Short on time? This page is the recap. For explanations, the cooling graph and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- A differential equation links a function, its independent variable and one or more of its derivatives, for example dP/dt = 0.03P.
- Its solution is a function, not a number. In Topic 7.1 you only write and read equations; you do not solve them.
- “The rate of change of A” is a derivative, dA/dt. It goes on the left.
- Whatever the rate depends on goes on the right, with a constant of proportionality k.
- Decide the sign from the story: decreasing quantity, negative rate.
- One given fact (a value of the quantity and its rate) lets you find k. Then you can find the rate at any other value.
Key relationships
| Words | Equation |
|---|---|
| A changes at a rate proportional to A | dA/dt = kA |
| … proportional to the square root of A | dA/dt = k√A |
| … inversely proportional to A | dA/dt = k/A |
| … jointly proportional to A and (M − A) | dA/dt = kA(M − A) |
| … proportional to the difference between A and a fixed value R | dA/dt = k(A − R) |
| rate in minus rate out | dA/dt = (in) − (out) |
| acceleration is proportional to velocity | d²s/dt² = k·ds/dt |
Units: both sides match, so units of k = (units of dA/dt) ÷ (units of what k multiplies). For dA/dt = kA with t in hours, k is per hour.
Assumptions
- k is a constant: it does not change with time or with the quantity.
- Say which sign k has. With dQ/dt = −kQ, take k > 0.
- The variables and their units must be stated (for example, t in minutes).
Mistakes to avoid
- Writing the quantity, not its rate: P = kP instead of dP/dt = kP.
- Double negative: dQ/dt = −kQ with k < 0.
- Wrong input: “proportional to time” (kt) versus “proportional to the amount” (kA).
- “Inversely” read as “negative”: inverse means k/A.
- Rate given in words as positive: “cooling at 3 °C per minute” means dT/dt = −3.
- Dropping the variable: dC/dt = 30 − 0.25 instead of 30 − 0.25C.
Quick self-check
- The mass m of a snowball shrinks at a rate proportional to its surface area S. Write the equation. (dm/dt = −kS, k > 0)
- dy/dt = −k(y − 15), k > 0, and y = 35 when dy/dt = −4. Find k. (k = 4/20 = 0.2)
- Using question 2, find dy/dt when y = 25. (−0.2 × 10 = −2)
- Put dW/dt = 2/W, with W > 0, into words. (W increases at a rate inversely proportional to W, constant 2.)
Next: practice questions.