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Gradient and Area Under Curve Calculate or estimate gradients of graphs and areas under graphs (including quadratic and other non- linear) interpreting results in cases such as distance-time, velocity-time and financial contexts Grade 9 If you have any questions regarding these resources or come across any errors, please contact [email protected]

Transcript of [PPT]PowerPoint Presentation - Home - Claydon High School, … Revision/CMAT... · Web viewGradient...

Page 1: [PPT]PowerPoint Presentation - Home - Claydon High School, … Revision/CMAT... · Web viewGradient and Area Under Curve Calculate or estimate gradients of graphs and areas under

Gradient and Area Under Curve

Calculate or estimate gradients of graphs and areas under graphs (including quadratic and other non-

linear) interpreting results in cases such as distance-time, velocity-time and financial contexts

Grade 9

If you have any questions regarding these resources or come across any

errors, please contact [email protected]

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Lesson OverviewObjective(s) Calculate or estimate gradients of graphs and areas under graphs (including quadratic and other non-linear) interpreting results in cases such

as distance-time, velocity-time and financial contextsGrade 9

Prior Knowledge Rearranging equationsy = mx + c

Duration Related to lesson on ‘Gradients and Rates of Change’ – plan to teach together. Content covered here approximate time = 50 minutes

Resources Print slides: 20 - 28 Equipment

Progression of Learning

What are the students learning? How are the students learning? (Activities & Differentiation)

Gradient from a linear graph Give students slide 20. Using slide 4 show how gradient is calculated graphically (this should be familiar). Look at the different wordings - but all the same. Students to practice finding the gradients of the lines on slide 5.

10

Gradient from a linear equation Give students slide 21. Demonstrate 2 examples of rearranging into the form y = mx + c. 4 further questions for students to practice. Answers on slide 7.

10

Gradient of a quadratic graph – at a particular point Give students slide 22. Show the initial method of considering points just below and just above the point given. Extend this to drawing a tangent and then finding the gradient of the tangent using the method for linear graphs. Students to complete a further question on slide 10. Then give students slide 23 – contextualised problem.

10

Interpreting gradient and gradients of zero Give students slide 24. Students to complete notes for this section using slides 12 and 13. 5

Velocity time graphs – draw and interpreting Give students slide 25 – students to make notes on this sheet. Teacher can pose additional questions related to the give graph.Give students slide 26 – practice question. Solution on slide 15. Extension questions on slide 16.

5

Gradients and area under curve in exam questions (from specimen papers)

Give students slide 27 and 28. This includes 3 exam questions related to objective. Students to use notes from lesson. Ensure that all steps are shown. Relate to mark scheme to show how the marks are allocated.

10

Next Steps Rates of change Assessment PLC/Reformed Specification/Target 9/Algebra/Gradients and area under curve

Lesson Plan

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Key VocabularyGradientEstimate LinearNon-linearQuadraticVelocity

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Change in yChange in x

Gradient from a linear graphA linear graph is a straight line graph with an equation which can be written in the form y = mx + cThe gradient is ‘m’ and can be found from the graph

RiseStep

y2 – y1

x2 – x1

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Find the gradients of these lines:

a = 2b = 1/3c = -3d = 1e = -2

f = -1/3g = 5h = -5i = 1/5j = -3/4

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Gradient from a linear equationRearrange this equation to make y the subject and identify the gradient of this linear graph.

2y = 5x - 8 y = x - 4

Gradient =

2y - 5x + 8 = 0

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Gradient from a linear equationRearrange this equation to make y the subject and identify the gradient of this linear graph.

