36_Cal_Solution of Calculus_6e

36_Cal_Solution of Calculus_6e - Answer: V A(V ) = 4 1 /3 V...

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T h x 2 x r = µ V π 1 / 3 and so A =4 π µ V π 2 / 3 . Answer: A ( V )=4 π µ V π 2 / 3 , 0 <V < + . It is permissible, and sometimes desirable, to use instead the domain 0 5 V< + . C01S0M.017: The following Fgure shows an equilateral triangle with sides of length 2 x and an altitude of length h . Because T is a right triangle, we see that x 2 + h 2 =(2 x ) 2 , so that h = x 3 . The area of this triangle is A = hx and its perimeter is P =6 x .So A = x 2 3 and x = P 6 . Therefore A ( P )= P 2 3 36 , 0 <P< . C01S0M.018: The square has perimeter x and thus edge length y = 1 4 x . The circle has circumference 100 x .Thu si f z is the radius of the circle, then 2 πz = 100 x , so that z = (100 x ) / (2 π ). The area of the square is y 2 and the area of the circle is πz
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This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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