Wemazi

Topic 7.1

Combine competing rates

Builds on writing a rate that is proportional to a driver. 7 minutes.

A tank that fills and leaks

A tank holds 100 liters. A pipe adds 30 liters each minute, and a leak drains 10% of the water each minute.

The graph shows the water \(W\). The green arrow is the rate in and the red arrow the rate out, each as wide as its rate.

Drag the slider. At the start the pipe adds 30 and the leak removes 10, so the water rises at \(30 - 10 = 20\) liters per minute. The orange tangent has this slope, the rate in minus the rate out.

As the water rises, the leak removes more and the slope becomes smaller. At 300 liters the leak would also remove 30, so the slope would be 0.

Rate in and rate out

The tank now holds any amount. Drag the moment along the graph. At every moment the slope is the rate in minus the rate out. It is positive where the green arrow is wider and negative where the red arrow is wider.

Change the rate in. The slope is 0 only where the two arrows have the same width, so the amount moves toward that value. A larger rate in makes that value higher.

Notation

  1. Let \(W\) be the water in liters and \(t\) the time in minutes. Both rates are then in liters per minute.

  2. In words, the rate of change of the amount equals the rate in minus the rate out.

    \[\cy{\dfrac{dW}{dt}} = \cg{\text{rate in}} - \cr{\text{rate out}}\]

  3. The pipe is steady, so its rate is 30. The leak takes 10% of the water, so its rate is \(0.1W\).

    \[\cg{30}, \qquad \cr{0.1\,W}\]

  4. Subtract the rate out from the rate in. The difference is the net rate.

    \[\cy{\dfrac{dW}{dt}} = \cg{30} - \cr{0.1\,W}\]

  5. The amount does not change when the two rates are equal, so \(30 = 0.1W\).

    \[30 - 0.1W = 0 \ \Rightarrow\ W = 300\]

When it fails

A tank starts with 100 liters. A pipe adds 30 liters per minute and a drain removes 40, so \(dW/dt = 30 - 40 = -10\) and \(W\) reaches 0 at \(t = 10\).

After that the equation gives a negative amount, the red dashed line. A tank cannot hold less than 0 liters, so the equation holds only while \(W > 0\).

Example 1: births and deaths

A town has 50 thousand people. Each year, births add 7% of the population, and deaths remove 3% of the population. Write the equation for the population \(P\), in thousands, with \(t\) in years.

  1. Births add 7% of \(P\), so the rate in is \(0.07P\), the green arrow. Deaths remove 3%, so the rate out is \(0.03P\), the red arrow.

    \[\cg{0.07\,P}, \qquad \cr{0.03\,P}\]

  2. Subtract the rate out from the rate in. Both terms contain \(P\), so combine them.

    \[\cy{\dfrac{dP}{dt}} = \cg{0.07\,P} - \cr{0.03\,P} = 0.04\,P\]

  3. At the start the births are \(0.07 \times 50 = 3.5\) and the deaths 1.5 thousand per year, so the slope is 2.

Example 2: sugar in a vat

A vat holds 400 liters of syrup with 10 kg of sugar. New syrup with 0.1 kg of sugar per liter flows in at 10 liters per minute. The stirred mixture flows out at 10 liters per minute. Write the equation for the sugar \(S\), in kilograms.

  1. The rate in is the concentration times the flow, so the pipe brings 1 kilogram of sugar each minute.

    \[\cg{0.1\ \tfrac{\text{kg}}{\text{L}} \times 10\ \tfrac{\text{L}}{\text{min}} = 1\ \tfrac{\text{kg}}{\text{min}}}\]

  2. A tempting wrong move: The rate out is the sugar in the vat times the flow, \(S \times 10\).

    \[\dfrac{dS}{dt} = 1 - 10\,S\]

    The units do not match: kilograms times liters per minute is not kilograms per minute. At \(S = 10\) it removes 100 kg each minute from a vat with 10 kg, so the red dashed curve drops to the axis at once.

  3. The stirred mixture leaves with the vat's concentration, \(S\) kilograms in 400 liters. Multiply \(S/400\) by the flow.

    \[\cr{\dfrac{S}{400} \times 10 = \dfrac{S}{40}}\]

  4. Subtract the rate out from the rate in. The slope is 0 at \(S = 40\), where the concentration \(40/400 = 0.1\) kg per liter equals that of the new syrup.

    \[\cy{\dfrac{dS}{dt}} = \cg{1} - \cr{\dfrac{S}{40}}\]

Practice

A lake has \(N\) fish. Each year, births add 9% of \(N\), deaths remove 4% of \(N\), and fishing removes 200 fish. Write the equation. Then find the number of fish for which \(N\) does not change.

Subtract the total of the red rates out from the green rate in.

Solution

  1. Births are the rate in. Deaths and fishing are both rates out, so add them.

    \[\cy{\dfrac{dN}{dt}} = \cg{0.09\,N} - \left(\cr{0.04\,N} + \cr{200}\right)\]

  2. Combine the terms that contain \(N\).

    \[\dfrac{dN}{dt} = 0.05\,N - 200\]

  3. The number of fish does not change when the slope is 0. With 4000 fish the green arrow is as wide as the two red arrows together, and the graph is horizontal.

    \[0.05\,N - 200 = 0 \ \Rightarrow\ N = 4000\]

Questions

A pool is filled by a pump at a steady rate and drained by a pipe. The graph shows the rate in (solid) and the rate out (dashed) in liters per minute, and below it the water in the pool.

  1. At t = 2, what is the rate of change of the amount?

    • 4 liters per minute
    • 12 liters per minute
    • 8 liters per minute
    Answer

    4 liters per minute

  2. At which time is the amount of water the greatest?

    • t = 4
    • t = 0
    • t = 8
    Answer

    t = 4

  3. A fund earns interest at a rate of 3% of its balance \(B\) each year. The owner takes out $900 each year. What is \(dB/dt\) when \(B = \$40{,}000\)?

    • $300 per year
    • −$300 per year
    • $2,100 per year
    Answer

    $300 per year

Summary

You write each rate in and each rate out on its own, and subtract the total out from the total in.

This works because the slope of the amount is the rate in minus the rate out at every moment.

You use it when a quantity gains and loses at the same time, and a net rate of 0 gives the steady amount.

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