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Example 4 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at May 6, 2021 by Teachoo
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Chapter 6 Class 12 Application of Derivatives
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Transcript
Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and the width y is increasing at the rate of 2 cm/minute. When x = 10 cm and y = 6 cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.Let Length of rectangle = π₯ cm & Width of rectangle = π¦ cm Given length π₯ is decreasing at the rate of 3 cm/minute π π/π π = β 3 cm/ min and width y is increasing at the rate of 2 cm/min π π/π π = 2 cm /min (i) Finding rate of change of Perimeter Let P be the perimeter of rectangle. We need to calculate rate of change of perimeter when π₯ = 10 cm & π¦ = 6 cm i.e. we need to calculate ππ/ππ‘ when π₯ = 10 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‘ = β3 & ππ¦/ππ‘ = 2 ππ/ππ‘= 2(β 3 + 2) 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(ii) Finding rate of change of Area Let A be the Area of rectangle. We need to calculate Rate of change of area when π₯ = 10cm & π¦ = 6 cm i.e. we need to calculate π π¨/π π when π₯=10 & 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‘ = β3 & ππ¦/ππ‘ = 2 dA/dt = (β3)π¦+2 (π₯) π π¨/π π = β ππ+ππ Putting π = 10 & π = 6 cm ππ΄/ππ‘ = β 3(6) + 2(10) = β 18 + 20 = 2 Since Area is in cm2 & time is in time in minute ππ΄/ππ‘ = 2 cm2/min Hence, Area is increasing at the rate of 2 cm2/min.
