Translate a rate statement into an equation
A patch of algae
A patch of algae floats on a pond. The graph shows its area \(A\), in square meters, \(t\) days after the count starts.
Drag the slider to change the day. The orange line is the tangent, which touches the curve at the dot. Its slope is how fast the area grows, in square meters per day.
At \(t = 0\) the area is 10 and the slope is 2. When the area is 20, the slope is 4.
The teal bracket is the area \(A\). On every day the slope is 0.2 times the area, and the next steps turn this rule into an equation.
Any quantity and its rule
The curve now shows any quantity \(y\) against time \(t\). Drag the moment along the curve.
The slope and the height both change, but the slope divided by the height is the same number at every moment. Call this number the multiple \(k\). A larger multiple bends the curve more.
A rate statement tells how the rate of change depends on the quantity, such as “the rate is \(k\) times the quantity”. To write the equation, replace each phrase by a symbol.
Notation
In the opening, the rate of change of the area is 0.2 times the area. Name the quantity and the time, with units. Here \(A\) is the area in square meters, and \(t\) is the time in days.
\[\cb{A(t)}\ \text{in m}^2, \qquad t \text{ in days}\]The phrase “the rate of change of the area” means the derivative, which is the slope of the tangent.
\[\cy{\dfrac{dA}{dt}}\]The word “is” becomes an equals sign, and “0.2 times the area” becomes \(0.2A\). An equation that contains a derivative is a differential equation.
\[\cy{\dfrac{dA}{dt}} = 0.2\,\cb{A}\]Both sides must have the same unit. The left side is in square meters per day, and the number 0.2 is per day, so \(0.2A\) is also in square meters per day.
\[\dfrac{\text{m}^2}{\text{day}} = \dfrac{1}{\text{day}} \cdot \text{m}^2\]
When it fails
The pond has an edge at 40 square meters, so the real area stops there and the real slope is 0. The red dashed curve follows the rule past the edge, where it still gives \(0.2 \times 40 = 8\).
An equation is true only while its assumptions hold. This rule assumes that the algae have room to grow, so state the assumption with the model.
Example 1: a barrel that drains
Water drains from a barrel. The depth \(D\), in centimeters, falls at a rate equal to 0.5 times the square root of \(D\). Time \(t\) is in minutes. Write the equation.
The rate of change of the depth is the derivative \(dD/dt\), the slope of the orange tangent.
\[\cy{\dfrac{dD}{dt}}\]The depth falls, so the tangent slopes down and the rate is negative. The word “falls” puts a minus sign on the right side.
\[\cy{\dfrac{dD}{dt}} = \cr{-}\ \cdots\]The red arrow shows how fast water leaves. The phrase “0.5 times the square root of \(D\)” becomes \(0.5\sqrt{D}\).
\[\cy{\dfrac{dD}{dt}} = \cr{-}0.5\sqrt{\cb{D}}\]At the start \(D = 16\), so \(\sqrt{D} = 4\) and the rate is \(-0.5 \times 4 = -2\) centimeters per minute, the slope of the tangent.
\[-0.5\sqrt{16} = -2\]
Example 2: a growing town
The population \(P\) of a town, in thousands of people, grows each year at a rate equal to 8% of \(P\). At this moment \(P = 30\). Write the equation, with \(t\) in years.
The rate of change of \(P\) is \(dP/dt\). By the statement, it equals 8% of the population.
\[\cy{\dfrac{dP}{dt}} = \ ?\]A tempting wrong move: The population is 30 now, and 8% of 30 is 2.4, so use 2.4 as the rate.
\[\dfrac{dP}{dt} = 2.4\]This gives the fixed slope of the red dashed line, but the real curve becomes steeper as \(P\) grows. When \(P = 60\), 8% of \(P\) is 4.8, not 2.4.
The statement gives 8% of the population at each moment, so \(P\) stays in the equation. The number 30 is the population at this moment only.
\[\cy{\dfrac{dP}{dt}} = 0.08\,\cb{P}\]
Practice
The radius \(r\), in centimeters, of a tree trunk grows at a rate equal to 6 divided by \(r\). Time is in years. Write the equation. Then find the rate when \(r = 12\).
Write the derivative of \(r\) first. Then write the phrase with symbols. Move the dot to the dashed line, where \(r = 12\).
Solution
The rate of change of \(r\) is \(dr/dt\), and by the statement it equals 6 divided by \(r\).
\[\cy{\dfrac{dr}{dt}} = \dfrac{6}{\cb{r}}\]When \(r = 12\), the rate is \(6 \div 12 = 0.5\) centimeter per year. The dot is now on the dashed line, and the tangent has this slope.
\[\dfrac{6}{12} = 0.5\]
Questions
The graph shows a quantity \(y\) against time \(t\), with an orange tangent at two moments.
Which equation does the graph follow?
- \(dy/dt = 0.25y\)
- \(dy/dt = 2\)
- \(dy/dt = 4y\)
Answer
\(dy/dt = 0.25y\)
What is the slope of the tangent where the curve is 12 high?
- 3
- 12
- 0.25
- 2
Answer
3
The number \(N\) of cells in a dish decreases at a rate equal to 0.01 times the square of \(N\). What is \(dN/dt\) when \(N = 20\)?
- \(-4\)
- 4
- \(-0.2\)
- \(-0.04\)
Answer
\(-4\)
Summary
You name the quantity, write its rate as a derivative, and replace each phrase of the statement by a symbol.
This works because the slope of the curve is the rate, so the equation tells how the slope depends on the quantity.
You use it when a problem describes how fast a quantity changes and asks for the differential equation.
This is one of four skills in topic 7.1 of Which & Why: AP Calculus.