DP IB Maths: AA HL

Revision Notes

5.8.2 Applications of Chain Rule

Test Yourself

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
    • 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 often time, represented by t
    • seconds is the standard unit for time but this will depend on context
  • e.g.  Water running into a large hemi-spherical bowl
    • both the height and volume of water in the bowl are changing with time
      • time is the linking parameter between the rate of change of height and the rate of change of volume

How do I solve problems involving related rates of change?

  • Use of chain rule and product rule are common in such problems
  • Be clear about which variables are representing which quantities

STEP 1
Write down any variables and derivatives involved in the problem
e.g. x comma space y space comma space t comma space fraction numerator straight d y over denominator straight d x end fraction comma space fraction numerator straight d x over denominator straight d t end fraction comma space fraction numerator straight d y over denominator straight d t end fraction

STEP 2
Use an appropriate differentiation rule to set up an equation linking ‘rates of change’
e.g.  Chain rule: fraction numerator straight d y over denominator straight d t end fraction equals fraction numerator straight d y over denominator straight d x end fraction cross times fraction numerator straight d x over denominator straight d t end fraction  

STEP 3
Substitute in known values
e.g.  If, when t equals 3, fraction numerator straight d x over denominator straight d t end fraction equals 2 and fraction numerator straight d y over denominator straight d t end fraction equals 8, then 8 equals fraction numerator straight d y over denominator straight d x end fraction cross times 2

STEP 4
Solve the problem and interpret the answer in context if required
e.g. fraction numerator straight d y over denominator straight d x end fraction equals 8 over 2 equals 4   ‘when t equals 3y changes at a rate of 4, with respect to x

Exam Tip

  • If you struggle to determine which rate to use 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

In a manufacturing process a metal component is heated such that it’s cross-sectional area expands but always retains the shape of a right-angled triangle.  At time t seconds the triangle has base b cm and height h cm.

At the time when the component’s cross-sectional area is changing at 4 cm s-1, the base of the triangle is 3 cm and its height is 6 cm. Also at this time, the rate of change of the height is twice the rate of change of the base.

Find the rate of change of the base at this point of time.

MR5QgbAY_5-8-1-ib-hl-aa-only-rel-roc-we1-soln-

Differentiating Inverse Functions

What is meant by an inverse function?

  • Some functions are easier to process with x (rather than y) as the subject
    • i.e.  in the form x equals f left parenthesis y right parenthesis
  • This is particularly true when dealing with inverse functions
    • e.g.  If y equals f left parenthesis x right parenthesis the inverse would be written as y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis
      • finding f to the power of negative 1 end exponent left parenthesis x right parenthesis can be awkward
      • so write x equals f left parenthesis y right parenthesis instead

How do I differentiate inverse functions?

  • Since x equals f left parenthesis y right parenthesis it is easier to differentiate “x with respect to y” rather than “y with respect to x
    • i.e.  find fraction numerator d x over denominator d y end fraction rather than fraction numerator d y over denominator d x end fraction
    • Note that fraction numerator d x over denominator d y end fraction will be in terms of y but can be substituted

STEP 1
For the function y equals f left parenthesis x right parenthesis, the inverse will be y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis

Rewrite this as x equals f left parenthesis y right parenthesis

STEP 2
From x equals f left parenthesis y right parenthesis find fraction numerator d x over denominator d y end fraction

STEP 3
Find fraction numerator d y over denominator d x end fraction using fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator begin display style fraction numerator d x over denominator d y end fraction end style end fraction - this will usually be in terms of y

  • If an algebraic solution in terms of x is required substitute f left parenthesis x right parenthesis for y in fraction numerator d y over denominator d x end fraction
  • If a numerical derivative (e.g. a gradient) is required then use the y-coordinate
    • If the y-coordinate is not given, you should be able to work it out from the orginal function and x-coordinate

Exam Tip

  • With x's and y's everywhere this can soon get confusing!
    • Be clear of the key information and steps - and set your wokring out accordingly
      • The orginal function,   y equals f left parenthesis x right parenthesis
      • Its inverse,  y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis
      • Rewriting the inverse,  x equals f left parenthesis y right parenthesis
      • Finding fraction numerator d x over denominator d y end fraction first, then finding its reciprocal for fraction numerator d y over denominator d x end fraction
  • Your GDC can help when numerical derivatives (gradients) are required

Worked example

a)
Find the gradient of the curve at the point where y equals 3 on the graph of y equals f to the power of negative 1 end exponent left parenthesis x right parenthesis where f left parenthesis x right parenthesis equals square root of left parenthesis 5 x plus 1 right parenthesis cubed end root.

5-8-1-ib-hl-aa-only-we2a-soltn

b)       Given that y equals e to the power of x show that the derivative of y equals ln space x is 1 over x.

5-8-1-ib-hl-aa-only-we2b-soltn-

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Paul

Author: Paul

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.