Ex 6.1,7

Transcript

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 (a) the perimeter, and (b) the area of the rectangle. 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 w. r. t time when 𝑥 = 8 cm & y = 6 cm i.e. find 𝑑𝑝﷮𝑑𝑡﷯ when 𝑥 = 8 cm & 𝑦 = 6 cm We know that Perimeter of rectangle = 2 (length + width) p = 2 (𝑥 + 𝑦) Different w. r. t time 𝑑𝑝﷮𝑑𝑡﷯ = 𝑑 2 𝑥+𝑦﷯﷯﷮𝑑𝑡﷯ 𝑑𝑝﷮𝑑𝑡﷯ = 2 𝑑 (𝑥 + 𝑦)﷮𝑑𝑡﷯ 𝑑𝑝﷮𝑑𝑡﷯ = 2. 𝑑𝑥﷮𝑑𝑡﷯ + 𝑑𝑦﷮𝑑𝑡﷯﷯ From (1) & (2) 𝑑𝑝﷮𝑑𝑡﷯ = 2 ( – 5 + 4 ) 𝑑𝑝﷮𝑑𝑡﷯ = 2 ( – 1) 𝑑𝑝﷮𝑑𝑡﷯ = – 2 Since perimeter is in cm & time is in sec. 𝑑𝑝﷮𝑑𝑓﷯ = − 2 𝑐𝑚﷮𝑠𝑒𝑐﷯ 𝑑𝑝﷮𝑑𝑡﷯ = – 2 cm/sec Hence, perimeter is decreasing at the rate of 2cm/sec when 𝑥 = 8 cm & y = 6 cm (ii) Let A be the Area of rectangle we need to find rate of change of area w.r.t time 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 Differentiate w.r.t time 𝑑𝐴﷮𝑑𝑡﷯ = 𝑑(𝑥 . 𝑦)﷮𝑑𝑡﷯ 𝑑𝐴﷮𝑑𝑡﷯ = 𝑑 𝑥﷯﷮𝑑𝑡﷯ . 𝑦 + 𝑑𝑦﷮𝑑𝑡﷯ . 𝑥. 𝑑𝐴﷮𝑑𝑡﷯ = – 5 . 𝑦 + 4 . 𝑥 𝑑𝐴﷮𝑑𝑡﷯ = 4𝑥 – 5𝑦 when 𝑥 = 8 cm & 𝑦 = 6 cm 𝑑𝐴﷮𝑑𝑡﷯ = 4 (8) – 5(6) 𝑑𝐴﷮𝑑𝑡﷯ = 32 – 30 𝑑𝐴﷮𝑑𝑡﷯ = 2 Since Area is in cm2 & time is in sec So 𝑑𝐴﷮𝑑𝑡﷯ = 2𝑐𝑚2﷮𝑠𝑒𝑐﷯ 𝑑𝐴﷮𝑑𝑡﷯ = 2 cm2/ sec Hence, Area is increasing at the rate of 2cm2/sec when 𝑥 = 80cm & 𝑦 = 60 cm