Finding rate of change
Ex 6.1, 1 Deleted for CBSE Board 2022 Exams
Ex 6.1,17 (MCQ) Deleted for CBSE Board 2022 Exams
Example 5 Deleted for CBSE Board 2022 Exams
Ex 6.1,15 Important Deleted for CBSE Board 2022 Exams
Example 6 Deleted for CBSE Board 2022 Exams
Ex 6.1,16 Deleted for CBSE Board 2022 Exams
Ex 6.1, 18 (MCQ) Important Deleted for CBSE Board 2022 Exams
Example 2 Deleted for CBSE Board 2022 Exams
Ex 6.1,2 Deleted for CBSE Board 2022 Exams
Example 49 Deleted for CBSE Board 2022 Exams
Example 3 Deleted for CBSE Board 2022 Exams
Ex 6.1,5 Important Deleted for CBSE Board 2022 Exams
Ex 6.1,3 Deleted for CBSE Board 2022 Exams
Ex 6.1,6 Deleted for CBSE Board 2022 Exams
Ex 6.1,12 Deleted for CBSE Board 2022 Exams
Ex 6.1,13 Important Deleted for CBSE Board 2022 Exams
Misc. 19 (MCQ) Deleted for CBSE Board 2022 Exams
Ex 6.1,14 Deleted for CBSE Board 2022 Exams
Example 43 Important Deleted for CBSE Board 2022 Exams
Ex 6.1,4 Important Deleted for CBSE Board 2022 Exams
Example 42 Important Deleted for CBSE Board 2022 Exams
Example 4 Important Deleted for CBSE Board 2022 Exams
Ex 6.1,7 Deleted for CBSE Board 2022 Exams
Ex 6.1,8 Deleted for CBSE Board 2022 Exams
Ex 6.1,9 Deleted for CBSE Board 2022 Exams
Ex 6.1,11 Important Deleted for CBSE Board 2022 Exams
Misc 3 Important You are here
Ex 6.1,10 Important Deleted for CBSE Board 2022 Exams
Example 44 Important Deleted for CBSE Board 2022 Exams
Finding rate of change
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