7.4 Exponential models

Achievable AP Calculus AB

7. Differential equations

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Exponential models

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What you’ll learn

Exponential models: Translate word problems describing proportional growth into differential equations.
Solving for constants: Use initial conditions to find the starting value and the growth constant $k$ .

When a quantity changes at a rate proportional to its current size, it is modeled by a differential equation where the independent variable is missing:

$\frac{d y}{d t} = k y$

Because this exact derivation is identical every single time, you do not need to show separation of variables for this specific form on the AP exam. You may simply state the exponential solution:

$\frac{d y}{d t} = k y ⟹ y = y_{0} e^{k t}$

where $y_{0}$ is the initial value at $t = 0$ .

The sign of $k$ determines whether the solution models exponential growth or decay:

Growth: $k > 0$ (quantity increasing)
Decay: $k < 0$ (quantity decreasing)

Sidenote

Separation of variables

Although you are not required to show these steps, the derivation is included.

Separate the variables.
Integrate both sides.
Solve for $y$ .

$\int \frac{1}{y} d y ln ∣ y ∣ y y = \int k d t = k t + C = e^{k t + C} = C e^{k t}$

Using the initial condition $y (0) = y_{0}$ , the initial population at $t = 0$ ,

$y_{0} = C e^{0} = C$

Then the particular solution becomes

$y = y_{0} e^{k t}$

Example 1: Given multiplier $k$

The population of a bacterial culture increases at a rate $4$ times its current size at any moment. If the culture starts with $50$ bacteria, write an expression for the population $P (t)$ at any time $t$ .

(spoiler)

Because the rate is explicitly stated as “4 times its current size,” the growth constant is given directly as $k = 4$ . So the differential equation modeling this behavior is:

$\frac{d P}{d t} = 4 P$

which leads to the exponential model solution:

$P (t) = P_{0} e^{4 t}$

The initial population at $t = 0$ is $P_{0} = 50$ . Substitute this value to complete the particular solution:

$P (t) = 50 e^{4 t}$

Example 2: Unknown multiplier

2a) A population grows at a rate proportional to its current size. If it doubles every $5$ years, how many years will it take to triple?

Solution

(spoiler)

The phrase “proportional to its current size” implies the general model:

$P (t) = P_{0} e^{k t}$

1. Find $k$ using the doubling time:

Since the population doubles every $5$ years, then the population at $t = 5$ is twice the starting population:

$P (5) = 2 P_{0}$

Use this to find $k$ :

$2 P_{0} 2 ln (2) k = P_{0} e^{5 k} = e^{5 k} = 5 k = \frac{ln ( 2 )}{5} \approx 0.139$

2. Find the time $t$ when the population triples:

Substitute $k$ into the general solution to find the time $t$ when $P (t) = 3 P_{0}$ :

$3 P_{0} 3 ln (3) t = P_{0} e^{0.139 t} = e^{0.139 t} = 0.139 t \approx 7.925$

So it will take roughly $8$ years for the population to triple.

A radioactive substance decays according to the model $A (t) = A_{0} e^{k t}$ . The half-life, or the time it takes for the substance to decay to half its original amount, is $10$ hours. If there are initially $100$ grams of the substance, how long will it take to decay to $20$ grams?

Solution

(spoiler)

1. Find $k$ :

A half-life of $10$ hours means that when $t = 10$ , the amount is half the initial amount of $100$ grams, i.e.

$A (10) = \frac{1}{2} A_{0} = 50$

Substitute into the model given:

$50 0.5 ln (0.5) k = 100 e^{10 k} = e^{10 k} = 10 k = \frac{ln ( 0.5 )}{10} \approx - 0.0693$

Then the model is

$A (t) = 100 e^{- 0.0693 t}$

2. Find $t$ when the amount is $20$ grams:

Substitute $A (t) = 20$ into the model:

$20 0.2 ln (0.2) t = 100 e^{- 0.0693 t} = e^{- 0.0693 t} = - 0.0693 t = 23.219$

Therefore, it will take a little over $23$ hours to decay from $100$ to $20$ grams.

FRQ: Shifting and separating variables

When the rate is proportional to a difference, you must show every step of the separation of variables. This is typically a free-response problem. Below is a question created to mimic #5 on the FRQ portion of the 2012 AB exam:

A tank is being filled with water, and the rate at which the water level rises is proportional to the difference between the tank’s maximum height and the current water level. At time $t = 0$ , the water is $10$ cm deep. $H (t)$ is the height of the water, in centimeters, at time $t$ . minutes after filling begins. The tank has a maximum height of $120$ cm. The rate of change of the water height is modeled by the differential equation

$\frac{d H}{d t} = \frac{1}{4} (120 - H)$

a) Is the water rising faster when the water level is at $30$ cm or when it is at $90$ cm?

b) Use separation of variables to write an expression for $H (t)$ , the particular solution to the differential equation with initial condition $H (0) = 10$ .

