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14: Stationary Points – another notation and applications © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules.

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Presentation on theme: "14: Stationary Points – another notation and applications © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules."— Presentation transcript:

1 14: Stationary Points – another notation and applications © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules

2 Stationary Points Module C1 AQA Edexcel OCRMEI/OCR Module C2 "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

3 Stationary Points An Alternative Notation The notation for a function of x can be used instead of y. When is used, instead of using for the 2nd derivative, we write

4 Stationary Points Solution: (a) or (b) Determine the nature of the stationary points and sketch the curve e.g. (a) Find the coordinates of the stationary points on the curve where We now need to find the y -coordinates of the st. pts. Stationary points:

5 Stationary Points min at (0, 0) is a st. pt. (b) We use the stationary points to sketch the curve. max at (2, 4) x x Tip: When curve sketching, always find the point on the y -axis. ( In this example, it is a stationary point so we already know it. ) We have and when and is a st. pt. Now,

6 Stationary Points SUMMARY  To find stationary points, solve the equation maximum minimum  Determine the nature of the stationary points either by finding the gradients on the left and right of the stationary points or by finding the value of the 2 nd derivative at the stationary points or

7 Stationary Points Exercises Find the coordinates of the stationary points of the following functions, determine the nature of each and sketch the functions is a max. is a min. Ans. is a min. is a max. Ans.

8 Stationary Points Suppose also that the y -variable is replaced by A where A represents an area measured in cm 2. Then, Suppose that x represents a length in cm. Parts of the graph have no meaning... e.g. Applications of Stationary Points minimum x x maximum

9 Stationary Points For values of x between 0 and 4, Suppose also that the y -variable is replaced by A where A represents an area measured in cm 2. Then, Suppose that x represents a length in cm. Parts of the graph have no meaning... since length and area cannot be negative. e.g. Applications of Stationary Points minimum x x maximum gives the largest area, 6cm 2 gives the smallest area, 2cm 2

10 Stationary Points minimum x x maximum The stationary points give the greatest and least values of the area, A, and the values of x at which they occur. If we can find a function that describes a situation involving greatest and/or least values of a variable, we can use stationary points to find these values. An equation describing a situation is called a model.

11 Stationary Points e.g.1 The number in a population of foxes in a certain country area increases and decreases over the years. A model for the number in the population, y, is given by the equation where x is the number of years since (a)How many foxes were in the population in 1990 ? (b)Use Calculus to find the year in which the population was least and give its size. Solution: (a) In 1990, x = 0, so y = 100 There were 100 foxes in 1990.

12 Stationary Points We want to find the minimum value of y. Since y is a cubic function, one of these values gives the minimum and one gives the maximum. (b) Use Calculus to find the year in which the population was least and give its’ size.

13 Stationary Points Since 42 > 0, x = 5 gives the min. The minimum value is given by y : The smallest number of foxes was 75 in The quickest way to see which one gives the minimum is to use the 2 nd derivative.

14 Stationary Points e.g.2 I want to make a fruit cage from 2 pieces of netting. The smaller piece will make the top and I’ll use the longer piece, which is 10 m long, for the sides. The cage is to be rectangular and to enclose the largest possible area. One side won’t need to be netted as there is a wall on that side. How long should the 3 sides be? wall Solution: We draw a diagram and choose letters for the unknown lengths and area. Solution: xx y A The lengths are x metres and y metres. The area is A m 2.

15 Stationary Points Solution: wall xx y A We want to find the maximum value of A, so we need an expression for A that we can differentiate. The area of a rectangle  length  breadth We can’t differentiate with 3 variables so we need to substitute for one of them. Since the length of netting is 10m, we know that Rearranging,

16 Stationary Points Solution: wall xx y A Substituting for y in ( 1 ), We can now find the stationary points. Substitute in 52  x Substitute in ( 1 ),

17 Stationary Points Solution: wall xx y A We have We know that we have found the maximum since and we recognise the shape of this quadratic.

18 Stationary Points

19 The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

20 Stationary Points An Alternative Notation The notation for a function of x can be used instead of y. When is used, instead of using for the 2nd derivative, we write

21 Stationary Points Solution: (a) or (b) Determine the nature of the stationary points and sketch the curve e.g. (a) Find the coordinates of the stationary points on the curve where We now need to find the y -coordinates of the st. pts. Stationary points:

22 Stationary Points min at (0, 0) is a st. pt. (b) We use the stationary points to sketch the curve. max at (2, 4) x x Tip: When curve sketching, always find the point on the y -axis. ( In this example, it is a stationary point so we already know it. ) We have and when and is a st. pt. Now,

23 Stationary Points SUMMARY  To find stationary points, solve the equation maximum minimum  Determine the nature of the stationary points either by finding the gradients on the left and right of the stationary points or by finding the value of the 2 nd derivative at the stationary points or

24 Stationary Points If we can find a function that describes a situation involving greatest and/or least values of a variable, we can use stationary points to find these values. An equation describing a situation is called a model. Applications of Stationary Points

25 Stationary Points e.g.1 The number in a population of foxes in a certain country area increases and decreases over the years. A model for the number in the population, y, is given by the equation where x is the number of years since (a)How many foxes were in the population in 1990 ? (b)Use Calculus to find the year in which the population was least and give its’ size. Solution: (a) In 1990, x = 0, so y = 100 There were 100 foxes in 1990.

26 Stationary Points We want to find the minimum value of y. Since y is a cubic function, one of these values gives the minimum and one gives the maximum. (b) Use Calculus to find the year in which the population was least and give its size.

27 Stationary Points Since 42 > 0, x = 5 gives the min. The minimum value is given by y : The smallest number of foxes was 75 in The quickest way to see which one gives the minimum is to use the 2 nd derivative.

28 Stationary Points Solution: e.g.2 I want to make a fruit cage from 2 pieces of netting. The smaller piece will make the top and I’ll use the longer piece, which is 10 m long, for the sides. The cage is to be rectangular and to enclose the largest possible area. One side won’t need to be netted as there is a wall on that side. How long should the 3 sides be? wall Solution: We draw a diagram and choose letters for the unknown lengths and area. xx y A The lengths are x metres and y metres. The area is A m 2.

29 Stationary Points Solution: wall xx y A We want to find the maximum value of A, so we need an expression for A that we can differentiate. The area of a rectangle  length  breadth We can’t differentiate with 3 variables so we need to substitute for one of them. Since the length of netting is 10m, we know that Rearranging,

30 Stationary Points Solution: wall xx y A Substituting for y in ( 1 ), We can now find the stationary points. Substitute in 52  x Substitute in ( 1 ),

31 Stationary Points Solution: wall xx y A We have We know that we have found the maximum since and we recognise the shape of this quadratic.


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