Example 4

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 = 10cm and y = 6cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle. Let length of rectangle = 𝑥 cm & width of rectangle = 𝑦 cm Given the length 𝑥 is decreasing at the rate of 3 cm/ minute i.e. 𝑥 is decreasing w.r.t time i.e. ﷐𝑑𝑥﷮𝑑𝑡﷯ = – 3 cm/ min & The width y is increasing at the rate of 2 cm/min i.e. 𝑦 is increasing w.r.t time ﷐𝑑𝑦﷮𝑑𝑡﷯ = 2 cm /min (i) Let P be the perimeter of rectangle. We need to calculate rate of change of perimeter when 𝑥 = 10 cm & 𝑦 = 6 cm i.e. we need to calculate ﷐𝑑𝑃﷮𝑑𝑡﷯ when 𝑥 = 10 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﷯ = –3 & ﷐𝑑𝑦﷮𝑑𝑡﷯ = 2 ﷐𝑑𝑃﷮𝑑𝑡﷯= 2(– 3 + 2) ﷐𝑑𝑃﷮𝑑𝑡﷯= 2 (–1) ﷐𝑑𝑃﷮𝑑𝑡﷯= –2 Since perimeter is in cm & time is in minute ﷐𝑑𝑝﷮𝑑𝑡﷯ = ﷐−2 𝑐𝑚﷮𝑚𝑖𝑛﷯ ﷐𝑑𝑝﷮𝑑𝑡﷯ = – 2 cm / min Therefore perimeter is decreasing at the rate of 2 cm / min (ii) Let A be the Area of rectangle. We need to calculate Rate of change of area w.r.t when 𝑥 = 10cm & 𝑦 = 6 cm i.e. we need to calculate ﷐𝑑𝐴﷮𝑑𝑡﷯ when 𝑥=10 & 𝑦=6 cm We know that Area of rectangle = length × width A = 𝑥 × 𝑦 Now, ﷐𝑑𝐴﷮𝑑𝑡﷯ = ﷐𝑑 (𝑥𝑦)﷮𝑑𝑡﷯ ﷐𝑑𝐴﷮𝑑𝑡﷯ = ﷐𝑑(𝑥)﷮𝑑𝑡﷯. y + ﷐𝑑(𝑦)﷮𝑑𝑡﷯ . x From (1) & (2) ﷐𝑑𝑥﷮𝑑𝑡﷯ = –3 & ﷐𝑑𝑦﷮𝑑𝑡﷯ = 2 ﷐dA﷮dt﷯ = (– 3)𝑦+2 (𝑥) ﷐𝑑𝐴﷮𝑑𝑡﷯ = – 3𝑦+2𝑥 Putting 𝑥 = 10 & 𝑦 = 6 cm ﷐﷐﷐𝑑𝐴﷮𝑑𝑡﷯﷯﷮𝑥 =10, 𝑦 = 6﷯ = – 3(6) + 2(10) ﷐﷐﷐𝑑𝐴﷮𝑑𝑡﷯﷯﷮𝑥 =10, 𝑦 = 6﷯ = – 18 + 20 ﷐﷐﷐𝑑𝐴﷮𝑑𝑡﷯﷯﷮𝑥 =10, 𝑦 = 6﷯ = 2 Since Area is in cm2 & time is in time in minute ﷐𝑑𝐴﷮𝑑𝑡﷯ = ﷐2 𝑐𝑚2﷮𝑚𝑖𝑛﷯ ﷐𝑑𝐴﷮𝑑𝑡﷯ = 2 cm2/min Hence, Area is increasing at the rate of 2 cm2/min.