Connect an exponential model with its rate law
A rate that follows the amount
A bacteria culture holds \(P\) thousand cells after \(t\) hours. The slope of the orange tangent at the dot is the rate \(\cy{\frac{dP}{dt}}\).
Drag the dot to the right. The tangent becomes steeper, because a larger culture grows faster.
The right graph plots the rate across and the amount up. The orange bar shows the rate at the amount of the dot and ends on the green line.
Green means that the rate is positive. At \(P = 50\) the rate is 10 and at \(P = 100\) it is 20, so the rate divided by the amount is 0.2 each time.
The equation \(\frac{dP}{dt} = 0.2P\) has the solution \(P = 50e^{0.2t}\).
The line through the origin
Let \(y\) be an amount and \(k\) a constant. The equation \(\frac{dy}{dt} = ky\) is called the rate law.
Drag the slider for k. The line turns about the origin.
The line is green for a positive rate and red for a negative one. The amount grows for \(k > 0\) and decays for \(k < 0\).
At every moment, the rate divided by the amount equals \(k\).
Notation
The rate of change of \(y\) is proportional to \(y\), with constant of proportionality \(k\).
\[\cy{\frac{dy}{dt}} = ky\]Test \(y = y_0e^{kt}\). The chain rule brings the factor \(k\) down from the exponent, so the derivative is \(k\) times \(y\).
\[\frac{d}{dt}\left(y_0e^{kt}\right) = k\cdot y_0e^{kt} = ky\]The number \(y_0\) is the amount at \(t = 0\).
Every solution has this form: separate the variables, integrate and exponentiate.
\[\begin{aligned} \frac{dy}{y} &= k\,dt \\ \ln y &= kt + C \\ y &= e^{C}e^{kt} = y_0e^{kt} \end{aligned}\]At \(t = 0\) the factor \(e^{kt}\) is 1, so \(e^{C}\) is the starting amount \(y_0\). The step assumes \(y > 0\), as it is for an amount.
Some rates are proportional to a difference \(y - A\). Let \(u = y - A\). Because \(A\) is fixed, \(\frac{du}{dt} = \frac{dy}{dt} = ku\), so \(u\) follows the rate law and starts at \(y_0 - A\).
\[\begin{aligned} \frac{dy}{dt} &= k(y - A) \\ y &= A + (y_0 - A)e^{kt} \end{aligned}\]The rate is 0 at \(A\), so the curve approaches the dashed line at \(A\) and the line on the right crosses 0 there.
When it fails
The rate law uses the amount at each moment. Suppose instead that the rate stays at its starting value \(ky_0\).
The amount then follows the red dashed straight line, and the red vertical line on the right is this constant rate. Press the button to move forward in time. The curve rises above the straight line, because the rate \(ky\) grows with the amount. The fixed rate divided by the amount, \(\frac{ky_0}{y}\), falls below \(k\).
Example 1: Caffeine leaves the blood
The amount \(A\) of caffeine in the blood, in mg, satisfies \(\dfrac{dA}{dt} = -0.25A\), where \(t\) is in hours and \(A(0) = 80\). Write \(A(t)\) and find the rate when \(A = 40\).
The rate is \(-0.25\) times the amount, a negative constant, so the amount decays and the line on the right is red. The orange bar shows the starting rate.
\[\frac{dA}{dt} = -0.25\cdot80 = -20\ \text{mg/h}\]The solution is \(A = A_0e^{kt}\) with the starting amount \(A_0 = 80\).
\[A = 80e^{-0.25t}\]When \(A = 40\), the rate is \(-0.25\) times 40, so the amount falls by 10 mg per hour.
\[\frac{dA}{dt} = -0.25\cdot40 = -10\ \text{mg/h}\]
Example 2: Coffee in a cool room
Coffee at \(90^\circ\text{C}\) stands in a room at \(20^\circ\text{C}\). Its temperature \(T\) satisfies \(\dfrac{dT}{dt} = -0.1(T - 20)\), where \(t\) is in minutes. Find \(T(t)\).
The rate is proportional to \(T - 20\), not to \(T\), so the line on the right crosses 0 at \(T = 20\). The rate is 0 there, so the curve approaches the dashed line at the room temperature.
A tempting wrong move: Treat \(T\) as the amount. The solution is then a plain exponential.
\[T = 90e^{-0.1t}\]The red dashed curve falls below the room temperature, to about 12.2 at \(t = 20\). The rate is 0 at \(20^\circ\text{C}\), so the coffee cannot cool below it.
Let \(u = T - 20\). Then \(u\) follows the rate law and starts at \(90 - 20 = 70\), shown by the orange bar.
\[\frac{du}{dt} = -0.1u \;\Rightarrow\; u = 70e^{-0.1t}\]Add 20 to return to \(T\). As \(t\) increases, \(70e^{-0.1t}\) approaches 0, so the temperature approaches \(20^\circ\text{C}\).
\[T = 20 + 70e^{-0.1t}\]
Practice
A fish population \(P\) satisfies \(\dfrac{dP}{dt} = 0.03P\), where \(t\) is in years and \(P(0) = 400\). Write \(P(t)\) and find the rate when \(P = 800\).
On the right, the rate is 0.03 times the amount. Move the dot to \(P = 800\) and read the length of the orange bar.
Solution
The rate is a constant times the amount, so the solution is an exponential with \(k = 0.03\) and \(P_0 = 400\).
\[P = 400e^{0.03t}\]At \(P = 800\), the rate is 0.03 times 800, the length of the orange bar.
\[\begin{aligned} \frac{dP}{dt} &= 0.03\cdot800 \\ &= 24\ \text{fish per year} \end{aligned}\]
Questions
Questions 1 and 2 use a quantity with \(\dfrac{dy}{dt} = ky\). At the orange dot the amount is 60 and the tangent has slope 6.
What is \(k\)?
- 0.1
- 6
- 10
- 60
Answer
0.1
The amount doubles from 60 to 120. What happens to the rate?
- It doubles, from 6 to 12.
- It stays at 6.
- It increases by 60, to 66.
Answer
It doubles, from 6 to 12.
A pan cools with \(T = 25 + 55e^{-0.05t}\). Which rate law does \(T\) follow?
- \(\frac{dT}{dt} = -0.05(T - 25)\)
- \(\frac{dT}{dt} = -0.05T\)
- \(\frac{dT}{dt} = -0.05(T + 25)\)
- \(\frac{dT}{dt} = -0.05(T - 55)\)
Answer
\(\frac{dT}{dt} = -0.05(T - 25)\)
Summary
You recognize a rate that is a constant times the amount, or times a difference, and write the exponential solution.
This works because the derivative of \(y_0e^{kt}\) is \(k\) times itself.
You use it when the rate divided by the amount is the same at every moment.
This is one of four skills in topic 7.8 of Which & Why: AP Calculus.