Miscellaneous

Misc 1 (a)
Deleted for CBSE Board 2022 Exams

Misc 1 (b) Important Deleted for CBSE Board 2022 Exams

Misc 2 Important

Misc 3 Important You are here

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16

Misc 17 Important

Misc 18 Important

Misc. 19 (MCQ) Deleted for CBSE Board 2022 Exams

Misc 20 (MCQ) Important

Misc 21 (MCQ) Important

Misc 22 (MCQ)

Misc. 23 (MCQ) Important

Misc 24 (MCQ) Important

Chapter 6 Class 12 Application of Derivatives (Term 1)

Serial order wise

Last updated at May 6, 2021 by Teachoo

Misc 3 The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ?Let x be the equal sides of isosceles triangle i.e. AB = AC = π And, Base = BC = b Given that equal side of Triangle decreasing at 3 cm per second i.e. ππ₯/ππ‘= β 3 cm/sec. We need to find how fast is the area decreasing when the two equal sides are equal to the base i.e. π π¨/π π when π = b Finding Area Letβs draw perpendicular AD to BC i.e. AD β₯ BC In Isosceles triangle, perpendicular from vertex to the side bisects the side i.e. D is the mid point of BC Thus, we can write BD = DC = π/π In β ADB Using Pythagoras theorem (π΄π΅)^2=(π΄π·)^2+(π΅π·)^2 (π₯)^2=(π΄π·)^2+ (π/2)^2 π₯2 β (π/2)^2=(π΄π·)^2 (π΄π·)^2 = π₯2 β (π/2)^2 π¨π«=β(ππβ(π/π)^π ) We know that Area of isosceles triangle = 1/2 Γ Base Γ Height A = 1/2 Γ b Γ β(π₯2β(π/2)^2 ) A = π/π Γ b Γ β(ππβπ^π/π) Finding π π¨/π π Differentiating w.r.t. t ππ΄/ππ‘= 1/2 π . π(β(π₯^2 β π^2/4))/ππ‘ ππ΄/ππ‘= 1/2 π . π(β(π₯^2 β π^2/4))/ππ‘ Γππ₯/ππ₯ ππ΄/ππ‘= 1/2 π . π(β(π₯^2 β π^2/4))/ππ₯ Γπ π/π π ππ΄/ππ‘= 1/2 π . π(β(π₯^2 β π^2/4))/ππ₯ Γ π ππ΄/ππ‘= 1/2 π [1/(2β(π₯2 β π^2/4)) Γ π(π₯^2 β π^2/4)/ππ₯]Γ 3" " ππ΄/ππ‘= 1/2 π [1/(2β(π₯2 β π^2/4)) Γ(2π₯β0)]Γ 3" " π π¨/π π= πππ/(πβ(ππ β π^π/π)) Finding π π¨/π π at π = b β ππ΄/ππ‘β€|_(π₯ = π)=(3π^2)/(2β(π^2 β π^2/4)) = (6π^2)/(4β((4π^2 β π^2)/4))= (6π^2)/(4β((3π^2)/4))= (6π^2)/((4β3 π)/2)= (6π^2)/(2β3 π)= 3π/β3 =πβ3 Since dimension of area is cm2 and time is seconds β΄ ππ΄/ππ‘ = πβπ cm2/s