The Derivative as a Rate of Change · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

The Derivative as a Rate of Change

Mathematics · Calculus & Analysis · ages 17-18
Name ______________________   Date ____________

  1. C is a cost in pounds and n is a number of items. What are the units of dC/dn?

    • pounds
    • pounds per item
    • items per pound
    • pounds times items

  2. A tank holds V = 400 - 5t litres after t seconds. What is dV/dt, with its units?

    • -5 litres per second
    • 5 litres per second
    • 400 litres
    • -5t litres per second

  3. A positive derivative means the quantity is growing as the input grows.

    Circle one:   True   False

  4. A quantity P has dP/dt negative at a given moment. What is P doing?

    • it is zero
    • it is standing still
    • it is falling
    • it is rising slowly

  5. A tank follows V = 600 - 8t + 0.2t² litres after t seconds. What is dV/dt at t = 10?

    • -8 litres per second
    • 4 litres per second
    • -4 litres per second
    • 540 litres per second

  6. A is an area in square centimetres and r is a radius in centimetres. What are the units of dA/dr?

    • square centimetres per centimetre
    • cubic centimetres
    • centimetres per square centimetre
    • square centimetres

  7. Ravi reads dV/dt = -6 and says the tank holds -6 litres. Where is the slip?

    • the sign should have been dropped, so it holds 6 litres
    • the units should have been seconds per litre
    • a derivative is a rate of change, so -6 measures how fast the volume is falling, not the volume
    • the reading is only valid once t is bigger than 6

  8. A firm's revenue is R = 30n - 0.05n² pounds from n items. What is dR/dn at n = 100?

    • 25 pounds per item
    • 20 pounds per item
    • 30 pounds per item
    • 2500 pounds per item

  9. A ball's height is h = 30t - 5t² metres after t seconds. At what time t is dh/dt zero?

    Answer: ______________

  10. Nina works out dV/dt = -4 for a tank and reports that it is filling at 4 litres per second. Where is the slip?

    • the units should have been seconds, not litres per second
    • the answer should have been -4t rather than -4
    • a rate cannot be negative, so the differentiation is wrong
    • a negative rate means the volume is falling, so it is emptying
LightMySky · lightmysky.comW1-mt_RqyLE3jrAW-s1

Answer key

For grown-ups. Fold this page away before handing over the rest.

The Derivative as a Rate of Change W1-mt_RqyLE3jrAW-s1

  1. pounds per item · Pounds sit on the top and items on the bottom, so the units are pounds per item: what one more item adds to the bill.
  2. -5 litres per second · Differentiating gives -5, and the units are litres over seconds. So 5 litres leave the tank every second.
  3. True · That is what a positive rate of change says. The size of the number says how fast, and the sign says which way.
  4. it is falling · A negative rate of change means the quantity goes down as time goes up. It says nothing about how big P is.
  5. -4 litres per second · dV/dt = -8 + 0.4t. At t = 10 that is -8 + 4, which is -4 litres per second.
  6. square centimetres per centimetre · The rule does not change: top unit over bottom unit. Square centimetres over centimetres gives square centimetres per centimetre.
  7. a derivative is a rate of change, so -6 measures how fast the volume is falling, not the volume · The derivative measures how fast the volume is changing, not how much there is. The volume is V, and it is a separate number that Ravi has not been given.
  8. 20 pounds per item · dR/dn = 30 - 0.1n. At n = 100 that is 30 - 10, which is 20 pounds per item.
  9. 3 · dh/dt = 30 - 10t. Setting that to zero gives 10t = 30, so t = 3 seconds.
  10. a negative rate means the volume is falling, so it is emptying · She read the size and dropped the sign. Negative means the quantity goes down as time goes up, so the tank is losing 4 litres a second.
Worksheet · LightMySky