Use initial conditions and units
Three savings accounts
Three people open savings accounts. Each account earns interest at a rate equal to 5% of its balance \(B\) each year, and the graph shows the balances for 20 years.
The three faint curves follow the same equation but start at different heights. Drag the orange dot to change the starting deposit.
At the start, the slope is 5% of the deposit. It is 25 dollars per year for a deposit of 500, 50 for 1,000, and 100 for 2,000.
The equation and the start
The faint curves all follow one equation. Drag the orange dot to choose where the curve starts.
The equation tells the slope at each height, but it does not tell the height at the start. This extra fact is the initial condition.
The initial condition picks one curve, and it stays out of the equation because the rate uses the amount at the current moment.
Notation
The initial condition gives the amount at the starting time. For the accounts, it is the deposit.
\[\cy{B(0)} = 1000\]The equation and the initial condition together are an initial value problem.
\[\dfrac{dB}{dt} = 0.05\,B, \qquad B(0) = 1000\]The deposit 1000 is in the condition only, because the equation uses the balance \(B\) at the current moment. At year 10, the slope comes from the balance at year 10.
Both sides of the equation must have the same unit. The left side is in dollars per year, and the constant 0.05 is per year, so \(0.05B\) is in dollars per year.
\[\dfrac{\$}{\text{year}} = \dfrac{1}{\text{year}} \cdot \$\]The constant depends on the time unit. If you measure time in months, the constant becomes \(0.05/12\) per month, because a month is 1/12 of a year.
Example 1: a radioactive sample
A sample has 120 mg of a radioactive substance. The amount \(m\) decays at a rate equal to 0.09 times \(m\), with \(t\) in days. Write the initial value problem. Then find the rate at the start.
The substance decays, so the rate is negative. The rate is proportional to the amount \(m\) now, shown by the teal bracket.
\[\dfrac{dm}{dt} = -0.09\,\cb{m}\]The starting amount goes in the initial condition, not in the equation.
\[\cy{m(0)} = 120\]The constant 0.09 is per day, so each side is in milligrams per day. At the start \(m = 120\), and the rate is in red because the sample loses mass.
\[\begin{aligned} \dfrac{dm}{dt} &= -0.09 \times 120 \\ &= \cr{-10.8}\ \text{mg/day} \end{aligned}\]
Example 2: cells in a culture
The number \(N\) of cells in a culture follows \(dN/dt = 0.6N\), with \(t\) in hours. At this moment \(N = 500\). About how many cells does the culture gain in the next minute?
The constant 0.6 is per hour, so the rate is in cells per hour.
\[0.6 \times 500 = 300\ \text{cells per hour}\]A tempting wrong move: The rate is 300, so the culture gains 300 cells in a minute.
\[300\ \text{cells per minute}\]The orange tangent rises 300 over a run of one full hour, so 300 is the gain in an hour. A minute is only a small part of that run.
A minute is 1/60 of an hour, so over a run of one minute the tangent rises only 1/60 of 300.
\[300 \div 60 = 5\ \text{cells per minute}\]
Practice
A pizza at 20 °C goes into an oven at 220 °C. Its temperature \(T\) rises at a rate equal to 0.04 times \(220 - T\), with \(t\) in minutes. Write the initial value problem. Find the rate at the start and the rate when \(T = 120\).
The starting temperature goes in the condition. In the equation, use the temperature at the moment you ask about.
Solution
The equation uses the gap between the oven and the pizza, the teal bracket. The condition gives the start.
\[\begin{aligned} \dfrac{dT}{dt} &= 0.04\,(220 - T) \\ T(0) &= 20 \end{aligned}\]At the start \(T = 20\), so the gap is 200 and the rate is 8 degrees Celsius per minute.
\[0.04\,(220 - 20) = 8\]When \(T = 120\), the gap is 100, so the rate is smaller and the orange tangent is less steep.
\[0.04\,(220 - 120) = 4\]
Questions
The graph shows three curves that follow \(dy/dt = -0.1y\), with \(t\) in hours. The orange tangent touches the middle curve at the start.
What is the slope of the middle curve at t = 0?
- −5
- −0.1
- 50
Answer
−5
Which statement about the three curves is correct?
- At t = 0, the top curve slopes down more steeply than the others.
- At t = 0, all three curves have the same slope.
- The top curve follows a different equation.
Answer
At t = 0, the top curve slopes down more steeply than the others.
With \(t\) in days, a quantity follows \(dy/dt = 0.12y\). The same quantity is recorded with \(w\) in weeks. Which equation is correct?
- \(dy/dw = 0.84\,y\)
- \(dy/dw = 0.12\,y\)
- \(dy/dw \approx 0.017\,y\)
Answer
\(dy/dw = 0.84\,y\)
Summary
The initial condition is one point on the graph, and it picks one curve because the equation gives the slope at every height.
Both sides of the equation have the same unit, so the constant 0.05 is per year and changes with the unit of time.
You use this when a problem gives a starting amount, or measures time in a unit different from the unit of the constant.
This is one of four skills in topic 7.1 of Which & Why: AP Calculus.