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Part (a)

30. 5. Part (a). This estimate of 30.8’ is HIGHER than the actual value because r(t) is CONCAVE DOWN. (r - ) = (t - ). 2. r = 2t + 20. r  2(5.4) + 20. r  30.8 feet. Part (b). Find dV/dt when t=5:. Volume = 4/3 p r 3. dV/dt = 4 p r 2 (dr/dt). dV/dt = 4 p (30) 2 (2).

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Part (a)

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  1. 30 5 Part (a) This estimate of 30.8’ is HIGHER than the actual valuebecause r(t) is CONCAVE DOWN. (r - ) = (t - ) 2 r = 2t + 20 r  2(5.4) + 20 r  30.8 feet

  2. Part (b) Find dV/dt when t=5: Volume = 4/3pr3 dV/dt = 4pr2(dr/dt) dV/dt = 4p(30)2(2) dV/dt = 7200p ft3/min

  3. (0,5.7) (2,4) (5,2) (11,0.6) (7,1.2) (12,0.5) 2 5 12 7 11 12 r’(t) dt  0 Part (c) Right Riemann sum: (2)(4)+(3)(2)+(2)(1.2)+(4)(.6)+(1)(.5) 8 + 6 + 2.4 + 2.4 + .5 This represents the total change in theballoon’s radius during the 12 minute interval.  19.3

  4. (0,5.7) (2,4) (5,2) (11,0.6) (7,1.2) (12,0.5) 2 5 12 7 11 12 r’(t) dt. 0 Part (d) (IGNORED) Our approximation of 19.3 feet was less than Since the curve was decreasing, we purposely ignoredthese white areas that should have been considered.

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