7.4 Exponential models

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is currently in development and is a work-in-progress.

Exponential models

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What you’ll learn:

How to derive the general solution to exponential differential equations
How the initial condition leads to the specific form $y = y_{0} e^{k t}$

When a quantity changes at a rate proportional to its current size, it can be modeled by the differential equation

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

Here, $k$ is a constant of proportionality. (If $k > 0$ , the quantity grows; if $k < 0$ , it decays.)

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

Start by separating separating the variables:

$\frac{1}{y} d y = k d t$

Now integrate both sides:

$\int \frac{1}{y} d y = \int k d t$

$ln ∣ y ∣ = k t + C$

Exponentiate to solve for $y$ :

$y = e^{k t + C}$

$y = e^{k t} \cdot e^{C}$

Since $e^{C}$ is a positive constant, we can replace it with a new constant (still called $C$ ):

$y = C e^{k t}$

This is the general solution for all differential equations worded as a rate that is proportional to the current amount.

Using an initial condition

If the initial condition is $y (0) = y_{0}$ , substitute $t = 0$ and $y = y_{0}$ into $y = C e^{k t}$ :

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

Then the particular solution becomes

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

This equation models exponential growth or decay where a quantity starts from $y_{0}$ at time $t = 0$ .

Examples

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

The differential equation that models this situation is

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

and the general solution is

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

where $P_{0}$ is the initial population and $P (t)$ is the population at time $t$ (in years).

Since the population doubles every $5$ years, then the population at $t = 5$ is twice the starting population i.e. $P (5) = 2 P_{0}$ . Use this to find $k$ :

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

Now find the time $t$ when the population is triple the initial population, or when $P (t) = 3 P_{0}$ :

$3 P_{0} 3 ln (3) t = P_{0} e^{(\frac{l n ( 2 )}{5} t)} = e^{(\frac{l n ( 2 )}{5} t)} = \frac{ln ( 2 )}{5} t = \frac{5 ln ( 3 )}{ln ( 2 )} \approx 7.925$

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.

a) What is the value of $k$ ?

b) If there are initially $100$ grams of the substance, how long will it take to decay to $20$ grams?

Solutions

a) Value of $k$

(spoiler)

The half-life of $10$ hours means that when $t = 10$ , the amount is half the initial amount, or that

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

Substitute into $A (t) = A_{0} e^{k t}$ :

$\frac{1}{2} A_{0} = A_{0} e^{10 k}$

Divide by $A_{0}$ and solve for $k$ :

$\frac{1}{2} ln (0.5) k = e^{(10 k)} = 10 k = \frac{ln ( 0.5 )}{10} \approx - 0.0693$

b) Time to decay from $100$ to $20$ grams

(spoiler)

Using $k = \frac{l n ( 0.5 )}{10}$ , the model is

$A (t) = A_{0} e^{l n (0.5) t /10}$

Now substitute $A_{0} = 100$ and $A (t) = 20$ :

$20 = 100 e^{l n (0.5) t /10}$

$\frac{20}{100} = e^{l n (0.5) t /10}$

$ln (20/100) = \frac{ln ( 0.5 )}{10} t$

$t = ln (20/100) \cdot \frac{10}{ln ( 0.5 )}$

$= 23.219$

It will take a little over $23$ hours to decay from $100$ to $20$ grams.

Here is a question similar to #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

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

at the two heights.

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)

Separate the variables:

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

Integrate both sides:

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

Plug in the initial condition $H (0) = 10$ to find $C$ :

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

$C = - ln (110)$

Now 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:

How to derive the general solution to exponential differential equations
How the initial condition leads to the specific form $y = y_{0} e^{k t}$

When a quantity changes at a rate proportional to its current size, it can be modeled by the differential equation

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

Here, $k$ is a constant of proportionality. (If $k > 0$ , the quantity grows; if $k < 0$ , it decays.)

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

Start by separating separating the variables:

$\frac{1}{y} d y = k d t$

Now integrate both sides:

$\int \frac{1}{y} d y = \int k d t$

$ln ∣ y ∣ = k t + C$

Exponentiate to solve for $y$ :

$y = e^{k t + C}$

$y = e^{k t} \cdot e^{C}$

Since $e^{C}$ is a positive constant, we can replace it with a new constant (still called $C$ ):

$y = C e^{k t}$

This is the general solution for all differential equations worded as a rate that is proportional to the current amount.

Using an initial condition

If the initial condition is $y (0) = y_{0}$ , substitute $t = 0$ and $y = y_{0}$ into $y = C e^{k t}$ :

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

Then the particular solution becomes

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

This equation models exponential growth or decay where a quantity starts from $y_{0}$ at time $t = 0$ .

Examples

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

The differential equation that models this situation is

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

and the general solution is

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

where $P_{0}$ is the initial population and $P (t)$ is the population at time $t$ (in years).

