



Finding rate of change
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Finding rate of change
Last updated at April 19, 2021 by Teachoo
Ex 6.1, 7 The length x of a rectangle is decreasing at the rate of 5 cm/minute & the width y is increasing at the rate of 4 cm/minute. When x = 8 cm & y = 6cm, find the rates of change of (a) the perimeter.Let Length of rectangle = π₯ & Width of rectangle = π¦ Given that Length of rectangle is decreasing at the rate of 5 cm/min i.e. π π/π π = β 5 cm/ min And width of rectangle is increasing at the rate of 4 cm/min I.e. π π/π π = 4 cm/min Let P be the perimeter of rectangle We need to find rate of change of perimeter when π₯ = 8 cm & y = 6 cm i.e. Finding ππ/ππ‘ when π₯ = 8 cm & π¦ = 6 cm We know that Perimeter of rectangle = 2 (Length + Width) P = 2 (π₯ + π¦) Now, ππ/ππ‘ = (π (2(π₯ + π¦)))/ππ‘ ππ/ππ‘= 2 [π(π₯ + π¦)/ππ‘] π π·/π π= 2 [π π/π π+ π π/π π] From (1) & (2) ππ₯/ππ‘ = β5 & ππ¦/ππ‘ = 4 ππ/ππ‘= 2(β5 + 4) ππ/ππ‘= 2 (β1) π π·/π π= β2 Since perimeter is in cm & time is in minute ππ/ππ‘ = β 2 cm/min Therefore, perimeter is decreasing at the rate of 2 cm/min Ex 6.1, 7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of change of (b) the area of the rectangle.Let A be the Area of rectangle We need to find Rate of change of area when π₯ = 8 & π¦ = 6 cm i.e. π π¨/π π when π₯ = 8 cm & π¦ = 6 cm We know that Area of rectangle = Length Γ Width A = π₯ Γ π¦ Now, ππ΄/ππ‘ = (π(π₯. π¦))/ππ‘ ππ΄/ππ‘ = ππ₯/ππ‘ . π¦ + ππ¦/ππ‘ . π₯. From (1) & (2) ππ₯/ππ‘ = β5 & ππ¦/ππ‘ = 4 ππ΄/ππ‘ = (β5)π¦ + (4)π₯ ππ΄/ππ‘ = 4π₯ β 5π¦ Putting π₯ = 8 cm & π¦ = 6 cm ππ΄/ππ‘ = 4 (8) β 5(6) Using product rule in x . y as (u.v)β = uβ v + vβ u ππ΄/ππ‘ = 32 β 30 π π¨/π π = 2 Since Area is in cm2 & time is in minute ππ΄/ππ‘ = 2 cm2/ min Hence, Area is increasing at the rate of 2cm2/min