Homework6

Homework6 - MA 115 Homework Solutions for Week 6 4.9#12 f x...

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Unformatted text preview: MA 115 Homework Solutions for Week 6 4.9 #12 f ( x ) = x 2 + x + 1 x = x + 1 + 1 x ⇒ F ( x ) = 1 2 x 2 + x + ln | x | + C 1 if x < 1 2 x 2 + x + ln | x | + C 2 if x > 4.9 #19 f ′ ( x ) = √ x (6 + 5 x ) = 6 x 1 / 2 + 5 x 3 / 2 ⇒ f ( x ) = 4 x 3 / 2 + 2 x 5 / 2 + C f (1) = 6 + C and f (1) = 10 ⇒ C = 4, so f ( x ) = 4 x 3 / 2 + 2 x 5 / 2 + 4. 4.9 #25 f ′′ ( θ ) = sin θ + cos θ ⇒ f ′ ( θ ) =- cos θ + sin θ + C . f ′ (0) =- 1+ C and f ′ (0) = 4 ⇒ C = 5, so f ′ ( θ ) =- cos θ +sin θ +5 and hence, f ( θ ) =- sin θ- cos θ +5 θ + D . f (0) =- 1 + D and f (0) = 3 ⇒ D = 4, so f ( θ ) =- sin θ- cos θ + 5 θ + 4. 4.9 #40 a ( t ) = v ′ ( t ) = 5 + 4 t- 2 t 2 ⇒ v ( t ) = 5 t + 2 t 2- 2 3 t 3 + C . v (0) = 3 ⇒ C = 3, so v ( t ) = 5 t + 2 t 2- 2 3 t 3 + 3. v ( t ) = s ′ ( t ) ⇒ s ( t ) = 5 2 t 2 + 2 3 t 3- 1 6 t 4 + 3 t + D . s (0) = 10 ⇒ D = 10, so the particle’s position after t seconds is given by s ( t ) = 5 2 t 2 + 2 3 t...
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This note was uploaded on 04/07/2008 for the course MA 115 taught by Professor Mahalanobis during the Fall '08 term at Stevens.

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Homework6 - MA 115 Homework Solutions for Week 6 4.9#12 f x...

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