Post by: Sanguinne on May 19, 2013, 05:34:06 pm
exercise 9J
The rate of flow of water into a hot water system during a 12h period is thought to be dv/dt = 10 + cos(t x pi/4), where V is in litres and t is the number of hours after 8 am.
a) sketch the graph of dv/dt against t
b) find the length for which the rate is above 10.5L/min
im having difficulty with part b. my answer is 9 hours and 20 minutes but the txtbook's answer is 4h. Can someone find out where i went wrong???
Post by: Alwin on May 19, 2013, 05:43:13 pm
q7
exercise 9J
The rate of flow of water into a hot water system during a 12h period is thought to be dv/dt = 10 + cos(t x pi/4), where V is in litres and t is the number of hours after 8 am.
a) sketch the graph of dv/dt against t
b) find the length for which the rate is above 10.5L/min
im having difficulty with part b. my answer is 9 hours and 20 minutes but the txtbook's answer is 4h. Can someone find out where i went wrong???
explain to me please why you called this topic "stupid integration" ? where does it ask to integrate..
Post by: BubbleWrapMan on May 19, 2013, 05:45:29 pm
The total length of time is thus
Also, what Alwin said.
Post by: Alwin on May 19, 2013, 05:48:38 pm
Clearly, after solving 10 + cos(t pi/4) = 10.5 where t from 0 to 4 you get t= 4/3, 20/3 and 28/3
Hence, dv/dt > 10.5 for 4/3 + (28/3 - 20/3) = 4 hours as required.
EDIT: Damn! shouldn't have bothered putting in the pretty picture, beaten but posting anyways :3
Post by: Sanguinne on May 19, 2013, 06:14:22 pm
this whole time i thought the equation was 10 + cos(pi x t/2) but it was actually divided by 4....
well thanks for the help :)