Solutions

a) $30$ vs. $90$ cm

(spoiler)

To compare how fast the water is rising, compare the rate of change at the two heights using:

$\frac{d H}{d t} = \frac{1}{4} (120 - H)$

When $H = 30$ :

$\frac{d H}{d t} = \frac{1}{4} (120 - 30) = \frac{1}{4} (90) = 22.5$

When $H = 90$ :

$\frac{d H}{d t} = \frac{1}{4} (120 - 90) = \frac{1}{4} (30) = 7.5$

Since $22.5 > 7.5$ , the water is rising faster when the water level is at $30$ cm.

b) Expression for $H (t)$

(spoiler)

1. Separate the variables:

$\frac{1}{120 - H} d H = \frac{1}{4} d t$

2. Integrate both sides:

$\int \frac{1}{120 - H} d H = \int \frac{1}{4} d t - ln ∣120 - H ∣ = \frac{t}{4} + C$

3. Use the initial condition $H (0) = 10$ to find $C$ :

$- ln (120 - 10) C = \frac{1}{4} (0) + C = - ln (110)$

4. Solve for $H$ :

$- ln ∣120 - H ∣ ln ∣120 - H ∣ ln ∣120 - H ∣ - ln (110) ln \frac{120 - H}{110} \frac{120 - H}{110} 120 - H H (t) = \frac{t}{4} - ln (110) = - \frac{t}{4} + ln (110) = - \frac{t}{4} = - \frac{t}{4} = e^{- t /4} = 110 e^{- t /4} = 120 - 110 e^{- t /4}$

Exponential differential equations: general solution

Modeled by $\frac{d y}{d t} = k y$
General solution: $y = C e^{k t}$
$k > 0$ : growth; $k < 0$ : decay

Separation of variables method

Separate variables: $\frac{1}{y} d y = k d t$
Integrate: $ln ∣ y ∣ = k t + C$
Exponentiate: $y = C e^{k t}$

Using initial condition

Substitute $y (0) = y_{0}$ into $y = C e^{k t}$
Particular solution: $y = y_{0} e^{k t}$

Population growth example

Model: $P (t) = P_{0} e^{k t}$
Doubling time: $k = \frac{l n ( 2 )}{5}$
Time to triple: $t = \frac{5 l n ( 3 )}{l n ( 2 )} \approx 7.93$ years

Radioactive decay example

Model: $A (t) = A_{0} e^{k t}$
Half-life $T$ : $k = \frac{l n ( 1/2 )}{T}$
Time to decay from $100$ to $20$ grams: $t = ln (0.2) \cdot \frac{10}{l n ( 0.5 )} \approx 23.22$ hours

Tank filling (non-standard exponential model)

Differential equation: $\frac{d H}{d t} = \frac{1}{4} (120 - H)$
Water rises faster at lower $H$ (e.g., $30$ cm vs $90$ cm)
Solution with $H (0) = 10$ : $H (t) = 120 - 110 e^{- t /4}$

Exponential models

What you’ll learn

Exponential models: Translate word problems describing proportional growth into differential equations.
Solving for constants: Use initial conditions to find the starting value and the growth constant $k$ .

When a quantity changes at a rate proportional to its current size, it is modeled by a differential equation where the independent variable is missing:

$\frac{d y}{d t} = k y$

Because this exact derivation is identical every single time, you do not need to show separation of variables for this specific form on the AP exam. You may simply state the exponential solution:

$\frac{d y}{d t} = k y ⟹ y = y_{0} e^{k t}$

where $y_{0}$ is the initial value at $t = 0$ .

The sign of $k$ determines whether the solution models exponential growth or decay:

Growth: $k > 0$ (quantity increasing)
Decay: $k < 0$ (quantity decreasing)

Sidenote

Separation of variables

Although you are not required to show these steps, the derivation is included.

Separate the variables.
Integrate both sides.
Solve for $y$ .

$\int \frac{1}{y} d y ln ∣ y ∣ y y = \int k d t = k t + C = e^{k t + C} = C e^{k t}$

Using the initial condition $y (0) = y_{0}$ , the initial population at $t = 0$ ,

$y_{0} = C e^{0} = C$

Then the particular solution becomes

$y = y_{0} e^{k t}$

Example 1: Given multiplier $k$

The population of a bacterial culture increases at a rate $4$ times its current size at any moment. If the culture starts with $50$ bacteria, write an expression for the population $P (t)$ at any time $t$ .

