Related Rates of Change (DP IB Maths: AI HL)

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Related Rates of Change

What is meant by rates of change?

  • A rate of change is a measure of how a quantity is changing with respect to another quantity
  • Mathematically rates of change are derivatives
    • space fraction numerator straight d V over denominator straight d r end fraction could be the rate at which the volume of a sphere changes relative to how its radius is changing
  • Context is important when interpreting positive and negative rates of change
    • A positive rate of change would indicate an increase
      • e.g. the change in volume of water as a bathtub fills
    • A negative rate of change would indicate a decrease
      • e.g. the change in volume of water in a leaking bucket

What is meant by related rates of change?

  • Related rates of change are connected by a linking variable or parameter
    • this is usually time, represented byspace t
    • seconds is the standard unit for time but this will depend on context
  • e.g.  Water running into a large bowl
    • both the height and volume of water in the bowl change with time
    • time is the linking parameter

How do I solve problems involving related rates of change?

  • Use of chain rule

   space y equals g left parenthesis u right parenthesis      space u equals f left parenthesis x right parenthesis space space rightwards double arrow space space fraction numerator straight d y over denominator straight d x end fraction equals fraction numerator straight d y over denominator straight d u end fraction cross times fraction numerator straight d u over denominator straight d x end fraction

  • Chain rule is given in the formula booklet in the format above
    • Different letters may be used relative to the context
      • e.g. space V for volume,space S for surface area,space h for height,space r for radius
  • Problems often involve one quantity being constant
    • so another quantity can be expressed in terms of a single variable
    • this makes finding a derivative a lot easier
  • For time problems at least, it is more convenient to use

space space fraction numerator straight d y over denominator straight d t end fraction equals fraction numerator straight d x over denominator straight d t end fraction cross times fraction numerator straight d y over denominator straight d x end fraction

and if it is more convenient to findspace fraction numerator straight d x over denominator straight d y end fraction thanspace fraction numerator straight d y over denominator straight d x end fractionthen use chain rule in the form
space fraction numerator straight d y over denominator straight d t end fraction equals fraction numerator straight d x over denominator straight d t end fraction divided by fraction numerator straight d x over denominator straight d y end fraction
  • Neither of these alternative versions of chain rule are in the formula booklet
STEP 1
Write down the rate of change given and the rate of change required
(If unsure of the rates of change involved, use the units given as a clue
e.g. space straight m space straight s to the power of negative 1 end exponent (metres per second) would be the rate of change of length, per time,space fraction numerator straight d l over denominator straight d t end fraction)

STEP 2
Use chain rule to form an equation connecting these rates of change with a third rate
The third rate of change will come from a related quantity such as volume, surface area, perimeter

STEP 3
Write down the formula for the related quantity (volume, etc) accounting for any fixed quantities
Find the third rate of change of the related quantity (derivative) using differentiation

STEP 4
Substitute the derivative and known rate of change into the equation and solve it

Examiner Tip

  • If you struggle to determine which rate to use in an exam then you can look at the units to help
    • e.g.  A rate of 5 cm3 per second implies volume per time so the rate would be fraction numerator d V over denominator d t end fraction

Worked example

A cuboid has a square cross-sectional area of side lengthspace x cm and a fixed height of 5 cm.
The volume of the cuboid is increasing at a rate of 20 cm3 s-1.
Find the rate at which the side length is increasing at the point when its side length is 3 cm.
5-2-3-ib-hl-ai-rel-roc-we-soltn

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Paul

Author: Paul

Expertise: Maths

Paul has taught mathematics for 20 years and has been an examiner for Edexcel for over a decade. GCSE, A level, pure, mechanics, statistics, discrete – if it’s in a Maths exam, Paul will know about it. Paul is a passionate fan of clear and colourful notes with fascinating diagrams – one of the many reasons he is excited to be a member of the SME team.