Example 2 - Chapter 6 Class 12 Application of Derivatives

Transcript

Example 2 The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 10 centimeters ? Let 𝑥 be length of side V be volume t be time per second Now , we know that Volume of cube = (side)3 V = 𝑥3 Also it is given that Volume of cube is increasing at rate of 9 cubic cm/sec. Therefore ﷐𝑑𝑉﷮𝑑𝑡﷯ = 9 Putting V = 𝑥3 ﷐﷐𝑑𝑥﷮3﷯﷮𝑑𝑡﷯ = 9 ﷐﷐𝑑𝑥﷮3﷯﷮𝑑𝑡﷯ . ﷐𝑑𝑥﷮𝑑𝑥﷯ = 9 ﷐﷐𝑑𝑥﷮3﷯﷮𝑑𝑥﷯ . ﷐𝑑𝑥﷮𝑑𝑡﷯ = 9 3𝑥2 . ﷐𝑑𝑥﷮𝑑𝑡﷯ = 9 ﷐𝑑𝑥﷮𝑑𝑡﷯ = ﷐9﷮﷐3𝑥﷮2﷯﷯ ﷐𝑑𝑥﷮𝑑𝑡﷯ = ﷐3﷮﷐𝑥﷮2﷯﷯ Also, We know that Surface area = 6 side2 S = 6𝑥2 Now, we need to find Increase in surface Area as compared to time i.e. ﷐𝑑𝑆﷮𝑑𝑡﷯ ﷐𝑑𝑆﷮𝑑𝑡﷯ = ﷐𝑑(6﷐𝑥﷮2﷯)﷮𝑑𝑡﷯ = ﷐𝑑6𝑥2﷮𝑑𝑡﷯ . ﷐𝑑𝑥﷮𝑑𝑥﷯ = 6 . ﷐𝑑(𝑥2)﷮𝑑𝑥﷯ . ﷐𝑑𝑥﷮𝑑𝑡﷯ = 6 . (2x) . ﷐𝑑𝑥﷮𝑑𝑡﷯ = 12𝑥 . ﷐𝒅𝒙﷮𝒅𝒕﷯ = 12𝑥 . ﷐𝟑﷮𝒙𝟐﷯ = ﷐36﷮𝑥﷯ Thus, ﷐𝑑𝑠﷮𝑑𝑡﷯ = ﷐36﷮𝑥﷯ When 𝑥= 10 cm ﷐𝑑𝑠﷮𝑑𝑡﷯ = ﷐36﷮10﷯ ﷐𝑑𝑠﷮𝑑𝑡﷯ = 3.6 Since surface area is in cm2 & time is in seconds, ﷐𝑑𝑠﷮𝑑𝑡﷯ = 3.6 ﷐𝑐𝑚2﷮𝑠﷯ ﷐𝒅𝒔﷮𝒅𝒕﷯ = 3.6 cm2 /s