stupid integration

[Sanguinne]

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???

[Alwin]

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..

[BubbleWrapMan]

when from the graph.

The total length of time is thus .

Also, what Alwin said.

[Alwin]

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

[Sanguinne]

wow....

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