Applications of Differentiation (Cambridge O Level Additional Maths)

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Finding Gradients

How do I use the derivative to find the gradient of a curve?

  • The gradient of a curve at a point is the gradient of the tangent to the curve at that point
  • To find the gradient of a curve, at any point on the curve
    • differentiate to find fraction numerator straight d y over denominator straight d x end fraction (unless fraction numerator straight d y over denominator straight d x end fractionis already known)
    • substitute the x‑coordinate of the point into the derivative fraction numerator d y over denominator d x end fraction and evaluate

Grad Tang Norm Illustr 1, A Level & AS Maths: Pure revision notes

How do I find the approximate change in y as x increases?

  • gradient equals fraction numerator change space in space y over denominator change space in space x end fractionso, for small changes you can write 
    • change space in space y equals gradient space cross times space change space in space x
  • For example, if the gradient of y equals x minus 1 over x at x equals 2 is 5 over 4 
    • what is the approximate change in y as x increases from 2 to 2 plus h, where h is small?
      • change space in space y equals gradient space cross times space change space in space x equals 5 over 4 h 

Examiner Tip

  • Read the question carefully; sometimes you are given fraction numerator d y over denominator d x end fraction and so don't need to differentiate initially - don't just automatically differentiate the first thing you see!
  • The following mean the same thing:
    • "Find the gradient of the curve at x equals 2"
    • "Find the gradient of the tangent at x equals 2"
      • the tangent gradient = curve gradient at that point
    • "Find the rate of change of y with respect to x at x equals 2"

Worked example

A curve has the equation y equals 4 over 3 x cubed plus 3 x minus 8.

(a)

Find the gradient of the curve when x equals 2.

y is already in a form that can be differentiated

fraction numerator straight d y over denominator straight d x end fraction equals 4 x squared plus 3

Substitute x equals 2 into fraction numerator straight d y over denominator straight d x end fraction

fraction numerator straight d y over denominator straight d x end fraction equals 4 cross times 2 squared plus 3 equals 19

The gradient of the curve at bold italic x bold equals bold 2 is 19

(b)

Work out the possible values of x for which the rate of change of y with respect to x is 4.

"Rate of change" is another way of describing the derivative

table row cell fraction numerator straight d y over denominator straight d x end fraction end cell equals 4 row cell 4 x squared plus 3 end cell equals 4 end table

Solve this equation to find x
Note that it is quadratic equation so it could have up to two solutions
The question refers to 'values' implying there is (or could be) more than one value for x

table row cell 4 x squared end cell equals 1 row cell x squared end cell equals cell 1 fourth end cell row x equals cell plus-or-minus square root of 1 fourth end root end cell row x equals cell plus-or-minus 1 half end cell end table

The possible values of bold italic x, that give a rate of change of 4, are bold italic x bold equals bold 1 over bold 2 and bold italic x bold equals bold minus bold 1 over bold 2

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Increasing & Decreasing Functions

What are increasing and decreasing functions?

  • A function is increasing when fraction numerator d y over denominator d x end fraction greater than 0 (the gradient is positive)
    • This means graph of a function goes up as bold italic x increases
  • A function is decreasing when fraction numerator d y over denominator d x end fraction less than 0 (the gradient is negative)
    • This means graph of a function goes down as bold italic x increases

 Incr Decr Illustr 1, A Level & AS Maths: Pure revision notes

 

How do I find where functions are increasing or decreasing?

  • To identify the intervals on which a function is increasing or decreasing 
STEP 1
Find the derivative f'(x)

STEP 2
Solve the inequalities
bold space bold italic f bold apostrophe bold left parenthesis bold italic x bold right parenthesis bold greater than bold 0 (for increasing intervals) and/or
bold space bold italic f bold apostrophe bold left parenthesis bold italic x bold right parenthesis bold less than bold 0 (for decreasing intervals)

