The rate at which water flows out of a pipe, in gallers per hour, is given by the differentiable function R of time t. The info below shows the rate as measured every 3 hours for a 24-hour period.
(t) hours -- R(t) gallons/hr
0 -- 9.6
3 -- 10.4
6 -- 10.8
9 -- 11.2
12 -- 11.4
15 -- 11.3
18 -- 10.7
21 -- 10.2
24 -- 9.6
a.) Use a midpint Riemann sum with 4 subdivisions of equal length to approximate (integral sign is ~) 0~24 R(t) dt. Using the correct units, how do you explain this answer in terms of water flow?
b.) Is there sometime t, 0<t<24, such that R'(t) = 0. How?
c.) The rate of water flow R(t) can be approximated by Q(t) = (1/79)(768+23t-(t^2)). Use Q(t) to approximate the average rate of water flow during the 24-hour time period.
*******I desperatly need help with this prob. My teacher is no help, and I don't know where to go for any help.
(t) hours -- R(t) gallons/hr
0 -- 9.6
3 -- 10.4
6 -- 10.8
9 -- 11.2
12 -- 11.4
15 -- 11.3
18 -- 10.7
21 -- 10.2
24 -- 9.6
a.) Use a midpint Riemann sum with 4 subdivisions of equal length to approximate (integral sign is ~) 0~24 R(t) dt. Using the correct units, how do you explain this answer in terms of water flow?
b.) Is there sometime t, 0<t<24, such that R'(t) = 0. How?
c.) The rate of water flow R(t) can be approximated by Q(t) = (1/79)(768+23t-(t^2)). Use Q(t) to approximate the average rate of water flow during the 24-hour time period.
*******I desperatly need help with this prob. My teacher is no help, and I don't know where to go for any help.