(spoiler)

Because the rate is explicitly stated as “4 times its current size,” the growth constant is given directly as $k = 4$ . So the differential equation modeling this behavior is:

$\frac{d P}{d t} = 4 P$

which leads to the exponential model solution:

$P (t) = P_{0} e^{4 t}$

The initial population at $t = 0$ is $P_{0} = 50$ . Substitute this value to complete the particular solution:

$P (t) = 50 e^{4 t}$

Example 2: Unknown multiplier

2a) A population grows at a rate proportional to its current size. If it doubles every $5$ years, how many years will it take to triple?

Solution

(spoiler)

The phrase “proportional to its current size” implies the general model:

$P (t) = P_{0} e^{k t}$

1. Find $k$ using the doubling time:

Since the population doubles every $5$ years, then the population at $t = 5$ is twice the starting population:

$P (5) = 2 P_{0}$

Use this to find $k$ :

$2 P_{0} 2 ln (2) k = P_{0} e^{5 k} = e^{5 k} = 5 k = \frac{ln ( 2 )}{5} \approx 0.139$

2. Find the time $t$ when the population triples:

Substitute $k$ into the general solution to find the time $t$ when $P (t) = 3 P_{0}$ :

$3 P_{0} 3 ln (3) t = P_{0} e^{0.139 t} = e^{0.139 t} = 0.139 t \approx 7.925$

So it will take roughly $8$ years for the population to triple.

A radioactive substance decays according to the model $A (t) = A_{0} e^{k t}$ . The half-life, or the time it takes for the substance to decay to half its original amount, is $10$ hours. If there are initially $100$ grams of the substance, how long will it take to decay to $20$ grams?

Solution

(spoiler)

1. Find $k$ :

A half-life of $10$ hours means that when $t = 10$ , the amount is half the initial amount of $100$ grams, i.e.

$A (10) = \frac{1}{2} A_{0} = 50$

Substitute into the model given:

$50 0.5 ln (0.5) k = 100 e^{10 k} = e^{10 k} = 10 k = \frac{ln ( 0.5 )}{10} \approx - 0.0693$

Then the model is

$A (t) = 100 e^{- 0.0693 t}$

2. Find $t$ when the amount is $20$ grams:

Substitute $A (t) = 20$ into the model:

$20 0.2 ln (0.2) t = 100 e^{- 0.0693 t} = e^{- 0.0693 t} = - 0.0693 t = 23.219$

Therefore, it will take a little over $23$ hours to decay from $100$ to $20$ grams.

FRQ: Shifting and separating variables

A tank is being filled with water, and the rate at which the water level rises is proportional to the difference between the tank’s maximum height and the current water level. At time $t = 0$ , the water is $10$ cm deep. $H (t)$ is the height of the water, in centimeters, at time $t$ . minutes after filling begins. The tank has a maximum height of $120$ cm. The rate of change of the water height is modeled by the differential equation

$\frac{d H}{d t} = \frac{1}{4} (120 - H)$

a) Is the water rising faster when the water level is at $30$ cm or when it is at $90$ cm?

b) Use separation of variables to write an expression for $H (t)$ , the particular solution to the differential equation with initial condition $H (0) = 10$ .

Solutions

a) $30$ vs. $90$ cm

(spoiler)

To compare how fast the water is rising, compare the rate of change at the two heights using:

$\frac{d H}{d t} = \frac{1}{4} (120 - H)$

When $H = 30$ :

$\frac{d H}{d t} = \frac{1}{4} (120 - 30) = \frac{1}{4} (90) = 22.5$

When $H = 90$ :

$\frac{d H}{d t} = \frac{1}{4} (120 - 90) = \frac{1}{4} (30) = 7.5$

Since $22.5 > 7.5$ , the water is rising faster when the water level is at $30$ cm.

b) Expression for $H (t)$

(spoiler)

1. Separate the variables:

$\frac{1}{120 - H} d H = \frac{1}{4} d t$

2. Integrate both sides:

$\int \frac{1}{120 - H} d H = \int \frac{1}{4} d t - ln ∣120 - H ∣ = \frac{t}{4} + C$

3. Use the initial condition $H (0) = 10$ to find $C$ :

$- ln (120 - 10) C = \frac{1}{4} (0) + C = - ln (110)$

4. Solve for $H$ :

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.