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**“Teach A Level Maths” Vol. 1: AS Core Modules**

14: Stationary Points – another notation and applications © Christine Crisp

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**Module C1 Module C2 AQA Edexcel OCR MEI/OCR**

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**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

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**e.g. (a) Find the coordinates of the stationary points on the curve where**

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

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**(b) We use the stationary points to sketch the curve.**

We have and when and Now, is a st. pt. is a st. pt. min at (0, 0) max at (2, 4) (b) We use the stationary points to sketch the curve. 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. ) x

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

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Exercises Find the coordinates of the stationary points of the following functions, determine the nature of each and sketch the functions. 1. is a max. is a min. Ans. 2. is a min. is a max. Ans.

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**Applications of Stationary Points**

e.g. Suppose that x represents a length in cm. Suppose also that the y-variable is replaced by A where A represents an area measured in cm2. Then, minimum x maximum Parts of the graph have no meaning . . .

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**Applications of Stationary Points**

e.g. Suppose that x represents a length in cm. Suppose also that the y-variable is replaced by A where A represents an area measured in cm2. Then, maximum x Parts of the graph have no meaning . . . x minimum since length and area cannot be negative. For values of x between 0 and 4, gives the largest area, 6cm2 gives the smallest area, 2cm2

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**An equation describing a situation is called a model.**

minimum 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.

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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 1990. How many foxes were in the population in 1990? 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.

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**(b) Use Calculus to find the year in which the population was least and give its’ size.**

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.

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**The quickest way to see which one gives the minimum is to use the 2nd derivative.**

Since 42 > 0, x = 5 gives the min. The minimum value is given by y: The smallest number of foxes was 75 in 1995.

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**e. g. 2 I want to make a fruit cage from 2 pieces of netting**

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 10m 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? Solution: We draw a diagram and choose letters for the unknown lengths and area. Solution: wall The lengths are x metres and y metres. x x A The area is A m2. y

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wall x 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. Solution: Since the length of netting is 10m, we know that Rearranging,

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**5 2 × = x x A y wall Substituting for y in (1),**

We can now find the stationary points. Solution: Substitute in 5 2 × = x Substitute in (1),

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**x A y wall We have We know that we have found the maximum since**

Solution: and we recognise the shape of this quadratic.

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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.

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**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

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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:

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**(b) We use the stationary points to sketch the curve. max at (2, 4)**

min at (0, 0) is a st. pt. (b) We use the stationary points to sketch the curve. max at (2, 4) 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 Now,

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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 2nd derivative at the stationary points or

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**Applications of 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.

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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 1990. How many foxes were in the population in 1990? 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.

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**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.

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**Since 42 > 0, x = 5 gives the min.**

The minimum value is given by y: The smallest number of foxes was 75 in 1995. The quickest way to see which one gives the minimum is to use the 2nd derivative.

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**e. g. 2 I want to make a fruit cage from 2 pieces of netting**

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 10m 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. x y A The lengths are x metres and y metres. The area is A m2. Solution:

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wall x 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, Solution:

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**5 2 × = x A y wall Substituting for y in (1),**

We can now find the stationary points. Substitute in 5 2 × = Substitute in (1), Solution:

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**x A y wall We have We know that we have found the maximum since**

and we recognise the shape of this quadratic. Solution:

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© Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 39: Connected Rates of Change.

© Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 39: Connected Rates of Change.

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