Implicit Differentiation (DP IB Maths: AA HL)

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Implicit Differentiation

What is implicit differentiation?

  • An equation connecting x and y is not always easy to write explicitly in the form y equals f left parenthesis x right parenthesis or x equals f left parenthesis y right parenthesis
    • In such cases the equation is written implicitly
      • as a function of x and y
      • in the form f left parenthesis x comma y right parenthesis equals 0
  • Such equations can be differentiated implicitly using the chain rule

fraction numerator straight d over denominator d x end fraction open square brackets f left parenthesis y right parenthesis close square brackets equals f apostrophe left parenthesis y right parenthesis fraction numerator straight d y over denominator straight d x end fraction

  • A shortcut way of thinking about this is that ‘y is a function of a x
    • when differentiating a function of y chain rule says “differentiate with respect to y, then multiply by the derivative of y” (which is fraction numerator straight d y over denominator straight d x end fraction)

Applications of Implicit Differentiation

What type of problems could involve implicit differentiation?

  • Broadly speaking there are three types of problem that could involve implicit differentiation
    • algebraic problems involving graphs, derivatives, tangents, normals, etc
      • where it is not practical to write y explicitly in terms of x
      • usually in such cases, fraction numerator straight d y over denominator straight d x end fraction will be in terms of x and y
    • optimisation problems that involve time derivatives
      • more than one variable may be involved too

e.g.  Volume of a cylinder, V equals pi r squared h

e.g.  The side length and (so) area of a square increase over time

    • any problem that involves differentiating with respect to an extraneous variable
      • e.g. y equals f left parenthesis x right parenthesis but the derivative fraction numerator straight d y over denominator straight d theta end fraction is required (rather than fraction numerator straight d y over denominator straight d x end fraction)

How do I apply implicit differentiation to algebraic problems?

  • Algebraic problems revolve around values of the derivative (gradient) open parentheses fraction numerator straight d y over denominator straight d x end fraction close parentheses
    • if not required to find this value it will either be given or implied
  • Particular problems focus on special case tangent values
    • horizontal tangents
      • also referred to as tangents parallel to the x-axis
      • this is when fraction numerator straight d y over denominator straight d x end fraction equals 0
    • vertical tangents
      • also referred to as tangents parallel to the y-axis
      • this is when fraction numerator straight d x over denominator straight d y end fraction equals 0
      • In such cases it may appear that fraction numerator 1 over denominator begin display style fraction numerator straight d y over denominator straight d x end fraction end style end fraction equals 0 but this has no solutions; this occurs when for nearby values of xfraction numerator straight d y over denominator straight d x end fraction rightwards arrow plus-or-minus infinity

(i.e.  very steep gradients, near vertical)

  • Other problems may involve finding equations of (other) tangents and/or normals
  • For problems that involve finding the coordinates of points on a curve with a specified gradient the method below can be used

STEP 1
Differentiate the equation of the curve implicitly

STEP 2
Substitute the given or implied value of fraction numerator straight d y over denominator straight d x end fraction to create an equation linking x and y

STEP 3
There are now two equations

    • the original equation
    • the linking equation

Solve them simultaneously to find the x and y coordinates as required

Examiner Tip

  • After some rearranging, fraction numerator d y over denominator d x end fraction will be in terms of both x and y
    • There is usually no need (unless asked to by the question) to write fraction numerator d y over denominator d x end fraction in terms of x (or y) only
  • If evalutaing derivatives, you'll need both x and y coordinates, so one may have to be found from the other using the original function

Worked example

The curve C has equation x squared plus 2 y squared equals 16.

a)
Find the exact coordinates of the points where the normal to curve C has gradient 2.

5-8-2-ib-hl-aa-only-we3a-soltn

b)
Find the equations of the tangents to the curve that are
(i)       parallel to the x-axis
(ii)      parallel to the y-axis.

5-8-2-ib-hl-aa-only-we3b-soltn

How do I apply implicit differentiation to optimisation problems?

  • For a single variable use chain rule to differentiate implicitly
    • e.g. A square with side length changing over time, A equals x squared

fraction numerator straight d A over denominator straight d t end fraction equals 2 x fraction numerator straight d x over denominator straight d t end fraction

  • For more than one variable use product rule (and chain rule) to differentiate implicitly
    • e.g. A square-based pyramid with base length and height changing over time, V equals 1 third x squared h

fraction numerator straight d V over denominator straight d t end fraction equals 1 third open square brackets x squared fraction numerator straight d h over denominator straight d t end fraction plus 2 x fraction numerator straight d x over denominator straight d t end fraction h close square brackets equals 1 third x open parentheses x fraction numerator straight d h over denominator straight d t end fraction plus 2 h fraction numerator straight d x over denominator straight d t end fraction close parentheses

  • After differentiating implicitly the rest of the question should be similar to any other optimisation problem
    • be aware of phrasing
      • “the rate of change of the height of the pyramid” (over time) is fraction numerator straight d h over denominator straight d t end fraction
    • when finding the location of minimum and maximum problems
      • there is not necessarily a turning point
      • the minimum or maximum could be at the start or end of a given or appropriate interval

Examiner Tip

  • If you are struggling to tell which derivative is needed for a question, writing all possibilities down may help you
    • You don't need to work them out at this stage but if you conisder them it may nudge you to the next stage of the solution
    • e.g.  For V equals straight pi r to the power of italic 2 h, possible derivatives are fraction numerator d V over denominator d r end fraction comma space fraction numerator d V over denominator d h end fraction and fraction numerator d V over denominator d t end fraction

Worked example

The radius, r cm, and height, h cm, of a cylinder are increasing with time.  The volume, V cm3, of the cylinder at time t seconds is given by V equals pi r squared h.

a)       Find an expression for fraction numerator straight d V over denominator straight d t end fraction.

5-8-2-ib-hl-aa-only-we2a-soltn

b)

At time T seconds, the radius of the cylinder is 4 cm, expanding at a rate of 2 cm s-1. At the same time, the height of the cylinder is 10 cm, expanding at a rate of 3 cm s-1.

Find the rate at which the volume is expanding at time T seconds.

5-8-2-ib-hl-aa-only-we2b-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.