solution_2.5 - ×-3 3 8 a f t y = 2 tan y cos(2 t b ∂f...

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Solutions to selected problems in Section 2.5 5.a) f ( t, y ) = tan t - ty 1 / 3 . b) ∂f ∂y = - 1 3 ty - 2 / 3 . c) f fails to be continuous when t is an odd multiple of π/ 2, and ∂f /∂y fails to be continuous at y = 0. We can use these values of t and y to draw vertical and horizontal lines. These lines form rectangles, from which we can use the point ( t 0 , y 0 ) = ( - 1 , 1) to pick out the largest open rectangle ( - π/ 2 , π/ 2) × (0 , ). 6.a) f ( t, y ) = t 2 - e - y y 2 - 9 . b) ∂f ∂y = e - y ( y 2 - 9 + 2 y ) ( y 2 - 9) 2 - 2 t 2 y ( y 2 - 9) 2 . c) Both f and ∂f /∂y are continuous for y 6 = ± 3. The largest rectangle is ( -∞ , )
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Unformatted text preview: × (-3 , 3). 8. a) f ( t, y ) = 2 + tan y cos(2 t ) . b) ∂f ∂y = 1 cos 2 y cos(2 t ) . c) f and ∂f /∂y fail to be continuous if y or 2 t is an odd multiple of π/ 2. The largest rectangle is (3 π/ 4 , 5 π/ 4) × (-π/ 2 , π/ 2). 11. The principle is that if y ( t ) is a solution, so is y ( t-1). So if the solution for y (0) = 2 is 2 / √ 1-t , then the solution for y (1) = 2 is 2 / q 1-( t-1) = 2 / √ 2-t . 1...
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