# Ex 6.3,5

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.3,5 Find the slope of the normal to the curve 𝑥=𝑎 cos3𝜃, 𝑦=𝑎 sin3 𝜃 at 𝜃= 𝜋4 Given 𝑥=𝑎 cos3𝜃 Differentiating w.r.t. θ 𝑑𝑥𝑑𝜃= 𝑑 a cos3𝜃𝑑𝜃 𝑑𝑥𝑑𝜃=𝑎 . 𝑑 cos3𝜃𝑑𝜃 𝑑𝑥𝑑𝜃=𝑎 . 3 cos2𝜃. − sin𝜃 𝑑𝑥𝑑𝜃=− 3𝑎 sin𝜃 cos2𝜃 We know that slope of tangent = 𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝜃= 𝑑𝑦𝑑𝜃÷ 𝑑𝑥𝑑𝜃 𝑑𝑦𝑑𝑥= 3 𝑎 sin2𝜃 cos𝜃− 3𝑎 sin𝜃 cos2𝜃 𝑑𝑦𝑑𝑥= − sin𝜃 cos𝜃 𝑑𝑦𝑑𝑥=− tan𝜃 Putting 𝜃= 𝜋4 𝑑𝑦𝑑𝑥𝜃 = 𝜋4=−𝑡𝑎𝑛 𝜋4 =−1 Now we know that Tangent is perpendicular to Normal Hence, Slope of tangent × Slope of Normal = −1 −1 × Slope of Normal = −1 Slope of Normal = −1−1 = 1 Hence Slope of Normal is 1

Ex 6.3

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Ex 6.3,5 Important You are here

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Ex 6.3,27

Chapter 6 Class 12 Application of Derivatives

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.