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The Derivative as a Rate of Change

Read dy/dx as how fast one quantity changes per unit of another, with the units to match, so a derivative answers questions about cost per item, litres per second or metres per second.

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What a learner can do afterwards

  • State the units of dV/dt when V is in litres and t in seconds
  • Interpret a derivative of zero in a real context
  • Say what a negative derivative means for a quantity that is being modelled

1 · Read

You can now differentiate a polynomial in one line. So far dy/dx has meant one thing: the steepness of a picture. It means something better than that. The Ridgeway drainage tank holds V litres of water t seconds after the valve opens, and V is a rule in t. Nobody on the crew cares about the steepness of that graph. They care how fast the volume is changing, which is the same number wearing different clothes. A derivative is a rate: how much one real quantity changes for each unit of another.

The units follow the notation exactly. dV/dt has V on the top and t on the bottom, so its units are the units of V divided by the units of t. Litres over seconds gives litres per second. Cost in pounds over items gives pounds per item. Distance in metres over time in seconds gives metres per second, which is speed, a rate you have measured since long before you met calculus. Read the two letters, take their units, and put one over the other. There is nothing else to remember.

V is in litrest is in secondsdV/dt is litres per second
Top unit over bottom unit. The notation is telling you the answer.
Try it together

Say the tank follows V = 500 - 30t + 0.5t², with V in litres and t in seconds. Differentiating gives dV/dt = -30 + t litres per second. At the moment the valve opens, t = 0, so dV/dt = -30: water is leaving at 30 litres every second. At t = 10 it is -20, so the flow has slowed. Setting -30 + t = 0 gives t = 30, and at that instant the volume is 500 - 900 + 450 = 50 litres and has stopped changing. One rule answers all three questions.

The sign carries the story. A positive derivative means the quantity is growing as the input grows. A negative one means it is shrinking, which is why every reading above came out negative: the tank was emptying. A derivative of zero means the quantity is momentarily still, neither rising nor falling at that instant. Two warnings about zero. It does not mean the quantity itself is zero, and unless it stays zero over a stretch, it does not mean the quantity has settled. It is one motionless instant.

A sign is not decoration. It says which way the quantity is going.

positive: growingnegative: shrinkingzero: still for an instant
Three signs, three stories, and none of them is about the size of the quantity.
Try it together

None of this is about water. A workshop's cost is C = 900 + 4n + 0.03n² pounds for n items, so dC/dn = 4 + 0.06n pounds per item. At n = 100 that is 10, meaning the next item adds about £10 to the bill. The sign is positive, so cost rises with output, which is what anyone would expect. Change the letters and the reading changes with them: for height h in metres against time t in seconds, dh/dt is metres per second and tells you how fast something is climbing.

A derivative is a rate of change, not just a steepness. Its units are the units of the top variable divided by the units of the bottom one, so dV/dt with litres and seconds is litres per second. A positive value means the quantity is growing, a negative value means it is shrinking, and zero means it is momentarily still. Zero does not mean the quantity is zero, and one instant of zero does not mean it has stopped for good.

2 · Watch

Take it off screen

Print a worksheetA4 with an answer key page for grown-ups. No screen, no internet.

Where it sits

Then practise

24 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.

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The Derivative as a Rate of Change · Mathematics, ages 17 to 18 · LightMySky