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Ex 6.3

Ex 6.3, 1
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Ex 6.3,2 Deleted for CBSE Board 2023 Exams

Ex 6.3,3 Important Deleted for CBSE Board 2023 Exams

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Ex 6.3,6 Deleted for CBSE Board 2023 Exams You are here

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Ex 6.3, 14 (i) Deleted for CBSE Board 2023 Exams

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Ex 6.3,26 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.3,27 (MCQ) Deleted for CBSE Board 2023 Exams

Last updated at April 14, 2021 by Teachoo

Ex 6.3, 6 Find the slope of the normal to the curve π₯=1βπ sinβ‘π , π¦ =π cos^2 π at π= π/2Slope of tangent is ππ¦/ππ₯ ππ¦/ππ₯=(ππ¦/ππ)/(ππ₯/ππ) π=πβπ πππβ‘π½ Differentiating w.r.t. ΞΈ ππ₯/ππ=π(1 β π sinβ‘π )/ππ ππ₯/ππ=0βπ cosβ‘π ππ₯/ππ=βπ cosβ‘π π=π γπππγ^πβ‘π½ Differentiating w.r.t. ΞΈ ππ¦/ππ=π(π cos^2β‘π )/ππ ππ¦/ππ=π . 2 cosβ‘γπ . π(cosβ‘π )/ππγ ππ¦/ππ=2π cosβ‘γπ . (βsinβ‘π )γ ππ¦/ππ=β2π sinβ‘γπ cosβ‘π γ Now, ππ¦/ππ₯=(ππ¦/ππ)/(ππ/ππ) ππ¦/ππ₯=(β 2π sinβ‘γπ cosβ‘π γ)/(β π cosβ‘π ) ππ¦/ππ₯=(2π sinβ‘π)/π We need to find Slope of tangent at π=π/2 Putting π=π/2 β ππ¦/ππ₯β€|_(π = π/2)=2π/π π ππ(π/2) =2π/π (1) We know that Tangent is Perpendicular to Normal Hence Slope of tangent Γ Slope of Normal =β1 Slope of Normal =(β 1)/(πππππ ππ π‘ππππππ‘ ) Slope of Normal =(β 1)/( ππ¦/ππ₯) Slope of Normal =β1 Γ π/2π Slope of Normal =(βπ)/2π Hence Slope of Normal at is (βπ)/ππ