the question is:
the rate of emission of carbon pollution C, in tonnes per year from a factory from 1st January 2006 is given by:
C=500-(10/1+t)^2 where t is the time in years
i) what was the rate of emission of carbon pollution C on 1st January 2006?
ii) what was the rate of emission of carbon pollution C on 1st January 2008?
iii) what value does C approach as time passes?
iv) draw a sketch of C as a function of t?
v) calculate the total amount of carbon pollution emitted from the factory from 1st January 2006 to 1st January 2012? Answer correct to the nearest tonne.
Please help and explain this question to me - im studying for exams :(
the rate of emission of carbon pollution C, in tonnes per year from a factory from 1st January 2006 is given by:
C=500-(10/1+t)^2 where t is the time in years
i) what was the rate of emission of carbon pollution C on 1st January 2006?
ii) what was the rate of emission of carbon pollution C on 1st January 2008?
iii) what value does C approach as time passes?
iv) draw a sketch of C as a function of t?
v) calculate the total amount of carbon pollution emitted from the factory from 1st January 2006 to 1st January 2012? Answer correct to the nearest tonne.
Please help and explain this question to me - im studying for exams :(
-
C=500-(10/1+t)^2 where t is the time in years
1)
Substitute t=0 into C=500- 100/(1+t)^2
C=500-100 = 400
2)
substitute t=2 into C=500- 100/(1+t)^2
C = 500-100/9
C = 488.87
3)
As t approaches infinity 100/(1+t)^2 approaches 0
Therefore, C = 500-100/(1+t)^2 approaches 500
5)
this is a 6-year period
substitute t=6 into C=500-(10/1+t)^2
C =500 -100 /49
C = 497.96 = 498 tonnes
4) try clicking the link
1)
Substitute t=0 into C=500- 100/(1+t)^2
C=500-100 = 400
2)
substitute t=2 into C=500- 100/(1+t)^2
C = 500-100/9
C = 488.87
3)
As t approaches infinity 100/(1+t)^2 approaches 0
Therefore, C = 500-100/(1+t)^2 approaches 500
5)
this is a 6-year period
substitute t=6 into C=500-(10/1+t)^2
C =500 -100 /49
C = 497.96 = 498 tonnes
4) try clicking the link
-
5) C(6) gives rate in 2012; integrate to get tonnes carbon emitted:
= ∫(0 to 6)500 - 100/(1+t)^2 dt
= ∫(0 to 6)500 dt - ∫(0 to 6)100/(1+t)^2 dt
= 500t - 100/(1+t) * (-1)|(0 to 6)
= 500t + 100/(1+t)|(0 to 6)
= 500(6) + 100/(1+6) - 500(0) - 100/(1+0)
= 20400/7
≈ 2914 tonnes carbon emitted 2006 to 2012
= ∫(0 to 6)500 - 100/(1+t)^2 dt
= ∫(0 to 6)500 dt - ∫(0 to 6)100/(1+t)^2 dt
= 500t - 100/(1+t) * (-1)|(0 to 6)
= 500t + 100/(1+t)|(0 to 6)
= 500(6) + 100/(1+6) - 500(0) - 100/(1+0)
= 20400/7
≈ 2914 tonnes carbon emitted 2006 to 2012
Report Abuse
-
You've just got to plug in the variables. The time passed on jan. 1 2006 is 0 years. So, you plug that in for t. So the equation is C=500-(10/1+(0))^2. Then solve.