Explanation Given: From Example 6, the desired differential equation is h ′ ( t ) = − k h , where t ≥ 0 , k = a A 2 g and g = 9.8 m / s 2 is the acceleration due to gravity. Initial condition is h ( 0 ) = H . The solution of the above initial value problem is h ( t ) = ( H − k t 2 ) 2 (1) Calculation: The water height function is obtained as follows, Given that H = 2.25 m , A = 2 m 2 , a = 0.5 m 2 . Here, k = a A 2 g = 0.5 2 2 ( 9.8 ) = 0.25 × 4.427 = 1.10 Substitute the values in equation (1), h ( t ) = ( 2

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Author: Briggs

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Chapter D1.1, Problem 34E

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To find: The water height function as well as determine the approximate time when the tank is first empty and graph the solution.

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Chapter D1.1, Problem 33E

Chapter D1.1, Problem 35E

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Chapter D1 Solutions

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Ch. D1.1 - Prob. 1E Ch. D1.1 - Prob. 2E Ch. D1.1 - Prob. 3E Ch. D1.1 - If the general solution of a differential equation...Ch. D1.1 - Does the function y(t) = 2t satisfy the...Ch. D1.1 - Does the function y(t) = 6e3t satisfy the initial...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...

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The water height function as well as determine the approximate time when the tank is first empty and graph the solution.

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