Explanation Given: The chemical reaction model in Example 7, is d y d t = k y 2 . And the initial condition is t = 0 , y = 45 grams and at t = 2 and y = 4 grams . Formula used: The integral if, ∫ y n d y = y n + 1 n + 1 + c Calculation: Consider the differential equation: d y d t = k y 2 This gives d y y 2 = k d t This differential equation is of variable separable form that is f ( y ) d y = g ( x ) d x . Now, integrate to obtain, ∫ d y y 2 = k ∫ d t − 1 y = k t + c y = − 1 k t + c So, the general solution is y = − 1 k t + c , where c is constant of integration. Now use the initial condition at t = 0 , y = 45 grams in the general solution y = − 1 k t + c . Then, y = − 1 k t + c 45 = − 1 k ⋅ 0 + c 45 = − 1 c c = − 1 45 Now substitute the value of c = − 1 45 , in the general solution y = − 1 k t + c

Calculus: Early Transcendental Functions

7th Edition

ISBN: 9781337552516

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 6.3, Problem 57E

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To calculate: The amount of y as a function of t using chemical reaction model in Example 7 which is dydt=ky2 and also sketch the graph of the function with the help of graphing utility.

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Chapter 6 Solutions

Calculus: Early Transcendental Functions

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The amount of y as a function of t using chemical reaction model in Example 7 which is d y d t = k y 2 and also sketch the graph of the function with the help of graphing utility.

Solution Summary: The author calculates the amount of y as a function of t using the chemical reaction model in Example 7 and sketched the graph of the function.

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To calculate: The amount of y as a function of t using chemical reaction model in Example 7 which is dydt=ky2 and also sketch the graph of the function with the help of graphing utility.

Chapter 6 Solutions