  • Most functions are a combination of increasing, decreasing and stationary
    • a range of values ofspace x (interval) is given where a function satisfies each condition
    • e.g.  The functionspace f begin mathsize 16px style stretchy left parenthesis x stretchy right parenthesis end style size 16px equals size 16px x to the power of size 16px 2 has derivativespace f to the power of apostrophe stretchy left parenthesis x stretchy right parenthesis equals 2 x so
      • space f left parenthesis x right parenthesis is decreasing for x less than 0
      • space f left parenthesis x right parenthesis is stationary at x equals 0
      • space f left parenthesis x right parenthesis is increasing for x greater than 0
  • To identify the intervals (the range of x values) for which a curve is increasing or decreasing you need to:
    1. Find the derivative fraction numerator d y over denominator d x end fraction
    2. Solve the inequalities fraction numerator d y over denominator d x end fraction greater than 0 (for increasing intervals) or fraction numerator d y over denominator d x end fraction less than 0 (for decreasing intervals)

Examiner Tip

  • In an exam, if you need to show a function is increasing or decreasing you can use either strict (<, >) or non-strict (≤, ≥) inequalities
    • You will get the marks either way in this course

Worked example

For what values of x is y equals 2 x cubed minus 3 x squared plus 5 a decreasing function?

The function is decreasing when its gradient open parentheses fraction numerator d y over denominator d x end fraction close parentheses is less than 0.

Find the derivative of the function by differentiating.

table row cell fraction numerator d y over denominator d x end fraction end cell equals cell space 6 x squared space minus space 6 x end cell end table 

Solve the inequality fraction numerator d y over denominator d x end fraction space less than space 0 to find the set of values where the gradient is negative.

table row cell 6 x squared space minus space 6 x space end cell less than cell space 0 end cell end table

Factorise.

table row cell 6 x open parentheses x space minus space 1 close parentheses space end cell less than cell space 0 end cell end table

The solutions to fraction numerator d y over denominator d x end fraction space equals space 0 are x space equals space 0 and  x space equals space 1. Find the correct way around for the inequalities by considering the graph of fraction numerator d y over denominator d x end fraction. The graph is a positive quadratic, so the function is negative between the values of 0 and 1 (where it is below the x-axis).

Considering a sketch of the graph of the gradient function may help you see this.

3CqZzydq_aqa-fm-increasing-and-decreasing-functions-rn-diagram-we-1 

You can check your answers by considering a sketch of the original function, it should be decreasing at the point where 0 space less than space x space less than space 1.

aqa-fm-increasing-and-decreasing-functions-rn-diagram-we-2

bold 0 bold space bold less than bold space bold italic x bold space bold less than bold space bold 1

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Tangents & Normals

What is a tangent?

  • At any point on the graph of a (non-linear) function, the tangent is the straight line that touches the graph at a point without crossing through it
  • Its gradient is given by the derivative function

Grad Tang Norm Illustr 2

How do I find the equation of a tangent?

  • To find the equation of a straight line, a point and the gradient are needed
  • The gradient, m, of the tangent to the function y equals straight f open parentheses x close parentheses at left parenthesis x subscript 1 comma blank y subscript 1 right parenthesis is
    • You can find this by differentiating the function, and then substituting the x-coordinate of the point into the derivative
  • Therefore find the equation of the tangent to the function y equals straight f left parenthesis x right parenthesis at the point left parenthesis x subscript 1 comma blank y subscript 1 right parenthesis by substituting the gradient, straight f apostrophe open parentheses x subscript 1 close parentheses, and point left parenthesis x subscript 1 comma blank y subscript 1 right parenthesis into y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses, giving:
    •  
  • (You could also substitute into y equals m x plus c)

What is a normal?

  • At any point on the graph of a (non-linear) function, the normal is the straight line that passes through that point and is perpendicular to the tangent

Grad Tang Norm Illustr 3

How do I find the equation of a normal?

