Question

Applied UMB RS LDC) 1. A cylindrical hot tub is being filled by a hose that pours out water at a constant rate of 0.25m/min at 8:00 a.m. The volume of water in the pool was 8m and the radius of the hot tub is 2m. a) Determine the average rate of change with respect to its height between 9:00 a.m. and 10:00 a.m. b) Find the instantaneous rate of change at 10:00 a.m.

Answer 1

Answer :-

A cylindical hot tub is being filled by a hose

Given that Rate of water = 0.25 m3/min Amount of water poured in 1 hour = 0.25X60 215 m² Ab 8.00 am solumn of water in the - pool = 8m3 of tub is am Radine volumne at 9:00 am = column of wates at 8:00 + Amount of wates poured in 1 hour =8+5=23m² Next hows ie volumne of water at 10:00 am

= amount of water ab 9:00+ Arneusit of water poured in a hown we know that radius of tub=2m We have volumn of culinder formulae 8= radius h= height Both the pasametexas h and v Oncrease with time and are functions of time.

calculating Height (h) at 9:00 am h =- 23 2 40 47 V is the volumne at 9:00 am calculating Height (m 2 at 10:00 am har tror = 32 ha 38 40 40 V is the volunn at 10:00 am (a) Average rate of change with relpect to its height between 9:00am and 10:00 am volumn at 10:00 am – volumn ab 9.00am hi-ha

38-22 = 1 = 40 (6) Instantaneous rake of change at 10 recall v= noh dr =no we r=2 (sadine) which is constant dv = 45 at any time. an Thue the instantancous sake of change of volumne = ar = 45 an