Since the population doubles every $5$ years, then the population at $t = 5$ is twice the starting population i.e. $P (5) = 2 P_{0}$ . Use this to find $k$ :

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

Now find the time $t$ when the population is triple the initial population, or when $P (t) = 3 P_{0}$ :

$3 P_{0} 3 ln (3) t = P_{0} e^{(\frac{l n ( 2 )}{5} t)} = e^{(\frac{l n ( 2 )}{5} t)} = \frac{ln ( 2 )}{5} t = \frac{5 ln ( 3 )}{ln ( 2 )} \approx 7.925$

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.

a) What is the value of $k$ ?

b) If there are initially $100$ grams of the substance, how long will it take to decay to $20$ grams?

Solutions

a) Value of $k$

(spoiler)

The half-life of $10$ hours means that when $t = 10$ , the amount is half the initial amount, or that

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

Substitute into $A (t) = A_{0} e^{k t}$ :

$\frac{1}{2} A_{0} = A_{0} e^{10 k}$

Divide by $A_{0}$ and solve for $k$ :

$\frac{1}{2} ln (0.5) k = e^{(10 k)} = 10 k = \frac{ln ( 0.5 )}{10} \approx - 0.0693$

b) Time to decay from $100$ to $20$ grams

(spoiler)

Using $k = \frac{l n ( 0.5 )}{10}$ , the model is

$A (t) = A_{0} e^{l n (0.5) t /10}$

Now substitute $A_{0} = 100$ and $A (t) = 20$ :

$20 = 100 e^{l n (0.5) t /10}$

$\frac{20}{100} = e^{l n (0.5) t /10}$

$ln (20/100) = \frac{ln ( 0.5 )}{10} t$

$t = ln (20/100) \cdot \frac{10}{ln ( 0.5 )}$

$= 23.219$

It will take a little over $23$ hours to decay from $100$ to $20$ grams.

Here is a question similar to #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

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

at the two heights.

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)

Separate the variables:

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

Integrate both sides:

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

Plug in the initial condition $H (0) = 10$ to find $C$ :

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

$C = - ln (110)$

Now solve for $H$ :

Achievable AP Calculus AB

7. Differential equations

Our AP Calculus AB course is currently in development and is a work-in-progress.

Exponential models

6 min read

Font

Discuss

Feedback

What you’ll learn:

How to derive the general solution to exponential differential equations
How the initial condition leads to the specific form $y = y_{0} e^{k t}$

When a quantity changes at a rate proportional to its current size, it can be modeled by the differential equation

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

Here, $k$ is a constant of proportionality. (If $k > 0$ , the quantity grows; if $k < 0$ , it decays.)

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

Start by separating separating the variables:

$\frac{1}{y} d y = k d t$

Now integrate both sides:

$\int \frac{1}{y} d y = \int k d t$

$ln ∣ y ∣ = k t + C$

Exponentiate to solve for $y$ :

$y = e^{k t + C}$

$y = e^{k t} \cdot e^{C}$

Since $e^{C}$ is a positive constant, we can replace it with a new constant (still called $C$ ):

$y = C e^{k t}$

This is the general solution for all differential equations worded as a rate that is proportional to the current amount.

Using an initial condition

If the initial condition is $y (0) = y_{0}$ , substitute $t = 0$ and $y = y_{0}$ into $y = C e^{k t}$ :

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

Then the particular solution becomes

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

This equation models exponential growth or decay where a quantity starts from $y_{0}$ at time $t = 0$ .

Examples

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

The differential equation that models this situation is

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

and the general solution is

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

where $P_{0}$ is the initial population and $P (t)$ is the population at time $t$ (in years).

Since the population doubles every $5$ years, then the population at $t = 5$ is twice the starting population i.e. $P (5) = 2 P_{0}$ . Use this to find $k$ :

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

Now find the time $t$ when the population is triple the initial population, or when $P (t) = 3 P_{0}$ :

$3 P_{0} 3 ln (3) t = P_{0} e^{(\frac{l n ( 2 )}{5} t)} = e^{(\frac{l n ( 2 )}{5} t)} = \frac{ln ( 2 )}{5} t = \frac{5 ln ( 3 )}{ln ( 2 )} \approx 7.925$

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.

a) What is the value of $k$ ?

b) If there are initially $100$ grams of the substance, how long will it take to decay to $20$ grams?

