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Ex 6.3, 14 - Find equations of tangent and normal to - Finding equation of tangent/normal when point and curve is given

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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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

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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