Question #158148

An environmental study of a certain community suggests that the average daily level of pollution in the air will be Q(p) = √ 0.6p + 20 units when the population is p thousand. It is estimated that after t years the population will be p(t) = 9 + 0.5t 2 thousand.

(a) Express the level of pollution in the air as a function of time.

(b) compute the level of pollution after 5 years from now.

(c) When will the pollution level reach 10 units? 


A certain bacterium grows in culture in a circular region. The radius of the circle, measured in centimeters, is given by r(t) = 6 − 5 (t 2+1) , where t is time measured in hours since a circle of a 1 cm radius of the bacterium was put into the culture.

(a) Express the area of the bacteria as a function of time.

(b) Find the exact and approximate area of the bacterial culture in 2 hours.

(c) Express the circumference of the bacteria as a function of time.

(d) After how long the circumference of the bacteria will be 10π cm.


1
Expert's answer
2021-01-26T01:29:00-0500

1)

a) Q(t)=0.6×(9+0.5×t2)+20Q(t)=\sqrt{0.6\times (9+0.5\times t^2)+20}

b) Q(5))=0.6×(9+0.5×52)+205.74Q(5) )=\sqrt{0.6\times (9+0.5\times 5^2)+20}\approx 5.74

c) 10=0.6×(9+0.5×t2)+2010=\sqrt{0.6\times (9+0.5\times t^2)+20} , therefore t=100200.690.515.77t=\sqrt{\frac{\frac{100-20}{0.6}-9}{0.5}}\approx15.77

2)

 a) S=π×r(t)2=π×(65(t2+1))2S=\pi\times r(t)^2 =\pi\times(6-\frac{5}{(t^2+1)})^2

b) S=π×(65(t2+1))2=15.7079616S=\pi\times(6-\frac{5}{(t^2+1)})^2=15.70796\approx 16

c) L=2×π×r(t)=2×π×(65(t2+1))L=2\times\pi\times r(t) =2\times\pi\times(6-\frac{5}{(t^2+1)})

d) 10×π=2×π×(65(t2+1))10\times\pi=2\times\pi\times(6-\frac{5}{(t^2+1)}) , therefore t=5561=2t=\sqrt{\frac{-5}{5-6}-1}=2

 


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