A research lab grows a type of bacterium in culture in a circular region. The radius of the circle, measured in 6 centimeters, is given by r(t) = 5 - 1²2+3 where t is time measured in hours passed since a circle of a 1cm radius of the bacterium was put into the culture. Express the area, A (t), of the bacteria as a function of time, and find the approximate area of the bacterial culture in 4 hours. A A(t) B A(t) = n(5- 6 1² +3 A(t) = 2n(5- 6 t²+3 6 ⒸA(₁) = 4√(3 - 12/²+3) 5 and Area after 4 hours = 14.72, cm² and Area after 4 hours = 29.43, cm² and Area after 4 hours = 68.93, cm² 6 2 ⒸA(1) = 24 (5 - 72 043) ² D t² +3 and Area after 4 hours = 137.86, cm²

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Question
6
centimeters, is given by r(t) = 5 -
t² +3
A research lab grows a type of bacterium in culture in a circular region. The radius of the circle, measured in
where t is time measured in hours passed since a circle of a 1cm
radius of the bacterium was put into the culture. Express the area, A (t), of the bacteria as a function of
time, and find the approximate area of the bacterial culture in 4 hours.
A
A(t) = n(5-
6
1² +3
B A(t) = 2n(5-
ⒸA(t)
A(1) = x (5
-
-) and Area after 4 hours =
t²+3
6
-) and Area after 4 hours = 29.43, cm²
6 2
1² +3
and Area after 4 hours =
14.72, cm²
6 2
ⒸA(1) = 24 (5 - 7² +43) ²
D)
25
2
t
68.93, cm²
and Area after 4 hours = 137.86, cm²
Transcribed Image Text:6 centimeters, is given by r(t) = 5 - t² +3 A research lab grows a type of bacterium in culture in a circular region. The radius of the circle, measured in where t is time measured in hours passed since a circle of a 1cm radius of the bacterium was put into the culture. Express the area, A (t), of the bacteria as a function of time, and find the approximate area of the bacterial culture in 4 hours. A A(t) = n(5- 6 1² +3 B A(t) = 2n(5- ⒸA(t) A(1) = x (5 - -) and Area after 4 hours = t²+3 6 -) and Area after 4 hours = 29.43, cm² 6 2 1² +3 and Area after 4 hours = 14.72, cm² 6 2 ⒸA(1) = 24 (5 - 7² +43) ² D) 25 2 t 68.93, cm² and Area after 4 hours = 137.86, cm²
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