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Chapter 6 Class 12 Application of Derivatives
Ex 6.3,7 Important Deleted for CBSE Board 2023 Exams
Ex 6.3,12 Deleted for CBSE Board 2023 Exams
Ex 6.3,15 Important Deleted for CBSE Board 2023 Exams
Ex 6.3,26 (MCQ) Important Deleted for CBSE Board 2023 Exams
Example 35 Important
Example 37 Important
Example 38 Important
Example 40 Important
Ex 6.5, 1 (i) Important
Ex 6.5, 5 (i)
Ex 6.5,7 Important
Ex 6.5,11 Important
Ex 6.5,18 Important
Ex 6.5, 20 Important
Ex 6.5,23 Important
Ex 6.5, 26 Important
Ex 6.5,28 (MCQ) Important
Example 46 Important Deleted for CBSE Board 2023 Exams
Example 47 Important
Misc 6 Important
Misc 11 Important
Misc 13 Important
Misc 17 Important
Misc 22 (MCQ)
Chapter 6 Class 12 Application of Derivatives
Last updated at April 14, 2021 by Teachoo
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