Question

pollution is removed from a lake at a rate of ( f(t)) kg/day on day ( t).
(a) explain the meaning of the statement ( f(12)=500).
on day choose one pollution is removed from the lake
at a rate of choose one kg/day.
(b) if ( int_{5}^{15} f(t) d t=4000), give the units of the 5, the 15, and the 4000.
5 choose one 15 choose one 4000 choose one
(c) give the meaning of ( int_{5}^{15} f(t) d t=4000).
from day 5 to day 15, pollution is removed at a rate of
choose one and the amount of pollution removed is
choose one

Explanation:

• Part (a):
• The function \(f(t)\) represents the rate of pollution removal. When \(t = 12\), \(f(12)=500\). Here, \(t\) represents the day. So, the input of the function \(t = 12\) means the 12th day. The output of the function \(y=f(t)\) gives the rate of pollution removal. So, \(500\) is the rate of pollution removal in \(kg/day\).
• Part (b):
• For the definite integral \(\int_{a}^{b}f(t)dt\), the variable of integration \(t\) has units. The limits of integration \(a\) and \(b\) (in this case \(a = 5\) and \(b=15\)) have the same units as the variable of integration \(t\). Since \(t\) represents the day, \(5\) and \(15\) have units of days.
• The integrand \(f(t)\) has units of \(kg/day\), and \(dt\) has units of days. Using the formula for the units of a definite integral \(\text{Units of}\int_{a}^{b}f(t)dt=\text{Units of}(f(t))\times\text{Units of}(dt)\). So, \(\text{Units of}(kg/day\times day)=kg\). Thus, \(4000\) has units of \(kg\).
• Part (c):
• The definite integral \(\int_{a}^{b}f(t)dt\) represents the net - change. The integrand \(f(t)\) is the rate of pollution removal (\(kg/day\)). The definite integral \(\int_{5}^{15}f(t)dt\) gives the total amount of pollution removed. The units of \(f(t)\) is \(kg/day\) (rate) and the value of the definite integral \(\int_{5}^{15}f(t)dt = 4000\) has units of \(kg\) (total amount).

Answer:

• (a) On day \(12\) pollution is removed from the lake at a rate of \(500\) \(kg/day\).
• (b) \(5\): days; \(15\): days; \(4000\): \(kg\).
• (c) From day \(5\) to day \(15\), pollution is removed at a rate of \(kg/day\) and the amount of pollution removed is \(4000\) \(kg\).