Page 185 - How To Solve Word Problems In Calculus

(b) The infection rate is Q (t).
10
Q(t) =
1 + 100e −1.5t

= 10(1 + 100e −1.5t −1
)
) (−150e
Q (t) =−10(1 + 100e −1.5t −2 −1.5t )

1500e −1.5t
=
(1 + 100e −1.5t 2
)
1500e −3

After 2 weeks, Q (2) = ≈ 2.089. The infection
−3 2
(1 + 100e )
rate is 2089 people per week.
(c) The rate of infection increases when Q (t) > 0 and decreases

when Q (t) < 0. The infection rate begins to decline when

Q (t) = 0.

1500e −1.5t

Q (t) =
)
(1 + 100e −1.5t 2
d d
)
)
(1 + 100e −1.5t 2 (1500e −1.5t ) − (1500e −1.5t ) (1 + 100e −1.5t 2
dt dt

Q (t) =
(1 + 100e −1.5t 4
)
(1 + 100e −1.5t 2 −1.5t ) − (1500e −1.5t )(2)(1+100e −1.5t )(−150e −1.5t )
) (−2250e
=
)
(1 + 100e −1.5t 4
−2250(1 + 100e −1.5t )e −1.5t [(1 + 100e −1.5t ) − 200e −1.5t ]
=
(1 + 100e −1.5t 4
)
−2250(1 + 100e −1.5t )e −1.5t (1 − 100e −1.5t )
=
)
(1 + 100e −1.5t 4
−2250e −1.5t (1 − 100e −1.5t )
0 =
)
(1 + 100e −1.5t 3
Since 2250e −1.5t = 0, it follows that
1 − 100e −1.5t = 0
1 = 100e −1.5t
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