Question 5 - Chapter 6 Class 12 Application of Derivatives (Important Question)
Last updated at April 16, 2024 by Teachoo
Chapter 6 Class 12 Application of Derivatives
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 12 Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams
Question 26 (MCQ) Important Deleted for CBSE Board 2024 Exams
Example 23 Important
Example 25 Important
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Ex 6.3, 1 (i) Important
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Ex 6.3,11 Important
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Ex 6.3, 20 Important
Ex 6.3,23 Important
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Ex 6.3,28 (MCQ) Important
Question 14 Important Deleted for CBSE Board 2024 Exams
Example 33 Important
Misc 3 Important
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Question 6 (MCQ) Deleted for CBSE Board 2024 Exams
Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Question 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