35_Cal_Solution of Calculus_6e

35_Cal_Solution of Calculus_6e - V ( S ) = Ã r S 6 ! 3 =...

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C01S0M.012: If 70 5 F 5 90, then 70 5 32 + 9 5 C 5 90. Hence 70 32 5 9 5 C 5 90 32; 38 5 9 5 C 5 58; 190 5 9 C 5 290; 190 9 5 C 5 290 9 . Answer: The Celsius temperature ranged from a low of about 21 . 1 C to a high of about 32 . 2 C. C01S0M.013: Because 25 <R< 50, 25 I<IR< 50 I , so that 25 I<E< 50 I ; 25 I< 100 < 50 I ; I< 4 < 2 I ; I< 4 and 2 <I. Therefore the current I lies in the range 2 <I< 4. C01S0M.014: Because 3 <L< 4, we see that 3 32 < L 32 < 4 32 ; r 3 32 < r L 32 < r 1 8 ; 2 π r 3 32 < 2 π r L 32 < 2 π r 1 8 ; π 2 r 3 2 <T<π r 1 2 . In approximate terms, 1 . 923825 <T< 2 . 221441. C01S0M.015: If a cube has edge length x , then its volume is V = x 3 and its total surface area is S =6 x 2 (because each of its six faces has area x 2 ). Hence x = p S/ 6, and therefore
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Unformatted text preview: V ( S ) = Ã r S 6 ! 3 = µ S 6 ¶ 3 / 2 , &lt; S &lt; + ∞ . Under certain circumstances it would be both permissible and desirable to let the domain of V be the interval [0 , + ∞ ). C01S0M.016: Let r denote the radius, and h the height, of the cylinder. Then its volume V and total surface area A are given by V = πr 2 h and A = 2 πrh + 2 πr 2 (look inside the front cover of the textbook). In this problem we are given h = r , so that V = πr 3 and A = 4 πr 2 . Therefore 2...
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