Ex 6.3,6 - Chapter 6 Class 12 Application of Derivatives

Transcript

Ex 6.3,6 Find the slope of the normal to the curve 𝑥=1−𝑎 sin⁡𝜃 , 𝑦 =𝑏 cos﷮2﷯ 𝜃 at 𝜃= 𝜋﷮2﷯ Slope of tangent is 𝑑𝑦﷮𝑑𝑥﷯ 𝑑𝑦﷮𝑑𝑥﷯= 𝑑𝑦﷮𝑑𝜃﷯﷮ 𝑑𝑥﷮𝑑𝜃﷯﷯ Now, 𝑑𝑦﷮𝑑𝑥﷯= 𝑑𝑦﷮𝑑𝜃﷯﷮ 𝑑𝜃﷮𝑑𝜃﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯= − 2𝑏 sin﷮𝜃 cos﷮𝜃﷯﷯﷮− 𝑎 cos﷮𝜃﷯﷯ 𝑑𝑦﷮𝑑𝑥﷯= 2𝑏 sin﷮𝜃﷯﷮𝑎﷯ We need to find Slope of tangent at 𝜃= 𝜋﷮2﷯ Putting 𝜃= 𝜋﷮2﷯ 𝑑𝑦﷮𝑑𝑥﷯﷯﷮𝜃 = 𝜋﷮2﷯﷯= 2𝑏﷮𝑎﷯ 𝑠𝑖𝑛 𝜋﷮2﷯﷯ = 2𝑏﷮𝑎﷯ 1﷯ We know that Tangent is Perpendicular to Normal Hence Slope of tangent × Slope of Normal =−1 Slope of Normal = − 1﷮𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 ﷯ Slope of Normal = − 1﷮ 𝑑𝑦﷮𝑑𝑥﷯﷯ Slope of Normal =−1 × 𝑎﷮2𝑏﷯ Slope of Normal = −𝑎﷮2𝑏﷯ Hence Slope of Normal at is −𝒂﷮𝟐𝒃﷯