Finding rate of change

Chapter 6 Class 12 Application of Derivatives
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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 = 𝑥 & 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. 𝒅𝒚/𝒅𝒕 = 4 cm/min Let P be the perimeter of rectangle We need to find rate of change of perimeter when 𝑥 = 8 cm & y = 6 cm i.e. Finding 𝑑𝑃/𝑑𝑡 when 𝑥 = 8 cm & 𝑦 = 6 cm We know that Perimeter of rectangle = 2 (Length + Width) P = 2 (𝑥 + 𝑦) Now, 𝑑𝑃/𝑑𝑡 = (𝑑 (2(𝑥 + 𝑦)))/𝑑𝑡 𝑑𝑃/𝑑𝑡= 2 [𝑑(𝑥 + 𝑦)/𝑑𝑡] 𝒅𝑷/𝒅𝒕= 2 [𝒅𝒙/𝒅𝒕+ 𝒅𝒚/𝒅𝒕] From (1) & (2) 𝑑𝑥/𝑑𝑡 = –5 & 𝑑𝑦/𝑑𝑡 = 4 𝑑𝑃/𝑑𝑡= 2(–5 + 4) 𝑑𝑃/𝑑𝑡= 2 (–1) 𝒅𝑷/𝒅𝒕= –2 Since perimeter is in cm & time is in minute 𝑑𝑃/𝑑𝑡 = – 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 𝑥 = 8 & 𝑦 = 6 cm i.e. 𝒅𝑨/𝒅𝒕 when 𝑥 = 8 cm & 𝑦 = 6 cm We know that Area of rectangle = Length × Width A = 𝑥 × 𝑦 Now, 𝑑𝐴/𝑑𝑡 = (𝑑(𝑥. 𝑦))/𝑑𝑡 𝑑𝐴/𝑑𝑡 = 𝑑𝑥/𝑑𝑡 . 𝑦 + 𝑑𝑦/𝑑𝑡 . 𝑥. From (1) & (2) 𝑑𝑥/𝑑𝑡 = –5 & 𝑑𝑦/𝑑𝑡 = 4 𝑑𝐴/𝑑𝑡 = (–5)𝑦 + (4)𝑥 𝑑𝐴/𝑑𝑡 = 4𝑥 – 5𝑦 Putting 𝑥 = 8 cm & 𝑦 = 6 cm 𝑑𝐴/𝑑𝑡 = 4 (8) – 5(6) Using product rule in x . y as (u.v)’ = u’ v + v‘ u 𝑑𝐴/𝑑𝑡 = 32 – 30 𝒅𝑨/𝒅𝒕 = 2 Since Area is in cm2 & time is in minute 𝑑𝐴/𝑑𝑡 = 2 cm2/ min Hence, Area is increasing at the rate of 2cm2/min 