Ex 6.1,2

Transcript

Ex 6.1,2 The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm? Let 𝑥 be the edge of a cube. Let V be the volume of cube. & S be the surface are of cube. Given that Volume of a cube is increasing at the rate of 8 cm3/ sec i.e. 𝑑𝑣﷮𝑑𝑡﷯ = 8 cm3/ sec We need to calculate how fast the surface Area is increasing when length of an edge is 12 cm i.e. We need to calculate 𝑑𝑠﷮𝑑𝑡﷯ when 𝑥 = 12 cm. Now, Volume of cube = (edge)3 𝑑(𝑥3)﷮𝑑𝑡﷯ = 8 𝑑(𝑥3)﷮𝑑𝑡﷯ × 𝑑𝑥﷮𝑑𝑥﷯= 8 𝑑(𝑥3)﷮𝑑𝑥﷯ × 𝑑𝑥﷮𝑑𝑡﷯= 8 3x2 . 𝑑𝑥﷮𝑑𝑡﷯= 8 𝑑𝑥﷮𝑑𝑡﷯ = 8﷮3𝑥2﷯ We know that Surface Area of cube = 6 (edge)2 S =6𝑥2 Different w.r.t 𝑡 𝑑𝑠﷮𝑑𝑡﷯ = 𝑑 6𝑥2﷯﷮𝑑𝑡﷯ 𝑑𝑠﷮𝑑𝑡﷯ = 6𝑑 𝑥2﷯﷮𝑑𝑡﷯ 𝑑𝑠﷮𝑑𝑡﷯ = 6 . 𝑑 𝑥2﷯﷮𝑑𝑡﷯ × 𝑑𝑥﷮𝑑𝑥﷯ 𝑑𝑠﷮𝑑𝑡﷯ = 6 . 𝑑 𝑥2﷯﷮𝑑𝑥﷯ × 𝑑𝑥﷮𝑑𝑡﷯ 𝑑𝑠﷮𝑑𝑡﷯ = 6 . 2𝑥 . 𝑑𝑥﷮𝑑𝑡﷯ 𝑑𝑠﷮𝑑𝑡﷯ = 12𝑥 . 𝒅𝒙﷮𝒅𝒕﷯ 𝑑𝑠﷮𝑑𝑡﷯ = 12 𝑥. 𝟖﷮𝟑𝒙𝟐﷯﷯ 𝑑𝑠﷮𝑑𝑡﷯ = 32﷮𝑥﷯ When 𝑥 = 12 𝑑𝑠﷮𝑑𝑡﷯﷯﷮𝑥 =12﷯ = 32﷮12﷯ 𝑑𝑠﷮𝑑𝑡﷯﷯﷮𝑥 =12﷯ = 8﷮3﷯ Since surface area is in cm2 & time is in sec 𝑑𝑠﷮𝑑𝑡﷯ = 8﷮3﷯ 𝑐𝑚2﷮𝑠𝑒𝑐﷯ 𝑑𝑠﷮𝑑𝑡﷯ = 𝟖﷮𝟑﷯ cm2/sec Hence, surface area is increasing at the rate of 8﷮3﷯ cm2/sec when edge is 12 cm