Model proportional growth or decay

A cooling cup of coffee

A cup of coffee at 90 °C stands in a room at 20 °C. The curve shows its temperature \(T\) over 30 minutes, and the dashed line marks the room temperature.

Drag the slider to change the minute. The teal bracket shows the gap, how much hotter the coffee is than the room. The orange tangent shows how fast the temperature falls.

At the start the gap is 70 and the slope is −7 degrees per minute. When the coffee reaches 55 °C, the gap is 35 and the slope is −3.5.

The slope is −0.1 times the gap, not −0.1 times the temperature, which would give −5.5 at 55 °C.

What the rate depends on

A rate statement names the quantity that the rate is proportional to. Call it the driver.

Here the driver is the gap between the amount \(y\) and a fixed value \(M\), the teal bracket. Move \(M\) down to 0, and the gap becomes the amount itself.

Either way, the slope at every moment is the same constant times the driver. To write a model, ask what the rate is proportional to.

Notation

1. A rate is proportional to its driver when it equals a constant \(k\) times the driver.

   \[\cy{\dfrac{dy}{dt}} = k \cdot (\text{driver})\]

2. If the driver is the amount \(y\), the rate is \(ky\). The amount grows when \(k > 0\) and decays when \(k < 0\).

   \[\cy{\dfrac{dy}{dt}} = k\,\cb{y}\]

3. If the driver is the gap between \(y\) and a fixed value \(M\), such as the room temperature, write it as \(y - M\), the teal bracket. With \(k < 0\) the rate is negative above \(M\) and positive below it, so \(y\) moves toward \(M\). Written as \(M - y\), the gap needs \(k > 0\).

   \[\cy{\dfrac{dy}{dt}} = k\,(\cb{y} - M)\]

4. If the driver is another quantity, write that quantity. Inversely proportional to \(y\) means that the driver is \(1/y\).

   \[\cy{\dfrac{dy}{dt}} = \dfrac{k}{\cb{y}}\]

5. To find \(k\), divide the rate by the driver at one moment when you know both.

   \[k = \dfrac{\text{rate}}{\text{driver}}\]

When it fails

The equation fixes the room at \(M = 20\). If a window opens and the room cools to 5 °C, the coffee cools toward 5 instead. The red dashed curve keeps \(M = 20\) and flattens there, so it is wrong.

The model needs \(M\) to stay fixed. If the room temperature changes, write a new equation.

Example 1: medicine in the blood

A patient has 80 mg of a medicine in the blood. The medicine leaves at a rate proportional to the amount \(Q\) present. At this moment it leaves at 9.6 mg per hour. Write the equation.

1. The rate is proportional to the amount, so the driver is \(Q\), the height of the teal bracket. The medicine leaves, so \(k\) is negative.

   \[\cy{\dfrac{dQ}{dt}} = k\,\cb{Q}\]

2. At this moment the orange tangent has slope \(-9.6\) and the driver is 80, so divide the rate by the driver.

   \[k = \dfrac{-9.6}{80} = -0.12\]

3. The rate is in mg per hour and the driver in mg, so \(k\) is \(-0.12\) per hour.

   \[\cy{\dfrac{dQ}{dt}} = -0.12\,\cb{Q}\]

4. As the amount falls, the medicine leaves more slowly, so the curve flattens. When \(Q = 40\), the rate is \(-0.12 \times 40 = -4.8\) mg per hour.

Example 2: ice on a lake

The thickness \(y\) of ice on a lake, in centimeters, grows at a rate inversely proportional to \(y\). When \(y = 2\), the ice grows 1.5 centimeters per day. Write the equation.

1. A tempting wrong move: Find \(k\) as in the medicine example: divide the rate 1.5 by the amount 2.

   \[\begin{aligned} k &= \dfrac{1.5}{2} = 0.75 \\ \dfrac{dy}{dt} &= 0.75\,y \end{aligned}\]

   Inversely proportional means that thicker ice grows more slowly, so the curve flattens. The red dashed curve follows this rule and climbs faster as the ice thickens. At \(y = 4\) it gives the rate 3, twice the rate at \(y = 2\).

2. The driver is \(1/y\), which is \(\tfrac{1}{2}\) at \(y = 2\), so \(k = 1.5 \div \tfrac{1}{2} = 3\).

   \[\cy{\dfrac{dy}{dt}} = \dfrac{3}{\cb{y}}\]

3. At \(y = 4\) the driver is \(\tfrac{1}{4}\), so the rate is \(3 \times \tfrac{1}{4} = 0.75\) centimeter per day, half the rate at \(y = 2\).

Practice

A car speeds up toward a top speed of 40 m/s. Its speed \(v\) increases at a rate proportional to \(40 - v\). At \(v = 10\), it gains 3 m/s each second. Write the equation and find the rate at \(v = 25\).

The driver is the teal gap between \(v\) and 40. Find \(k\) from the moment when \(v = 10\).

Questions

A cold drink warms in a room at 30 °C. The graph shows its temperature \(T\), and the dashed line marks the room temperature.

1. Which equation fits the graph at the marked moment?

   • \(dT/dt = 0.15\,(30 - T)\)
   • \(dT/dt = 0.15\,T\)
   • \(dT/dt = 1.5\,(30 - T)\)
   Answer

   \(dT/dt = 0.15\,(30 - T)\)

2. The dot moves toward the dashed line. What happens to the slope of the tangent?

   • It moves closer to 0.
   • It stays at 1.5.
   • It becomes steeper.
   Answer

   It moves closer to 0.

3. The amount \(C\) of chlorine in a pool decreases at a rate proportional to \(C\). At \(C = 6\) mg, it decreases 1.2 mg per day. What is the rate when \(C = 4\)?

   • −0.8 mg per day
   • −1.2 mg per day
   • −0.2 mg per day
   • 0.8 mg per day
   Answer

   −0.8 mg per day

Summary

You ask what the rate is proportional to, write that driver on the right side, and find the constant from one known rate.

This works because the slope is one constant times the driver, whether that is the amount, a gap, or another quantity.

You use it when a problem says that a rate is proportional to something or inversely proportional to it.

This is one of four skills in topic 7.1 of Which & Why: AP Calculus.