Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

The presentation includes all the necessary material for review of Chapter 6.2 The Gradient of the Tangent as a Limit for A-level Pure Mathematics Cambridge International Examinations. It will be useful for students as the review materials and also for teachers as the teaching tips and presentations in the class. The presentation was made by Oleksii Khlobystin (Alex) - Mathematics Teacher at Depu Foreign Language School, Chongqing City, China. Personal website: www.visualcv.com/o-khlobystin Personal email: [email protected] POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709 Made by Oleksii Khlobystin (Alex) - Mathematics Teacher at Depu Foreign Language School – Cambridge International Centre, Chongqing City, China. Personal website: www.visualcv.com/o-khlobystin Personal email: [email protected]

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

The Gradient of a The Gradient of a Tangent Tangent We found the rule for differentiating by noticing a pattern in results found by measuring gradients of tangents. However, if we want to prove the rule or find a rule for some other functions we need a method based on algebra. This presentation shows you how this is done. The emphasis in this presentation is upon understanding ideas rather than doing calculations. The Gradient of a The Gradient of a TangentTangentThe emphasis in this presentation is upon understanding ideas rather than doing calculations.

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

 (2,4) 2xy A(1,1) 2xyConsider the tangent at the point A( 1, Tangent at A 1 ) on As an approximation to the gradient of the tangent we can use the gradient of a chord from A to a point close to A. e.g. we can use the chord to the point ( 2, 4 ). ( We are going to use several points, so we’ll call this point B1 ).

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

A(1,1)  (2,4) 2xyTangent at 2xy B11212xxyym31214mConsider the tangent at the point A( 1, The gradient of the chord AB1 is given by Chord AB1 We can see this gradient is larger than the gradient of the tangent. 1 ) on A We can see this gradient is larger than the gradient of the tangent.

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

A(1,1) )252,51(2BTangent at point B2 that is closer to A( 1, 1 ), e.g. )252,51(2B B12xyChord 521511252mTo get a better estimate we can take a The gradient of the chord AB2 is AB2 A

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

use the point .)211,11(3B A(1,1) 2BB12xy Tangent at A3BChord We can get an even better estimate if we AB3 We need to zoom in to the curve to see more clearly.

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

We can get an even better estimate if we use the point .)211,11(3B A(1,1) 2xyTangent at 121111211m)211,11(3B The gradient of AB3 is Chord AB3 A

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

Continuing in this way, moving B closer and closer to A( 1, 1 ), and collecting the results in a table, we get 51110112425221102011002001132512100201000200101501001000110010x nt 3B5B)(2xy 4B2B1BPoi tangent at A.352120120012 x  1 Gradie nt of AB y  1 As B gets closer to A, the gradient approaches 2. This is the gradient of the

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

) chord the ofgradient ( limas ABAB As B gets closer to A, the gradient of the The gradient of the tangent at A is “ the limit of the gradient of the chord AB as B approaches A ” chord AB approaches the gradient of the We write that the gradient of the tangent at A tangent. As B gets closer to A, the gradient of the chord AB approaches the gradient of the tangent.

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

a curve given by )(xfy We will generalize the result above to find a formula for the gradient at any point on

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

)(xfy)(xfWe use  )(,xfxBxxy can be used for as standing for “difference”. Let A be the point ( x, ) on the  )(,xfxBxx)(,xfxAxWe need a general notation for the B.x is the Greek letter d so we can think of values. y curve coordinates of B that suggests it is near to A. So, is the small difference in x as we move from A to the difference in y

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

 )(,xfxBxx)(,xfxAand where xxxfxfm)()(xx)()(xfxfmxx) chord the ofgradient ( limas ABAB So, since the 1212xxyymWe have )()(lim0xfxfxxx We can’t put x = 0 in this as we would get which is undefined. 00 So, the gradient of the chord AB is m gradient of the tangent  the gradient of the tangent 

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

the gradient of the tangent  )()(lim0xfxfxxx of ) x y)()(xfxfyx( the letter h is sometimes used instead tangent  0limxxy the gradient of the curve, so0limxxydxdyIf we use for the difference in y- But, the gradient of the tangent gives values, We then get the gradient of the

Presentation POWER POINT Chapter 6.2 The Gradient of the Tangent as a Limit, A-level Pure Mathematics CIE 9709

curve where is given by )(xfy2)(xxfxdxdy222)(xxx2)(xxfFor , by the gradient of the tangent, so )()(lim0xfxfdxdyxxx2222xxxx22xx)2(lim0xxxSo, as ,x0e.g. Prove that the gradient function of the zero.xx)()(xfxf xdxdy2 )2(lim0xxxxxSolution: The gradient of the curve is given So, 202limxdxdyxxxSince has cancelled, we will not be dividing by

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17.05.2018