Gradient =

3y + 4x - 9 = 0 3y = -4x + 9 y = x + 3

(a) 4x – 2y = 12

(b) y – 4x = 8

(c) y – 8 = -1/2(x + 4)

(d) 3x – 4y = 8

PRACTICE

y = 2x - 6

y = 4x + 8

y = -1/2x + 6

y = 3/4x -2

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Gradient of a quadratic graphThis is a quadratic graph with equation y = x2

x

y We can estimate the gradient of this graph at the point (2,4) by considering points just above and just below x= 2ie x = 2.1 and x = 1.9

Between x = 2.1 and x = 1.9Gradient = 4.41 – 3.612.1 – 1.9

= 4

We can do this more efficiently and accurately by drawing a tangent at the point (2,4) and then finding the gradient of this line.

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Gradient of a quadratic graph

x

y

41 = 4

We can do this more efficiently and accurately by drawing a tangent at the point (2,4) and then finding the gradient of this line.

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Gradient of a quadratic graphThis is a quadratic graph with equation y = x2- 3x + 1

Estimate the gradient of this graph at the point (0,1)

x

y

-3 1 = -3

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Gradient of a quadratic graph - Contextualised

A container is filled with water in 5 seconds. The graph shows the depth of the water, d cm, at the time, t seconds.

Use the graph to estimate the rate at which the depth of water is increasing at 3 seconds.

8

2

82 = 4

Note that exam mark schemes will allow for some degree of error (3.5 – 4.5 in this case)

4

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Interpreting gradient

In this conversion graph the gradient is 2. This means for every 1 unit of x, you need 2 units of y or the value of y is double the value of x.

Y axis

X axis

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Gradients of zero

x

y At these points on the graph the gradient is zero (flat)

x = 0

x = -8/3

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Velocity time graphsThe gradient of a velocity-time graph represents acceleration or deceleration at that point.The units are m/s-2

0 1 2 3 4 5 6 7 8 9

The area under a velocity-time graph represents the distance travelled.The units here are metres.This can be estimated by splitting the area under the graph into sections, calculating each one and then adding them together.

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Velocity time graphs - PracticeA toy car is placed on the floor of a sports hall.

It moves in a straight line starting from rest.It travels with constant acceleration for 4 seconds reaching a velocity of 5m/s.It then slows down with constant deceleration of 1m/s2 for 2 seconds.It then hits a wall and stops.

• Draw the velocity – time graph• Work out the total distance travelled by the toy car

0.5 (4 x 5) + 0.5 (5 + 3) x 2

= 18

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Problem Solving and Reasoning•Show how points on a line all have the same gradient between them.

•Convince me that parallel lines have the same gradient.

•Demonstrate that if a gradient is m then a perpendicular to that line will have a gradient of -1/m.

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Exam Question – Specimen Papers

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Exam Question – Specimen Papers

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Exam Question – Specimen Papers

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Gradient from a linear graph

Student Sheet 1

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Gradient from a linear equation

2y - 5x + 8 = 0

3y + 4x - 9 = 0

DEMO PRACTICE

Student Sheet 2

(a) 4x – 2y = 12

(b) y – 4x = 8

(c) y – 8 = -1/2(x + 4)

(d) 3x – 4y = 8

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Gradient of a quadratic graphThis is a quadratic graph with equation y = x2

Student Sheet 3

x

y

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Gradient of a quadratic graph - Contextualised

A container is filled with water in 5 seconds. The graph shows the depth of the water, d cm, at the time, t seconds.

Use the graph to estimate the rate at which the depth of water is increasing at 3 seconds.

Student Sheet 4

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Interpreting gradient

Y axis

X axis

Student Sheet 5

Gradients of zero

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Velocity time graphs

0 1 2 3 4 5 6 7 8 9

Student Sheet 6

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Velocity time graphs - PracticeA toy car is placed on the floor of a sports hall.

It moves in a straight line starting from rest.It travels with constant acceleration for 4 seconds reaching a velocity of 5m/s.It then slows down with constant deceleration of 1m/s2 for 2 seconds.It then hits a wall and stops.

• Draw the velocity – time graph• Work out the total distance travelled by the toy car

Student Sheet 7

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Exam Question – Specimen Papers

Student Sheet 8

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Exam Question – Specimen Papers

Student Sheet 9