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**Method for finding tangent lines**

Pierre de Fermat ( ) Method for finding tangent lines

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**Life and work Interesting facts: Contribution to calculus development:**

He was a French lawyer He pursued maths as a hobby He is known as the “Prince of Amateurs” His contribution to calculus was less well known. Contribution to calculus development: determining maxima, minima finding tangents to various curves Evaluating area under a graph

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**Fermat’s Similar Triangles**

Tangent line at point T

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**Fermat’s Similar Triangles**

T(x,y) A(x,0) s y O(x-s,0) Fermat used 𝑇𝐴 𝑂𝐴 to calculate the gradient at T 𝑇𝐴 𝑂𝐴 = 𝑦−0 𝑥−(𝑥−𝑠) = 𝑓(𝑥) 𝑠

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**Fermat’s Similar Triangles**

𝑇 To calculate value of s, he used a technique based on similar triangles.

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**Fermat’s Similar Triangles**

O A He starts drawing tangent line at point T Then create triangle ∆OAT.

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**Fermat’s Similar Triangles**

P O B A Then extend the length E to create a new triangle ∆ OPB

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**Fermat’s Similar Triangles**

P O B A s E Set the length OA =s, OB = s + E

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**Fermat’s Similar Triangles**

O A T P B s E Properties of similar triangle: Perimeters of similar triangles are in the same ratio as their corresponding sides 𝑠 𝑠+𝐸 = 𝑇𝐴 𝑃𝐵

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**Fermat’s Similar Triangles**

O A T P B s 𝐸→0 O A T P B s E Assume T has a coordinate (x,y) where y = f(x) At this point, Fermat claims that error disappears when 𝐸→0 When 𝐸→0, PB →𝑓 𝑥 + 𝐸 𝑠 𝑠+𝐸 = 𝑇𝐴 𝑃𝐵 = 𝑓(𝑥) 𝑓 𝑥 + 𝐸

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**Fermat’s Similar Triangles**

O A T P B s 𝐸→0 By rearranging the equation , s= 𝑓(𝑥) 𝑓 𝑥 + 𝐸 −𝑓 𝑥 /𝐸

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**Fermat’s Similar Triangles**

O A T P B s 𝐸→0 So, the gradient of the tangent at point T can be found as follows:- 𝑓(𝑥) 𝑠

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Example Let’s find the tangent of the curve at (2,4)

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**Example Let find the tangent of the curve y= 𝑥 2 at (2,4)**

s= 𝑥 2 ((𝑥+𝐸) 2 − 𝑥 2 )/𝐸 = 𝑥 2 ( 𝑥 2 +𝐸 2 +2𝑥𝐸− 𝑥 2 )/𝐸

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**s= 𝑥 2 ( 𝑥 2 +𝐸 2 +2𝑥𝐸− 𝑥 2 )/𝐸 = 𝑥 2 𝐸+2𝑥 When 𝐸→0, then S = 𝑥 2**

Example s= 𝑥 2 ( 𝑥 2 +𝐸 2 +2𝑥𝐸− 𝑥 2 )/𝐸 = 𝑥 2 𝐸+2𝑥 When 𝐸→0, then S = 𝑥 2

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Example Gradient of the tangent = 𝑓(𝑥) 𝑠 = 𝑥 2 𝑥/2 = 2𝑥 At point (2,4), Gradient of the tangent = 2 x 2 = 4

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**Example Let y = 4x+c be the equation of the tangent at point (2,4)**

By substituting (2,4) in the equation, 4=4 (2) + C C = -4

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Example Therefore, the equation of the tangent is Y = 4 x -4

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