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Finding rate of change
Ex 6.1, 1
Ex 6.1,17 (MCQ)
Example 5
Ex 6.1,15 Important
Example 6
Ex 6.1,16
Ex 6.1, 18 (MCQ) Important
Example 2
Ex 6.1,2
Example 35
Example 3
Ex 6.1,5 Important You are here
Ex 6.1,3
Ex 6.1,6
Ex 6.1,12
Ex 6.1,13 Important
Misc 16 (MCQ)
Ex 6.1,14
Example 31 Important
Ex 6.1,4 Important
Example 30 Important
Example 4 Important
Ex 6.1,7
Ex 6.1,8
Ex 6.1,9
Ex 6.1,11 Important
Misc 2 Important
Ex 6.1,10 Important
Example 32 Important
Finding rate of change
Last updated at May 29, 2023 by Teachoo
Ex 6.1, 5 A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?Let r be the radius of circle & A be the Area of circle Given that When stone is dropped into a lake waves move in a circle at speed of 5 cm/sec i.e. Radius of circle increasing at a rate of 4 cm / sec. i.e. 𝒅𝒓/𝒅𝒕 = 5 cm/sec We need find how fast area increasing when radius is 8 cm i.e. we need to find 𝒅𝑨/𝒅𝒕 when r = 8 cm. We know that Area of circle = πr2 Now, 𝒅𝑨/𝒅𝒕 = (𝒅(𝝅𝒓𝟐))/𝒅𝒕 𝑑𝐴/𝑑𝑡 = π (𝑑(𝑟2))/𝑑𝑡 𝑑𝐴/𝑑𝑡 = π (𝑑(𝑟2))/𝑑𝑡 × 𝑑𝑟/𝑑𝑟 𝑑𝐴/𝑑𝑡 = π (𝑑(𝑟2))/𝑑𝑟 × 𝑑𝑟/𝑑𝑡 𝑑𝐴/𝑑𝑡 = π × 2r × 𝒅𝒓/𝒅𝒕 𝑑𝐴/𝑑𝑡 = 2πr × 5 𝑑𝐴/𝑑𝑡 = 10πr When r = 8 cm ├ 𝑑𝐴/𝑑𝑡┤|_(𝑟 = 8) = 10 × π × 8 ├ 𝑑𝐴/𝑑𝑡┤|_(𝑟 = 8) = 80π Since Area is in cm2 & time is in sec ├ 𝑑𝐴/𝑑𝑡┤|_(𝑟 = 8)= 80π cm2/sec Hence Area is increasing at the rate of 80π cm2/sec when r = 8 cm