# Ex 6.3, 5 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 14, 2021 by Teachoo

Last updated at April 14, 2021 by Teachoo

Transcript

Ex 6.3, 5 Find the slope of the normal to the curve π₯=π cos^3β‘π, π¦=π sin3 π at π=π/4Given π₯=π cos^3β‘π Differentiating w.r.t. ΞΈ ππ₯/ππ=π(γa cosγ^3β‘π )/ππ ππ₯/ππ=π .π(cos^3β‘π )/ππ ππ₯/ππ=π . 3 cos^2β‘π. (βsinβ‘π ) ππ₯/ππ=β 3π sinβ‘γπ cos^2β‘π γ Similarly π¦=π sin3 π Differentiating w.r.t. ΞΈ ππ¦/ππ=π(π sin3 π" " )/ππ ππ¦/ππ=π .π(sin3 π)/ππ ππ¦/ππ=π . 3 sin^2β‘π. (cosβ‘π ) ππ¦/ππ= 3π sin^2β‘γπ .πππ β‘π γ We know that Slope of tangent = ππ¦/ππ₯ =ππ¦/ππΓ·ππ₯/ππ =(3π sin^2β‘γπ cosβ‘π γ)/(β 3π sinβ‘γπ cos^2β‘π γ ) =(βsinβ‘π)/cosβ‘π =βtanβ‘π Putting π=π/4 β ππ¦/ππ₯β€|_(π = π/4)=βπ‘ππ(π/4) =β1 Now we know that Tangent is perpendicular to Normal Hence, 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 Normal is 1

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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.