Example 2 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2018 by Teachoo
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 𝑑𝑉𝑑𝑡 = 9 Putting V = 𝑥3 𝑑𝑥3𝑑𝑡 = 9 𝑑𝑥3𝑑𝑡 . 𝑑𝑥𝑑𝑥 = 9 𝑑𝑥3𝑑𝑥 . 𝑑𝑥𝑑𝑡 = 9 3𝑥2 . 𝑑𝑥𝑑𝑡 = 9 𝑑𝑥𝑑𝑡 = 93𝑥2 𝑑𝑥𝑑𝑡 = 3𝑥2 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. 𝑑𝑆𝑑𝑡 𝑑𝑆𝑑𝑡 = 𝑑(6𝑥2)𝑑𝑡 = 𝑑6𝑥2𝑑𝑡 . 𝑑𝑥𝑑𝑥 = 6 . 𝑑(𝑥2)𝑑𝑥 . 𝑑𝑥𝑑𝑡 = 6 . (2x) . 𝑑𝑥𝑑𝑡 = 12𝑥 . 𝒅𝒙𝒅𝒕 = 12𝑥 . 𝟑𝒙𝟐 = 36𝑥 Thus, 𝑑𝑠𝑑𝑡 = 36𝑥 When 𝑥= 10 cm 𝑑𝑠𝑑𝑡 = 3610 𝑑𝑠𝑑𝑡 = 3.6 Since surface area is in cm2 & time is in seconds, 𝑑𝑠𝑑𝑡 = 3.6 𝑐𝑚2𝑠 𝒅𝒔𝒅𝒕 = 3.6 cm2 /s
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Example 5
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Example 9
Example 10
Example 11
Example 12
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Example 14
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Example 17
Example 18
Example 19
Example 20
Example 21
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Example 35 Important
Example 36
Example 37 Important
Example 38 Important
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Example 40 Important
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Example 46 Important
Example 47 Important
Example 48
Example 49
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Chapter 6 Class 12 Application of Derivatives
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.