     1. Chapter 6 Class 12 Application of Derivatives (Term 1)
2. Serial order wise
3. Ex 6.3

Transcript

Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (v) 𝑥=cos⁡𝑡, 𝑦=sin⁡𝑡 𝑎𝑡 𝑡= 𝜋/4At 𝒕= 𝝅/𝟒 x = cos 𝜋/4 = 1/√2 y = sin 𝜋/4 = 1/√2 ∴ At 𝑡=𝜋/4 , the point is (1/√2 " ," 1/√2) Finding 𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=(𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) Now, Hence, 𝑑𝑦/𝑑𝑥=(𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡) 𝒙=𝒄𝒐𝒔⁡𝒕 Differentiating w.r.t.𝑡 𝑑𝑥/𝑑𝑡=𝑑(cos⁡𝑡 )/𝑑𝑡 𝑑𝑥/𝑑𝑡=−sin⁡𝑡 𝒚=𝒔𝒊𝒏⁡𝒕 Differentiating w.r.t. 𝑡 𝑑𝑦/𝑑𝑡=𝑑(sin⁡𝑡 )/𝑑𝑡 𝑑𝑦/𝑑𝑡=cos⁡𝑡 𝑑𝑦/𝑑𝑥=cos⁡𝑡/(−sin⁡𝑡 ) 𝑑𝑦/𝑑𝑥=−cot⁡𝑡 At 𝒕=𝝅/𝟒 〖𝑑𝑦/𝑑𝑥│〗_(𝑡=𝜋/4)=−𝑐𝑜𝑡(𝜋/4) =−1 ∴ Slope of tangent at (1/√2 " ," 1/√2) is – 1 We know that Slope of tangent × Slope of Normal =−1 −1 × Slope of Normal =−1 Slope of Normal =(−1)/(−1) Slope of Normal =1 Hence Slope of tangent is – 1 & Slope of Normal is 1 Finding equation of tangent & normal Now Equation of line at (𝑥1 , 𝑦1) & having Slope m is 𝑦−𝑦1=𝑚(𝑥−𝑥1) Equation of tangent at (1/√2 " ," 1/√2) & having Slope −1 is 𝑦 −1/√2 =−𝑥+ 1/√2 𝑥+𝑦 =1/√2+ 1/√2 𝑥+𝑦 =2/√2 𝑥+𝑦 =√2 𝒙+𝒚−√𝟐=𝟎 Equation of Normal at (1/√2 " ," 1/√2) & having Slope 1 is (𝑦−1/√2)=1(𝑥−1/√2) 𝑦 −1/√2 =𝑥− 1/√2 𝑦 =𝑥−1/√2+ 1/√2 𝑦 =𝑥 𝒙 =𝒚 