

Behavior when k<0k < 0 in dydt=ky\frac{dy}{dt} = ky; the quantity decreases quickly at first and then levels off toward 0 (for y>0y > 0).

7
New cards

Instantaneous relative growth rate

1ydydt\frac{1}{y}\frac{dy}{dt}; in an exponential model, it is constant and equals kk (an “instantaneous percent rate” as a decimal).

8
New cards

Constant percent change (continuous)

Another description of exponential behavior: the relative growth rate stays constant over time, so 1ydydt=k\frac{1}{y}\frac{dy}{dt}=k.

9
New cards

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., 1ydy=kdt\frac{1}{y}dy = k dt).

10
New cards

Natural logarithm (ln)

The log base ee; used when integrating 1y\frac{1}{y} (giving lny\ln|y|) and when solving exponential equations for kk or tt.

11
New cards

Constant of integration (C)

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

12
New cards

General solution to dydt=ky\frac{dy}{dt} = ky

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

13
New cards

Initial condition

A given value of the function at a specific time (e.g., y(0)=y0y(0)=y_0) used to determine the constant CC in the exponential solution.

14
New cards

Finding C from y(t0)=y0y(t_0)=y_0

If y=Cekty = Ce^{kt} and y(t0)=y0y(t_0) = y_0, then C=y0×ekt0C = y_0 \times e^{-kt_0} (so CC equals y(0)y(0) only when t0=0t_0=0).

15
New cards

Using two points to find k (ratio method)

From y(t)=Cekty(t)=Ce^{kt}, using y(t1)=y1y(t_1)=y_1 and y(t2)=y2y(t_2)=y_2 gives k=1(t2t1)ln(y2y1)k = \frac{1}{(t_2-t_1)} \cdot \ln\left( \frac{y_2}{y_1} \right).

16
New cards

Finding k from instantaneous rate data

If dydt=ky\frac{dy}{dt} = ky and at some time you know both yy and dydt\frac{dy}{dt}, then k=dydtyk = \frac{\frac{dy}{dt}}{y} (with units “per time”).

17
New cards

Doubling time (T_d)

The time it takes an exponential growth model to multiply by 2; Td=ln(2)kT_d = \frac{\ln(2)}{k} (for k>0k > 0).

18
New cards

Half-life (T_{1/2})

The time it takes an exponential decay model to multiply by 12\frac{1}{2}; T1/2=ln(2)kT_{1/2} = \frac{-\ln(2)}{k} (positive when k<0k < 0).

19
New cards

Time-to-reach formula

For y(t)=Cekty(t)=Ce^{kt}, the time to reach yy^* is t=1kln(yC)t = \frac{1}{k} \cdot \ln\left( \frac{y^*}{C} \right) (requires logarithms to solve).

20
New cards

Equilibrium (limiting) value (L)

The long-term value that a quantity approaches in models like dydt=k(yL)\frac{dy}{dt} = k(y−L); solutions level off at y=Ly=L.

21
New cards

Exponential approach to equilibrium

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

22
New cards

Shifted exponential solution

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

23
New cards

Newton’s Law of Cooling/Heating

A temperature model stating dTdt\frac{dT}{dt} is proportional to the difference between object temperature and ambient temperature: dTdt=k(TTs)\frac{dT}{dt} = k(T−T_s), usually with k<0k < 0.

24
New cards

Ambient (surrounding) temperature (T_s)

The constant surrounding temperature in Newton’s Law of Cooling/Heating; the temperature difference TTsT−T_s decays toward 0.

25
New cards

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

Since d2ydt2=k2y\frac{d^2y}{dt^2} = k^2y, exponential solutions with y>0y > 0 are concave up even during decay (the slope is negative but becomes less negative).