Unit 7 · Topic 7.1 Beta

Writing Differential Equations

A differential equation states how fast a quantity changes.

Practice 2: Connecting RepresentationsPractice 4: Communication and Notation

Part 1 · Hook

Why this matters

A hot cup of coffee cools fast at first, then more and more slowly as it nears room temperature. Nobody hands you a formula for its temperature. What you can say is how fast it cools: the hotter it is compared with the room, the faster it drops. That sentence, written with a derivative, is a differential equation.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $P(t)$ is a number of cells and $t$ is in hours. In what units is $P'(t)$ measured?

cells per hour
cells
hours per cell
cells per hour per hour

Show the answer

A derivative is a rate: the units of the output per unit of the input.

Correct: cells per hour:
cells:
hours per cell:
cells per hour per hour:

2. A function is increasing on an interval where its derivative is

positive
negative
zero
increasing

Show the answer

If $f'(x) > 0$ on an interval, $f$ is increasing there.

Correct: positive:
negative:
zero:
increasing:

Part 4 · See it

See it first

The graph of T = 20 + 70e^(−0.05t) for t from 0 to 60 minutes falls from 90 °C toward the dashed room-temperature line at 20 °C. Tangent segments at T = 90, 55 and 30 have slopes −3.5, −1.75 and −0.5 °C per minute, each equal to −0.05 times (T − 20). — Coffee cooling in a 20 °C room. Each tangent segment has slope $-0.05(T - 20)$ : the farther $T$ is from 20, the steeper the graph. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

A quantity changes at a rate that depends on its current valuewrite the rate as a derivative and the dependence as a formula: $\frac{dP}{dt} = kP$
"Proportional to" means "a constant times" $\frac{dy}{dt}$ $= k \cdot (\text{that quantity})$ ; "inversely proportional" puts the quantity in the denominator
The signs of $k$ and of the other factor are knownthe sign of $\frac{dy}{dt}$ is known, so you know where the quantity grows and where it shrinks
One rate and the matching value are givensolve for $k$ , then predict the rate at any other value

Part 6 · Key ideas

Key ideas

A differential equation relates a derivative to the variables: $\frac{dy}{dt} = 0.4(50 - y)$ .
"The rate of change of $y$ is proportional to $y$ " is $\frac{dy}{dt} = ky$ . The derivative is on the left; the quantity is on the right.
"Proportional to the difference between $T$ and 20" is $\frac{dT}{dt} = k(T - 20)$ .
The sign of $\frac{dy}{dt}$ says where $y$ increases or decreases; you can read it before knowing any formula for $y$ .
$k$ has units: for $\frac{dP}{dt} = kP$ with $t$ in hours, $k$ is "per hour".

Part 7 · Misconception

A common mistake

The wrong idea: "A population grows at a rate proportional to its size" means $P = kt$ .

What actually happens: The rate is proportional: $\frac{dP}{dt} = kP$ . The sentence is about $\frac{dP}{dt}$ , and the size it mentions is $P$ , not $t$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A growing culture of cells

The number of cells P and its rate of change, measured at four moments
$P$ (cells)	200	350	500	800
$\frac{dP}{dt}$ (cells per hour)	52	91	130	208

1. Which differential equation is consistent with every column of the table?

$\frac{dP}{dt} = 0.26P$
$\frac{dP}{dt} = 52$
$\frac{dP}{dt} = 0.26t$
$P = 0.26\frac{dP}{dt}$

Show the answer

Divide each rate by its population: every ratio is $0.26$ , so the rate is proportional to $P$ with $k = 0.26$ .

Correct: $\frac{dP}{dt} = 0.26P$ : $\frac{52}{200}$ $= \frac{91}{350}$ $= \frac{130}{500}$ $= \frac{208}{800}$ $= 0.26$ : the rate is always $0.26$ times the size.
$\frac{dP}{dt} = 52$ : A constant rate would give one rate for every column, but the rate grows with P.
$\frac{dP}{dt} = 0.26t$ : The table gives no times; the rate is tied to the number of cells, not to the clock.
$P = 0.26\frac{dP}{dt}$ : The ratio is upside down: $0.26 \cdot 52 = 13.52$ , not 200.

2. The table fits $\frac{dP}{dt} = kP$ . Find the constant of proportionality $k$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$k$ $= \frac{dP/dt}{P}$ $= \frac{52}{200}$ $= 0.26$ per hour; every column gives the same value.

Answer: $0.26$ ≈ 0.260

3. What does the entry 91 in the table mean?

After 350 hours, there are 91 cells.
When there are 350 cells, the population is growing at 91 cells per hour.
From now on, the population grows by 91 cells each hour.
The population has grown by 91 cells since the first measurement.

Show the answer

The second row is $\frac{dP}{dt}$ : at the moment $P = 350$ cells, the population increases at 91 cells per hour.

After 350 hours, there are 91 cells.: The 350 is a number of cells, not a time, and 91 is a rate, not a count.
Correct: When there are 350 cells, the population is growing at 91 cells per hour.: It is the derivative, a rate, at the moment P = 350.
From now on, the population grows by 91 cells each hour.: The rate changes as P changes; 91 holds only at that moment.
The population has grown by 91 cells since the first measurement.: It grew by 150 cells between the first two columns; 91 is a rate, not a change in size.

4. The rate of change of y with respect to x is inversely proportional to the square of y. Which equation says this?

$\frac{dy}{dx} = ky^2$
$\frac{dy}{dx} = \frac{k}{x^2}$
$y = \frac{k}{x^2}$
$\frac{dy}{dx} = \frac{k}{y^2}$

Show the answer

"Inversely proportional to $y^2$ " means $\frac{k}{y^2}$ ; "the rate of change of $y$ with respect to $x$ " is $\frac{dy}{dx}$ .

$\frac{dy}{dx} = ky^2$ : That is directly proportional to the square of y.
$\frac{dy}{dx} = \frac{k}{x^2}$ : The sentence is about the square of y, not of x.
$y = \frac{k}{x^2}$ : The sentence describes the rate of change, so a derivative belongs on the left.
Correct: $\frac{dy}{dx} = \frac{k}{y^2}$ : Inversely proportional to $y^2$ : a constant divided by $y^2$ .

5. A quantity satisfies $\frac{dy}{dt} = 0.4(50 - y)$ . Find $\frac{dy}{dt}$ when $y = 30$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{dy}{dt}$ $= 0.4(50 - 30)$ $= 0.4(20)$ $= 8$ .

Answer: 8

6. Water drains from a tank. The depth $h$ , in feet, changes at a rate equal to $-0.6$ times the square root of the depth. Write $\frac{dh}{dt}$ in terms of $h$ .

Type an expression for $\frac{dh}{dt} =$ .

Show the answer

$\frac{dh}{dt} = -0.6\sqrt{h}$ : negative because the depth falls, and proportional to $\sqrt{h}$ .

Answer: $\frac{dh}{dt} =$ $-0.6\,\sqrt{h}$

Part 9 · Summary

Summary

A differential equation states how fast a quantity changes. Translate words carefully: "the rate of change of

y

" is

\frac{dy}{dt}

, "proportional to" is "

k

times", and the sign of the right side tells you where

y

rises or falls. One known rate fixes

k

Part 10 · Up next

What comes next

Part 11 · Connections