Advertising. A company is trying to expose a new product to as many people as possible through TV ads. Suppose that the rate of exposure to new people is proportional to the number of those who have not seen the product out of L possible viewers (limited growth). No one is aware of the product at the start of the campaign, and after 10 days, 33% of L are aware of the product. Mathematically, d N d t = k ( L − N ) N ( 0 ) = 0 N ( 10 ) = 0.33 L (A) Solve the differential equation. (B) How many days will it take to expose 66% of L ? (C) Graph the solution found in part (A) for 0 ≤ t ≤ 90.
Advertising. A company is trying to expose a new product to as many people as possible through TV ads. Suppose that the rate of exposure to new people is proportional to the number of those who have not seen the product out of L possible viewers (limited growth). No one is aware of the product at the start of the campaign, and after 10 days, 33% of L are aware of the product. Mathematically, d N d t = k ( L − N ) N ( 0 ) = 0 N ( 10 ) = 0.33 L (A) Solve the differential equation. (B) How many days will it take to expose 66% of L ? (C) Graph the solution found in part (A) for 0 ≤ t ≤ 90.
Advertising. A company is trying to expose a new product to as many people as possible through TV ads. Suppose that the rate of exposure to new people is proportional to the number of those who have not seen the product out of L possible viewers (limited growth). No one is aware of the product at the start of the campaign, and after 10 days, 33% of L are aware of the product. Mathematically,
d
N
d
t
=
k
(
L
−
N
)
N
(
0
)
=
0
N
(
10
)
=
0.33
L
(A) Solve the differential equation.
(B) How many days will it take to expose 66% of L?
(C) Graph the solution found in part (A) for 0 ≤ t ≤ 90.
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