Check sibling questions

Finding rate of change

Ex 6.1, 7 - The length x of a rectangle is decreasing at rate

Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 3

Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 4

Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 5
Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 6


Transcript

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 = 𝑦 Given that Length of rectangle is decreasing at the rate of 5 cm/min i.e. 𝒅𝒙/𝒅𝒕 = – 5 cm/ min And width of rectangle is increasing at the rate of 4 cm/min I.e. π’…π’š/𝒅𝒕 = 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 𝑑𝑃/𝑑𝑑 when π‘₯ = 8 cm & 𝑦 = 6 cm We know that Perimeter of rectangle = 2 (Length + Width) P = 2 (π‘₯ + 𝑦) Now, 𝑑𝑃/𝑑𝑑 = (𝑑 (2(π‘₯ + 𝑦)))/𝑑𝑑 𝑑𝑃/𝑑𝑑= 2 [𝑑(π‘₯ + 𝑦)/𝑑𝑑] 𝒅𝑷/𝒅𝒕= 2 [𝒅𝒙/𝒅𝒕+ π’…π’š/𝒅𝒕] From (1) & (2) 𝑑π‘₯/𝑑𝑑 = –5 & 𝑑𝑦/𝑑𝑑 = 4 𝑑𝑃/𝑑𝑑= 2(–5 + 4) 𝑑𝑃/𝑑𝑑= 2 (–1) 𝒅𝑷/𝒅𝒕= –2 Since perimeter is in cm & time is in minute 𝑑𝑃/𝑑𝑑 = – 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 & 𝑦 = 6 cm i.e. 𝒅𝑨/𝒅𝒕 when π‘₯ = 8 cm & 𝑦 = 6 cm We know that Area of rectangle = Length Γ— Width A = π‘₯ Γ— 𝑦 Now, 𝑑𝐴/𝑑𝑑 = (𝑑(π‘₯. 𝑦))/𝑑𝑑 𝑑𝐴/𝑑𝑑 = 𝑑π‘₯/𝑑𝑑 . 𝑦 + 𝑑𝑦/𝑑𝑑 . π‘₯. From (1) & (2) 𝑑π‘₯/𝑑𝑑 = –5 & 𝑑𝑦/𝑑𝑑 = 4 𝑑𝐴/𝑑𝑑 = (–5)𝑦 + (4)π‘₯ 𝑑𝐴/𝑑𝑑 = 4π‘₯ – 5𝑦 Putting π‘₯ = 8 cm & 𝑦 = 6 cm 𝑑𝐴/𝑑𝑑 = 4 (8) – 5(6) Using product rule in x . y as (u.v)’ = u’ v + vβ€˜ u 𝑑𝐴/𝑑𝑑 = 32 – 30 𝒅𝑨/𝒅𝒕 = 2 Since Area is in cm2 & time is in minute 𝑑𝐴/𝑑𝑑 = 2 cm2/ min Hence, Area is increasing at the rate of 2cm2/min

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.