42SAP Calculus

Tuesday, March 28, 2006

Chapter 8 Question 13 a & b

Chapter 8 Question 13

a.) v(t) = 3t2- 6t
x(t) = t3- 3t2 + C
x(2) = 4
4 = (2)3 + 3(2)2 + C
4 = -4 + C
C = 8
x(t) = t3 – 3t2 + 8

b.) M(t) = 1/b-a v(t) dt
M(t) = 1/4-1 3t2 – 6t dt
M(t) = 1/3 [ t3 - 3t2 ]
M(t) = 1/3 [ ((4)3 – 3(4)2) – ((1)3-3(1)2)]
M(t) = 1/3 [64 – 48 - 1 + 3]
M(t) = 1/3 [18]
M(t) = 6
6 = 3t2 - 6t
3(t2 - 2t - 2) = 0
(-b √ (b2-4ac)) / 2a
=(-(-2) √((-2)2 – 4(1)(2))) / 2(1)
=(2 √(4 + 8)) / 2
=(2 2√3) / 2
=1 + √3, 1- √3 , [1,4]
therefore, 1 + √3

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