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Ex 6.3
Ex 6.3,2
Ex 6.3,3 Important
Ex 6.3,4
Ex 6.3, 5 Important
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12
Ex 6.3,13
Ex 6.3, 14 (i)
Ex 6.3, 14 (ii) Important
Ex 6.3, 14 (iii) You are here
Ex 6.3, 14 (iv) Important
Ex 6.3, 14 (v)
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19
Ex 6.3,20
Ex 6.3,21 Important
Ex 6.3,22
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25
Ex 6.3,26 (MCQ) Important
Ex 6.3,27 (MCQ)
Last updated at Aug. 19, 2021 by Teachoo
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