Forming Differential Equations

Read the sentence for its two sides

A sentence about how something changes is a differential equation written in words. "A population grows at a rate proportional to its size" names both sides. "The rate" at which the population P grows is $\frac{dP}{dt}$dP/dt, the change in P per unit time. "Its size" is P. "Proportional to" means equal to a constant times, so the sentence becomes $\frac{dP}{dt}=$dP/dt = kP.

The constant k is positive here, because the population grows. Nothing in the sentence says how big k is; a reading fixes that. If the population is growing by 50 a year when it is 1000, then 50 = k × 1000 and k = 0.05 per year.

The sign says which way

A hot drink cools toward the temperature R of the room, faster when it is much hotter than the room and slower as it gets close. What drives the change is the gap T − R, and the temperature falls, so $\frac{dT}{dt}=-k\left(T-R\right)$dT/dt = −k(T − R), with k > 0.

Check the sign both ways. For tea at 90 degrees in a room at 20 with k = 0.1 per minute, the gap is 70 and $\frac{dT}{dt}=-0.1\times 70=-7$dT/dt = −0.1 × 70 = −7 degrees per minute: the tea cools. At 50 degrees the rate is −0.1 × 30 = −3 degrees per minute, slower. A cold drink at 5 degrees has a gap of −15, so $\frac{dT}{dt}=+1.5$dT/dt = +1.5 degrees per minute and it warms. One equation covers both, because the sign of the gap does the work.

Writing −kT instead would drive the temperature toward 0, not toward the room.

Rate, driver, constant

Every translation takes the same three moves. Name the rate: $\frac{d\left(\text{what changes}\right)}{dt}$d(what changes)/dt. Name what drives it. Join the two with a constant k, and give it the sign that makes the quantity rise or fall as the sentence says.

"Water leaks from a tank at a rate proportional to the volume left" gives $\frac{dV}{dt}=-kV$dV/dt = −kV: V is what changes, V drives it, and leaking makes it fall.

"The radius of a crystal grows at a rate inversely proportional to its radius" gives $\frac{dr}{dt}=\frac{k}{r}$dr/dt = k/r: inversely proportional means a constant divided by the driver.

"A rumor spreads through N people at a rate proportional to the product of those who have heard it and those who have not" gives $\frac{dx}{dt}=kx\left(N-x\right)$dx/dt = kx(N − x), where x is the number who have heard it.

"The depth of water falls at a rate proportional to the square root of the depth" gives $\frac{dy}{dt}=-k\sqrt{y}$dy/dt = −k√y, with y the depth.

More than one thing changing

When water runs into a tank at 5 liters a minute and leaks out at a rate proportional to the volume, the rate is what comes in minus what goes out: $\frac{dV}{dt}=5-kV$dV/dt = 5 − kV. With k = 0.1 per minute the volume stops changing when 5 = 0.1V, at V = 50 liters.

Sometimes the rate in the sentence belongs to a different quantity from the one asked about. A balloon’s volume grows at a steady 20 cubic centimeters per second, and the question is about its radius. The chain rule links them: $\frac{dV}{dt}=\left(\frac{dV}{dr}\right)\left(\frac{dr}{dt}\right)$dV/dt = (dV/dr)(dr/dt), and for a sphere $\frac{dV}{dr}=4\pi r^2$dV/dr = 4πr². So $\frac{dr}{dt}=\frac{20}{4\pi r^2}$dr/dt = 20/(4πr²). When r = 5 centimeters, $\frac{dr}{dt}=\frac{1}{5\pi }\approx 0.0637$dr/dt = 1/(5π) ≈ 0.0637 centimeters per second.

Then solve it

Once the equation is written, separating the variables solves it. $\frac{dT}{dt}=-0.1\left(T-20\right)$dT/dt = −0.1(T − 20) separates to $\left(\frac{1}{T-20}\right)dT=-0.1dt$(1/(T − 20)) dT = −0.1 dt, which integrates to ln|T − 20| = −0.1t + c, so $T-20=A{e}^{-0.1t}$T − 20 = Ae^(−0.1t).

The tea starts at 90, so A = 70 and $T=20+70e^{-0.1t}$T = 20 + 70e^(−0.1t). After 10 minutes $T=20+70e^{-1}\approx 45.75$T = 20 + 70e^(−1) ≈ 45.75 degrees, and the derivative there, from a difference quotient, is −2.575, which is −0.1 × 25.75, as the equation says. The tea reaches 30 degrees when $e^{-0.1t}=\frac{1}{7}$e^(−0.1t) = 1/7, at $t=10\ln 7\approx 19.5$t = 10 ln 7 ≈ 19.5 minutes.

Temperature T against time t in minutes, both solving $\frac{dT}{dt}=-0.1\left(T-20\right)$dT/dt = −0.1(T − 20). The gold curve starts at 90 and falls, passing 45.75 at t = 10. The dashed curve starts at 5 and rises. Both close in on the dashed horizontal line T = 20, the room, from opposite sides.

The usual mistakes

Writing a constant rate. "Grows at a rate proportional to its size" is $\frac{dP}{dt}=$dP/dt = kP, not $\frac{dP}{dt}=k$dP/dt = k: the size has to appear.

Leaving out the minus sign. A leak is $\frac{dV}{dt}=-kV$dV/dt = −kV with k > 0; without the minus, the volume would grow.

Using the temperature instead of the gap. Cooling toward a room at R is $\frac{dT}{dt}=-k\left(T-R\right)$dT/dt = −k(T − R), not −kT.

Writing a solution instead of a rate. P = kt is a formula for P, not a statement about $\frac{dP}{dt}$dP/dt.

Reading inversely proportional as a product. $\frac{dr}{dt}=\frac{k}{r}$dr/dt = k/r, not kr.

A sentence about a water butt

In the application below, the first step is the translation: the depth falls at a rate proportional to its square root, with a minus sign because it falls. The rest is separating and integrating.