                1. Chapter 6 Class 12 Application of Derivatives
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: (i) 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 𝑎𝑡 (0, 5) 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 Differentiating w.r.t. 𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯=4﷐𝑥﷮3﷯−18﷐𝑥﷮2﷯+26𝑥−10 Now Point Given is ﷐0 ,5﷯ Hence 𝑥=0 , 𝑦=5 Putting 𝑥=0 in (1) Slope of tangent at ﷐0 , 5﷯ ﷐﷐𝑑𝑦﷮𝑑𝑥﷯│﷮﷐0 , 5﷯﷯=4﷐﷐0﷯﷮3﷯−18﷐﷐0﷯﷮2﷯+26﷐0﷯−10 ﷐﷐𝑑𝑦﷮𝑑𝑥﷯│﷮﷐0 , 5﷯﷯=0−0+0−10 ﷐𝑑𝑦﷮𝑑𝑥﷯=−10 Hence, Slope of tangent =−10 We know that Slope of tangent × Slope of Normal =−1 −10 ×Slope of Normal =−1 Slope of Normal =﷐−1﷮−10﷯=﷐1﷮10﷯ Hence Slope of tangent at ﷐0 , 5﷯=−10 & Slope of Normal at ﷐0 , 5﷯=﷐1﷮10﷯ Finding equation of tangent & normal Ex 6.3,14 Find the equations of the tangent and normal to the given curves at the indicated points: (ii) 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 𝑎𝑡 (1, 3) Given Curve is 𝑦=𝑥4 −6𝑥3+13𝑥2 −10𝑥+5 Differentiating w.r.t x ﷐𝑑𝑦﷮𝑑𝑥﷯=4﷐𝑥﷮3﷯−18﷐𝑥﷮2﷯+26𝑥−10 Now Point Given is ﷐1 , 3﷯ Finding slope of tangent at (1,3) ﷐﷐𝑑𝑦﷮𝑑𝑥﷯│﷮﷐1 , 3﷯﷯=4﷐﷐1﷯﷮3﷯−18﷐﷐1﷯﷮2﷯+26﷐1﷯−10 =4−18+26−10 =2 ∴ Slope of tangent at ﷐1 , 3﷯ =2 Also, We know that Slope of tangent × Slope of Normal =−1 2× Slope of Normal =−1 Slope of Normal = ﷐−1﷮ 2﷯ Hence Slope of tangent at ﷐1 , 3﷯=2 & Slope of Normal at ﷐1 , 3﷯=﷐−1﷮ 2﷯ Finding equation of tangent & normal Ex 6.3,14 Find the equations of the tangent and normal to the given curves at the indicated points: (iii) 𝑦=𝑥3 𝑎𝑡 (1, 1) Given Curve is 𝑦=𝑥3 Differentiating w.r.t.𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯=3﷐𝑥﷮2﷯ We know that Slope of tangent is ﷐𝑑𝑦﷮𝑑𝑥﷯ Given point is ﷐1 , 3﷯ Slope of tangent ﷐𝑑𝑦﷮𝑑𝑥﷯ at ﷐1 , 3﷯ is ﷐﷐𝑑𝑦﷮𝑑𝑥﷯│﷮﷐1 , 3﷯﷯=3﷐﷐1﷯﷮2﷯ =3 We know that Slope of tangent × Slope of Normal =−1 3 × Slope of Normal =−1 Slope of Normal = ﷐−1﷮ 3﷯ Hence Slope of tangent at ﷐1 , 1﷯=3 Slope of Normal at ﷐ 1 , 1﷯=﷐−1﷮ 3﷯ Finding equation of tangent & normal Ex 6.3,14 Find the equations of the tangent and normal to the given curves at the indicated points: (iv) 𝑦=𝑥2 𝑎𝑡 (0, 0) Given Curve is 𝑦=﷐𝑥﷮2﷯ Differentiating w.r.t.𝑥 ﷐𝑑𝑦﷮𝑑𝑥﷯=2𝑥 We know that Slope of tangent is ﷐𝑑𝑦﷮𝑑𝑥﷯ Given point is ﷐0 , 0﷯ Slope of tangent at ﷐0 , 0﷯ ﷐﷐𝑑𝑦﷮𝑑𝑥﷯│﷮﷐0 , 0﷯﷯=2﷐0﷯ =0 We know that Slope of tangent × Slope if Normal =−1 0 × Slope if Normal =−1 Slope if Normal =﷐−1﷮ 0﷯ Equation of tangent at ﷐0 , 0﷯ & having Slope zero is ﷐𝑦−0﷯=0﷐𝑥−0﷯ Finding equation of tangent & normal 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﷮4﷯ At 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﷯﷯ ﷐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

Ex 6.3 