Ex 6.3, 14 - Find equations of tangent and normal to - Ex 6.3

Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 6 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 7 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 8 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 9 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 10 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 11 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 12 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 13 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 14 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 15 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 16 Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives - Part 17

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

Transcript

Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (i) 𝑦=π‘₯4 βˆ’6π‘₯3+13π‘₯2 βˆ’10π‘₯+5 π‘Žπ‘‘ (0, 5) 𝑦=π‘₯4 βˆ’6π‘₯3+13π‘₯2 βˆ’10π‘₯+5 Differentiating w.r.t. π‘₯ 𝑑𝑦/𝑑π‘₯=4π‘₯^3βˆ’18π‘₯^2+26π‘₯βˆ’10 Now Point Given is (0 ,5) Hence π‘₯=0 , 𝑦=5 Putting π‘₯=0 in (1) Slope of tangent at (0 , 5) 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((0, 5) )=4(0)^3βˆ’18(0)^2+26(0)βˆ’10 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((0, 5) )=0βˆ’0+0βˆ’10 𝑑𝑦/𝑑π‘₯=βˆ’10 Hence, Slope of tangent =βˆ’10 We know that Slope of tangent Γ— Slope of Normal =βˆ’1 βˆ’10 Γ—"Slope of Normal "=βˆ’1 "Slope of Normal" =(βˆ’1)/(βˆ’10)=1/10 Hence Slope of tangent at (0, 5)=βˆ’10 & Slope of Normal at (0, 5)=1/10 Finding equation of tangent & normal Now Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (0, 5) & Slope –10 is (π‘¦βˆ’5)=βˆ’10(π‘₯βˆ’0) π‘¦βˆ’5=βˆ’10π‘₯ 10π‘₯+π‘¦βˆ’5=0 πŸπŸŽπ’™+π’š=πŸ“ Equation of Normal at (0, 5) & Slope 1/10 is (π‘¦βˆ’5)=1/10 (π‘₯βˆ’0) π‘¦βˆ’5=1/10 π‘₯ 10(π‘¦βˆ’5)=π‘₯ 10π‘¦βˆ’50=π‘₯ π’™βˆ’πŸπŸŽπ’š+πŸ“πŸŽ=𝟎 Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (ii) 𝑦=π‘₯4 βˆ’6π‘₯3+13π‘₯2 βˆ’10π‘₯+5 π‘Žπ‘‘ (1, 3) Given Curve is 𝑦=π‘₯4 βˆ’6π‘₯3+13π‘₯2 βˆ’10π‘₯+5 Differentiating w.r.t x 𝑑𝑦/𝑑π‘₯=4π‘₯^3βˆ’18π‘₯^2+26π‘₯βˆ’10 Now Point Given is (1 , 3) Finding slope of tangent at (1,3) 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((1 , 3) )=4(1)^3βˆ’18(1)^2+26(1)βˆ’10 =4βˆ’18+26βˆ’10 =2 ∴ Slope of tangent at (1, 3) =2 Also, We know that Slope of tangent Γ— Slope of Normal =βˆ’1 2Γ— Slope of Normal =βˆ’1 Slope of Normal = (βˆ’1)/( 2) Hence Slope of tangent at (1 , 3)=2 & Slope of Normal at (1 , 3)=(βˆ’1)/( 2) Finding equation of tangent & normal Now Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (1, 3) & Slope 2 is (π‘¦βˆ’3)=2(π‘₯βˆ’1) π‘¦βˆ’3=2π‘₯βˆ’2 𝑦=2π‘₯βˆ’2+3 π’š=πŸπ’™+𝟏 Equation of Normal at (1, 3) & Slope (βˆ’1)/2 is (π‘¦βˆ’3)=(βˆ’1)/( 2) (π‘₯βˆ’1) 2(π‘¦βˆ’3)=βˆ’1(π‘₯βˆ’1) 2π‘¦βˆ’6=βˆ’π‘₯+1 2𝑦+π‘₯βˆ’7=0 𝒙+πŸπ’šβˆ’πŸ•=𝟎 Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iii) 𝑦=π‘₯3 π‘Žπ‘‘ (1, 1)Given Curve is 𝑦=π‘₯3 Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=3π‘₯^2 We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ Given point is (1 , 3) Slope of tangent 𝑑𝑦/𝑑π‘₯ at (1 , 3) is 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((1 , 3) )=3(1)^2 =3 We know that Slope of tangent Γ— Slope of Normal =βˆ’1 3 Γ— Slope of Normal =βˆ’1 Slope of Normal = (βˆ’1)/( 3) Hence Slope of tangent at (1 , 1)=3 Slope of Normal at ( 1 , 1)=(βˆ’1)/( 3) Finding equation of tangent & normal Now Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (1, 1) & Slope 3 is (𝑦 βˆ’1)=3(π‘₯βˆ’1) 𝑦 βˆ’1=3π‘₯βˆ’3 𝑦=3π‘₯βˆ’3+1 π’š=πŸ‘π’™βˆ’πŸ Equation of Normal at (1, 1) & Slope (βˆ’1)/3 is (π‘¦βˆ’1)=(βˆ’1)/( 3) (π‘₯βˆ’1) 3(π‘¦βˆ’1)=βˆ’(π‘₯βˆ’1) 3π‘¦βˆ’3=βˆ’π‘₯+1 3𝑦+π‘₯βˆ’3βˆ’1=0 𝒙+πŸ‘π’šβˆ’πŸ’=𝟎 Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iv) 𝑦=π‘₯2 π‘Žπ‘‘ (0, 0) Given Curve is 𝑦=π‘₯^2 Differentiating w.r.t.π‘₯ 𝑑𝑦/𝑑π‘₯=2π‘₯ We know that Slope of tangent is 𝑑𝑦/𝑑π‘₯ Given point is (0 , 0) Slope of tangent at (0 , 0) 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_((0 , 0) )=2(0) =0 We know that Slope of tangent Γ— Slope if Normal =βˆ’1 0 Γ— Slope if Normal =βˆ’1 Slope if Normal =(βˆ’1)/( 0) Equation of tangent at (0 , 0) & having Slope zero is (π‘¦βˆ’0)=0(π‘₯βˆ’0) We know that Equation of line at (π‘₯1 , 𝑦1)& having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Finding equation of tangent & normal Now Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (0, 0) & Slope 0 is (π‘¦βˆ’0)=0(π‘₯βˆ’0) π‘¦βˆ’0=0 π’š=𝟎 Equation of Normal at (0, 0) & Slope (βˆ’1)/0 is (π‘¦βˆ’0)=1/0 (π‘₯βˆ’0) 0 Γ— (π‘¦βˆ’0)=1(π‘₯βˆ’0) 0=π‘₯βˆ’0 𝒙=𝟎 Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (v) π‘₯=cos⁑𝑑, 𝑦=sin⁑𝑑 π‘Žπ‘‘ 𝑑= πœ‹/4At 𝒕= 𝝅/πŸ’ x = cos πœ‹/4 = 1/√2 y = sin πœ‹/4 = 1/√2 ∴ At 𝑑=πœ‹/4 , the point is (1/√2 " ," 1/√2) Finding 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=(𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) Now, Hence, 𝑑𝑦/𝑑π‘₯=(𝑑𝑦/𝑑𝑑)/(𝑑π‘₯/𝑑𝑑) 𝒙=𝒄𝒐𝒔⁑𝒕 Differentiating w.r.t.𝑑 𝑑π‘₯/𝑑𝑑=𝑑(cos⁑𝑑 )/𝑑𝑑 𝑑π‘₯/𝑑𝑑=βˆ’sin⁑𝑑 π’š=π’”π’Šπ’β‘π’• Differentiating w.r.t. 𝑑 𝑑𝑦/𝑑𝑑=𝑑(sin⁑𝑑 )/𝑑𝑑 𝑑𝑦/𝑑𝑑=cos⁑𝑑 𝑑𝑦/𝑑π‘₯=cos⁑𝑑/(βˆ’sin⁑𝑑 ) 𝑑𝑦/𝑑π‘₯=βˆ’cot⁑𝑑 At 𝒕=𝝅/πŸ’ 〖𝑑𝑦/𝑑π‘₯β”‚γ€—_(𝑑=πœ‹/4)=βˆ’π‘π‘œπ‘‘(πœ‹/4) =βˆ’1 ∴ Slope of tangent at (1/√2 " ," 1/√2) is – 1 We know that Slope of tangent Γ— Slope of Normal =βˆ’1 βˆ’1 Γ— Slope of Normal =βˆ’1 Slope of Normal =(βˆ’1)/(βˆ’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 (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) Equation of tangent at (1/√2 " ," 1/√2) & having Slope βˆ’1 is 𝑦 βˆ’1/√2 =βˆ’π‘₯+ 1/√2 π‘₯+𝑦 =1/√2+ 1/√2 π‘₯+𝑦 =2/√2 π‘₯+𝑦 =√2 𝒙+π’šβˆ’βˆšπŸ=𝟎 Equation of Normal at (1/√2 " ," 1/√2) & having Slope 1 is (π‘¦βˆ’1/√2)=1(π‘₯βˆ’1/√2) 𝑦 βˆ’1/√2 =π‘₯βˆ’ 1/√2 𝑦 =π‘₯βˆ’1/√2+ 1/√2 𝑦 =π‘₯ 𝒙 =π’š

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.