# Ex 6.3,14 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (i) π¦=π₯4 β6π₯3+13π₯2 β10π₯+5 ππ‘ (0, 5) π¦=π₯4 β6π₯3+13π₯2 β10π₯+5 Differentiating w.r.t. π₯ ππ¦/ππ₯=4π₯^3β18π₯^2+26π₯β10 Now Point Given is (0 ,5) Hence π₯=0 , π¦=5 Putting π₯=0 in (1) Slope of tangent at (0 , 5) γππ¦/ππ₯βγ_((0, 5) )=4(0)^3β18(0)^2+26(0)β10 γππ¦/ππ₯βγ_((0, 5) )=0β0+0β10 ππ¦/ππ₯=β10 Hence, Slope of tangent =β10 We know that Slope of tangent Γ Slope of Normal =β1 β10 Γ"Slope of Normal "=β1 "Slope of Normal" =(β1)/(β10)=1/10 Hence Slope of tangent at (0, 5)=β10 & Slope of Normal at (0, 5)=1/10 Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (0, 5) & Slope β10 is (π¦β5)=β10(π₯β0) π¦β5=β10π₯ 10π₯+π¦β5=0 πππ+π=π Equation of Normal at (0, 5) & Slope 1/10 is (π¦β5)=1/10 (π₯β0) π¦β5=1/10 π₯ 10(π¦β5)=π₯ 10π¦β50=π₯ πβπππ+ππ=π Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (ii) π¦=π₯4 β6π₯3+13π₯2 β10π₯+5 ππ‘ (1, 3) Given Curve is π¦=π₯4 β6π₯3+13π₯2 β10π₯+5 Differentiating w.r.t x ππ¦/ππ₯=4π₯^3β18π₯^2+26π₯β10 Now Point Given is (1 , 3) Finding slope of tangent at (1,3) γππ¦/ππ₯βγ_((1 , 3) )=4(1)^3β18(1)^2+26(1)β10 =4β18+26β10 =2 β΄ Slope of tangent at (1, 3) =2 Also, We know that Slope of tangent Γ Slope of Normal =β1 2Γ Slope of Normal =β1 Slope of Normal = (β1)/( 2) Hence Slope of tangent at (1 , 3)=2 & Slope of Normal at (1 , 3)=(β1)/( 2) Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (1, 3) & Slope 2 is (π¦β3)=2(π₯β1) π¦β3=2π₯β2 π¦=2π₯β2+3 π=ππ+π Equation of Normal at (1, 3) & Slope (β1)/2 is (π¦β3)=(β1)/( 2) (π₯β1) 2(π¦β3)=β1(π₯β1) 2π¦β6=βπ₯+1 2π¦+π₯β7=0 π+ππβπ=π Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iii) π¦=π₯3 ππ‘ (1, 1) Given Curve is π¦=π₯3 Differentiating w.r.t.π₯ ππ¦/ππ₯=3π₯^2 We know that Slope of tangent is ππ¦/ππ₯ Given point is (1 , 3) Slope of tangent ππ¦/ππ₯ at (1 , 3) is γππ¦/ππ₯βγ_((1 , 3) )=3(1)^2 =3 We know that Slope of tangent Γ Slope of Normal =β1 3 Γ Slope of Normal =β1 Slope of Normal = (β1)/( 3) Hence Slope of tangent at (1 , 1)=3 Slope of Normal at ( 1 , 1)=(β1)/( 3) Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (1, 1) & Slope 3 is (π¦ β1)=3(π₯β1) π¦ β1=3π₯β3 π¦=3π₯β3+1 π=ππβπ Equation of Normal at (1, 1) & Slope (β1)/3 is (π¦β1)=(β1)/( 3) (π₯β1) 3(π¦β1)=β(π₯β1) 3π¦β3=βπ₯+1 3π¦+π₯β3β1=0 π+ππβπ=π Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (iv) π¦=π₯2 ππ‘ (0, 0) Given Curve is π¦=π₯^2 Differentiating w.r.t.π₯ ππ¦/ππ₯=2π₯ We know that Slope of tangent is ππ¦/ππ₯ Given point is (0 , 0) Slope of tangent at (0 , 0) γππ¦/ππ₯βγ_((0 , 0) )=2(0) =0 We know that Slope of tangent Γ Slope if Normal =β1 0 Γ Slope if Normal =β1 Slope if Normal =(β1)/( 0) Equation of tangent at (0 , 0) & having Slope zero is (π¦β0)=0(π₯β0) We know that Equation of line at (π₯1 , π¦1)& having Slope m is π¦βπ¦1=π(π₯βπ₯1) Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (0, 0) & Slope 0 is (π¦β0)=0(π₯β0) π¦β0=0 π=π Equation of Normal at (0, 0) & Slope (β1)/0 is (π¦β0)=1/0 (π₯β0) 0 Γ (π¦β0)=1(π₯β0) 0=π₯β0 π=π Ex 6.3, 14 Find the equations of the tangent and normal to the given curves at the indicated points: (v) π₯=cosβ‘π‘, π¦=sinβ‘π‘ ππ‘ π‘= π/4 At π= π /π x = cos π/4 = 1/β2 y = sin π/4 = 1/β2 β΄ At π‘=π/4 , the point is (1/β2 " ," 1/β2) Finding ππ¦/ππ₯ ππ¦/ππ₯=(ππ¦/ππ‘)/(ππ₯/ππ‘) Now, Hence, ππ¦/ππ₯=(ππ¦/ππ‘)/(ππ₯/ππ‘) π₯=cosβ‘π‘ Differentiating w.r.t.π‘ ππ₯/ππ‘=π(cosβ‘π‘ )/ππ‘ ππ₯/ππ‘=βsinβ‘π‘ π¦=sinβ‘π‘ Differentiating w.r.t. π‘ ππ¦/ππ‘=π(sinβ‘π‘ )/ππ‘ ππ¦/ππ‘=cosβ‘π‘ ππ¦/ππ₯=cosβ‘π‘/(βsinβ‘π‘ ) ππ¦/ππ₯=βcotβ‘π‘ At π=π /π γππ¦/ππ₯βγ_(π‘=π/4)=βπππ‘(π/4) =β1 β΄ Slope of tangent at (1/β2 " ," 1/β2) is β 1 We know that Slope of tangent Γ Slope of Normal =β1 β1 Γ Slope of Normal =β1 Slope of Normal =(β1)/(β1) Slope of Normal =1 Hence Slope of tangent is β 1 & Slope of Normal is 1 Finding equation of tangent & normal Now Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (1/β2 " ," 1/β2) & having Slope β1 is π¦ β1/β2 =βπ₯+ 1/β2 π₯+π¦ =1/β2+ 1/β2 π₯+π¦ =2/β2 π₯+π¦ =β2 π+πββπ=π Equation of Normal at (1/β2 " ," 1/β2) & having Slope 1 is (π¦β1/β2)=1(π₯β1/β2) π¦ β1/β2 =π₯β 1/β2 π¦ =π₯β1/β2+ 1/β2 π¦ =π₯ π =π

Finding equation of tangent/normal when point and curve is given

Chapter 6 Class 12 Application of Derivatives

Concept wise

- Finding rate of change
- To show increasing/decreasing in whole domain
- To show increasing/decreasing in intervals
- Find intervals of increasing/decreasing
- Finding slope of tangent/normal
- Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
- Finding equation of tangent/normal when slope and curve are given
- Finding approximate value of numbers
- Finding approximate value of function
- Finding approximate value- Statement questions
- Finding minimum and maximum values from graph
- Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
- Absolute minima/maxima

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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.