Explanation Given: • Water flows at a rate of 3000 + 20 t liters per hour • t is in hours Key Concepts Used: In general, Quantity of water is defined as the product of rate of flow of water and the elapsed time. Quantity of Water=Flow rate × Time elapsed. Many applications are based on the following principle: The net change in a quantity s ( t ) is equal to the integral of its rate of change: s ( t 2 ) − s ( t 1 ) = ∫ s ' ( t ) ⋅ d t t 1 t 2 . where s ' ( t ) is the rate at which the quantity of water is changing. In the above equation, the term ∫ s ' ( t ) ⋅ d t t 1 t 2 is called the Integral of the rate of change and the term s ( t 2 ) − s ( t 1 ) is called the Net change over [ t 1 , t 2 ] . Here in the problem it is given that waterflows into an empty reservoir at a rate of 3000 + 20 t liters per hour where t is in hours. They ask us to find the quantity of water in the reservoir after 5 hour. The initial time in this problem is t 1 = 0 and the final time is t 2 = 5 . Thenet flow rate of water is given as s ' ( t ) = 3000 + 20 t . Substitute these values in the given equation to find the quantity of water in the reservoir after 5 hour. Calculation: Description: s ( t 2 ) − s ( t 1 ) = ∫ s ' ( t ) ⋅ d t t 1 t 2 Step 1 : Write net flow rate of water into the reservoir...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 5.6, Problem 1E

To determine

Quantity of water in the reservoir after $5$ hour

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Chapter 5.6, Problem 3PQ

Chapter 5.6, Problem 2E

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Chapter 5 Solutions

Calculus - Standalone book

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