pollution is removed from a lake at a rate of f(t) kg/day on day t. (a) explain the meaning of the statement…

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 ∫₅¹⁵ f(t) dt = 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 ∫₅¹⁵ f(t) dt = 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

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 ∫₅¹⁵ f(t) dt = 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 ∫₅¹⁵ f(t) dt = 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

Answer

Explanation:

Step1: Interpret function - value meaning

The function $f(t)$ gives the rate of pollution - removal in kg/day on day $t$. So, when $t = 12$ and $f(12)=500$, it means on day 12, pollution is removed from the lake at a rate of 500 kg/day.

Step2: Analyze units in definite - integral

The definite integral $\int_{a}^{b}f(t)dt$ represents the net change in the amount of pollution removed. The variable $t$ has units of days, and $f(t)$ has units of kg/day. For $\int_{5}^{15}f(t)dt = 4000$, the 5 and 15 are the lower and upper limits of integration with respect to time, so their units are days. The integrand $f(t)$ has units of kg/day and $dt$ has units of days. When we integrate $\int f(t)dt$, the days in the denominator of the units of $f(t)$ and the $dt$ cancel out, so the units of the result 4000 are kg.

Step3: Interpret definite - integral meaning

The definite integral $\int_{5}^{15}f(t)dt$ represents the total amount of pollution removed from day 5 to day 15. Since $\int_{5}^{15}f(t)dt = 4000$, it means from day 5 to day 15, pollution is removed at a varying rate (given by $f(t)$) and the amount of pollution removed is 4000 kg.

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 varying rate and the amount of pollution removed is 4000 kg.