  • The gradients of two perpendicular lines are negative reciprocals
    • This means that if m subscript 1 is the gradient of the first line and m subscript 2 is the gradient of a line perpendicular to the first line, then m subscript 2 equals fraction numerator negative 1 over denominator m subscript 1 end fraction
    • Rearranging the formula above, m subscript 1 cross times m subscript 2 equals negative 1 is a useful way to test whether two lines are perpendicular
  • Therefore gradient of the normal to the function y equals straight f open parentheses x close parenthesesat left parenthesis x subscript 1 comma blank y subscript 1 right parenthesis is Error converting from MathML to accessible text.
  • Find the equation of the normal to the function y equals straight f left parenthesis x right parenthesis at the point left parenthesis x subscript 1 comma blank y subscript 1 right parenthesis by using y minus y subscript 1 equals fraction numerator negative 1 over denominator straight f to the power of straight apostrophe stretchy left parenthesis x subscript 1 stretchy right parenthesis end fraction left parenthesis x minus x subscript 1 right parenthesis
    • (or y equals fraction numerator negative 1 over denominator straight f to the power of apostrophe open parentheses x subscript 1 close parentheses end fraction x plus c)

Examiner Tip

  • To be successful in this topic, first make sure you are confident with finding the equation of a straight line!

Worked example

The function straight f left parenthesis x right parenthesis is defined by

 straight f stretchy left parenthesis x stretchy right parenthesis equals 2 x to the power of 4 plus 3 over x squared blank x not equal to 0

a)
Find an equation for the tangent to the curve y equals straight f left parenthesis x right parenthesis at the point where x equals 1, giving your answer in the form y equals m x plus c.


First find the derivative straight f apostrophe open parentheses x close parentheses by differentiating
Start by rewriting straight f open parentheses x close parenthesesas powers of x

straight f open parentheses x close parentheses equals 2 x to the power of 4 plus 3 x to the power of negative 2 end exponent

Now differentiate
straight f apostrophe open parentheses x close parentheses equals 8 x cubed minus 6 x to the power of negative 3 end exponent

Now substitute x equals 1 into straight f apostrophe open parentheses x close parentheses to find the gradient of the tangent

table attributes columnalign right center left columnspacing 0px end attributes row cell straight f apostrophe open parentheses 1 close parentheses end cell equals cell 8 open parentheses 1 close parentheses cubed minus 6 open parentheses 1 close parentheses to the power of negative 3 end exponent end cell row blank equals cell 8 minus 6 end cell row blank equals 2 end table

We also need the y-coordinate, so substitute x equals 1 into straight f open parentheses x close parentheses also

table row cell straight f open parentheses 1 close parentheses end cell equals cell 2 open parentheses 1 close parentheses to the power of 4 plus 3 open parentheses 1 close parentheses to the power of negative 2 end exponent end cell row blank equals cell 2 plus 3 end cell row blank equals 5 end table

Now we can substitute the point (1, 5) and the gradient, 2, into y minus y subscript 1 equals m open parentheses x subscript 1 close parentheses open parentheses x minus x subscript 1 close parentheses

table row cell y minus 5 end cell equals cell 2 stretchy left parenthesis x minus 1 stretchy right parenthesis end cell end table

Note that we are asked for the final answer in the form y equals m x plus c, so rearrange to this form

table attributes columnalign right center left columnspacing 0px end attributes row cell y minus 5 end cell equals cell 2 x minus 2 end cell end table
bold italic y bold equals bold 2 bold italic x bold plus bold 3

b)
Find an equation for the normal at the point where x equals 1, giving your answer in the form a x plus b y plus d equals 0, where a, b and d are integers.


We already have the gradient of the tangent; the gradient of the normal is fraction numerator negative 1 over denominator m end fraction where m is the gradient of the tangent

gradient of normal = fraction numerator negative 1 over denominator 2 end fraction

Substitute the point (1, 5) from part (a) and the gradient of the normal into y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses

table row cell y minus 5 end cell equals cell open parentheses fraction numerator negative 1 over denominator 2 end fraction close parentheses open parentheses x minus 1 close parentheses end cell end table
And rearrange into the form required
table row cell y minus 5 end cell equals cell negative 1 half x plus 1 half end cell row cell 1 half x plus y minus 5 1 half end cell equals 0 end table

Note that ab and d must be integers so multiply by 2 to clear the fractions
table row cell bold italic x bold plus bold 2 bold italic y bold minus bold 11 end cell bold equals bold 0 end table

 

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Amber

Author: Amber

Expertise: Maths

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.