Watson is filling a pool with water. The flow rate r(t) gives the gallons per second at which the...

Question:

Watson is filling a pool with water. The flow rate {eq}r(t) {/eq} gives the gallons per second at which the water is flowing into the pool at time {eq}t {/eq} seconds after he turns on the faucet. We are going to make a new function, {eq}\displaystyle f (\hat t) = \int_{t = 5}^{\hat t} r(t)dt {/eq}. First, let's try to figure out why this is even a function.

(a) What does the function {eq}f {/eq} input? .

(b) What does the function {eq}f {/eq} output? .

(c) Suppose {eq}f(10) = 26 {/eq} and {eq}f(18) = 26 {/eq}. What would account for this?

Integration:

In this question, the function f integrates the rate of change over the desired time interval. What happens due to this integration is that the rate of change, which gives the amount of water added to the pool at any moment of time, gets summed up over the entire interval we are interested in.

a) The function f inputs the value of t between 5 and some value {eq}\hat t {/eq} into the function r(t).

b) The function f outputs the amount of water in gallons that is present in the pool after {eq}\hat t {/eq} seconds.

c) The value of the function will remain the same if no water is filled between 10th and the 18th seconds. This would also mean that rate of change r is zero between 10 and 18, and therefore, does not add any values to the function f. 