Integrals and rate word problem: a tank of heating oil

[xtrmk]

A tank contains 125 gallons of heating oil at time t = 0. During the time interval 0 <t<12 hours, heating oil is pumped into the tank at a rate
h(t) = 2 + [10 / (1+ln(t+1)] gallons per hour. During the same interval, heating oil is removed from the tank at rate R(t) = 12sin(t^2/47) gallons per hour.

b) is the level of heating oil rising or falling at t=6?
c) how many gallons of heating oil are in the tank at t=12
d) at what time is the volume of heating oil the least?

can someone guide me or tip me on how to start.
for b would I just compare the values of h(6) and r(6) ?

for c would i do the integral of h(t) - the integral of r(t) ?

help on d

any help is appreciated

 

[skeeter]

\(\displaystyle \L V(t) = 125 + \int_0^t h(x) - R(x) dx\)

\(\displaystyle \L \frac{dV}{dt} = h(t) - R(t)\)

the sign of dV/dt will tell you whether the oil level is rising (increasing) or falling (decreasing).

(b) determine the sign of dV/dt at t = 6.

(c) calculate V(12)

(d) graph dV/dt ... a minimum occurs when dV/dt changes sign from negative to positive.

 

[xtrmk]

\(\displaystyle \L V(t) = 125 + \int_0^t h(x) - R(x) dx\)

\(\displaystyle \L \frac{dV}{dt} = h(t) - R(t)\)

the sign of dV/dt will tell you whether the oil level is rising (increasing) or falling (decreasing).

this is just to clarify but is that the integral of h(t) minus the integral of r(t) or the integral of h(t) minus r(t)?

 

[skeeter]

there is no difference. you should already know this property of integrals ...

\(\displaystyle \L \int_a^b f(x) - g(x) dx = \int_a^b f(x) dx - \int_a^b g(x) dx\)