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  1. Chapter 6 Class 12 Application of Derivatives
  2. Concept 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‹/4 At 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) π‘₯=cos⁑𝑑 Differentiating w.r.t.𝑑 𝑑π‘₯/𝑑𝑑=𝑑(cos⁑𝑑 )/𝑑𝑑 𝑑π‘₯/𝑑𝑑=βˆ’sin⁑𝑑 𝑦=sin⁑𝑑 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 9 years. He provides courses for Maths and Science at Teachoo.