Solutions

a) Value of $k$

(spoiler)

The half-life of $10$ hours means that when $t = 10$ , the amount is half the initial amount, or that

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

Substitute into $A (t) = A_{0} e^{k t}$ :

$\frac{1}{2} A_{0} = A_{0} e^{10 k}$

Divide by $A_{0}$ and solve for $k$ :

$\frac{1}{2} ln (0.5) k = e^{(10 k)} = 10 k = \frac{ln ( 0.5 )}{10} \approx - 0.0693$

b) Time to decay from $100$ to $20$ grams

(spoiler)

Using $k = \frac{l n ( 0.5 )}{10}$ , the model is

$A (t) = A_{0} e^{l n (0.5) t /10}$

Now substitute $A_{0} = 100$ and $A (t) = 20$ :

$20 = 100 e^{l n (0.5) t /10}$

$\frac{20}{100} = e^{l n (0.5) t /10}$

$ln (20/100) = \frac{ln ( 0.5 )}{10} t$

$t = ln (20/100) \cdot \frac{10}{ln ( 0.5 )}$

$= 23.219$

It will take a little over $23$ hours to decay from $100$ to $20$ grams.

Here is a question similar to #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

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

at the two heights.

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)

Separate the variables:

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

Integrate both sides:

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

Plug in the initial condition $H (0) = 10$ to find $C$ :

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

$C = - ln (110)$

Now solve for $H$ :

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:

How to derive the general solution to exponential differential equations
How the initial condition leads to the specific form $y = y_{0} e^{k t}$

When a quantity changes at a rate proportional to its current size, it can be modeled by the differential equation

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

Here, $k$ is a constant of proportionality. (If $k > 0$ , the quantity grows; if $k < 0$ , it decays.)

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

Start by separating separating the variables:

$\frac{1}{y} d y = k d t$

Now integrate both sides:

$\int \frac{1}{y} d y = \int k d t$

$ln ∣ y ∣ = k t + C$

Exponentiate to solve for $y$ :

$y = e^{k t + C}$

$y = e^{k t} \cdot e^{C}$

Since $e^{C}$ is a positive constant, we can replace it with a new constant (still called $C$ ):

$y = C e^{k t}$

This is the general solution for all differential equations worded as a rate that is proportional to the current amount.

Using an initial condition

If the initial condition is $y (0) = y_{0}$ , substitute $t = 0$ and $y = y_{0}$ into $y = C e^{k t}$ :

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

Then the particular solution becomes

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

This equation models exponential growth or decay where a quantity starts from $y_{0}$ at time $t = 0$ .

Examples

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

The differential equation that models this situation is

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

and the general solution is

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

where $P_{0}$ is the initial population and $P (t)$ is the population at time $t$ (in years).

Since the population doubles every $5$ years, then the population at $t = 5$ is twice the starting population i.e. $P (5) = 2 P_{0}$ . Use this to find $k$ :

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

Now find the time $t$ when the population is triple the initial population, or when $P (t) = 3 P_{0}$ :

$3 P_{0} 3 ln (3) t = P_{0} e^{(\frac{l n ( 2 )}{5} t)} = e^{(\frac{l n ( 2 )}{5} t)} = \frac{ln ( 2 )}{5} t = \frac{5 ln ( 3 )}{ln ( 2 )} \approx 7.925$

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.

a) What is the value of $k$ ?

b) If there are initially $100$ grams of the substance, how long will it take to decay to $20$ grams?

Solutions

a) Value of $k$

(spoiler)

The half-life of $10$ hours means that when $t = 10$ , the amount is half the initial amount, or that

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

Substitute into $A (t) = A_{0} e^{k t}$ :

$\frac{1}{2} A_{0} = A_{0} e^{10 k}$

Divide by $A_{0}$ and solve for $k$ :

$\frac{1}{2} ln (0.5) k = e^{(10 k)} = 10 k = \frac{ln ( 0.5 )}{10} \approx - 0.0693$

b) Time to decay from $100$ to $20$ grams

(spoiler)

Using $k = \frac{l n ( 0.5 )}{10}$ , the model is

$A (t) = A_{0} e^{l n (0.5) t /10}$

Now substitute $A_{0} = 100$ and $A (t) = 20$ :

$20 = 100 e^{l n (0.5) t /10}$

$\frac{20}{100} = e^{l n (0.5) t /10}$

$ln (20/100) = \frac{ln ( 0.5 )}{10} t$

$t = ln (20/100) \cdot \frac{10}{ln ( 0.5 )}$

$= 23.219$

It will take a little over $23$ hours to decay from $100$ to $20$ grams.

Here is a question similar to #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

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

at the two heights.

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)

Separate the variables:

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

Integrate both sides:

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

Plug in the initial condition $H (0) = 10$ to find $C$ :

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

$C = - ln (110)$

Now solve for $H$ :

Exponential models

What you’ll learn:

Solving dtdy​=ky

Using an initial condition

Examples

Solution

Solutions

Solutions

Exponential models

What you’ll learn:

Solving dtdy​=ky

Using an initial condition

Examples

Solution

Solutions

Solutions

Exponential models

What you’ll learn:

Solving dtdy​=ky

Using an initial condition

Examples

Solution

Solutions

Solutions

Exponential models

What you’ll learn:

Solving dtdy​=ky

Using an initial condition

Examples

Solution

Solutions

Solutions

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

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

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

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