1. ## calc II problem

Can anyone help me with this problem?

a) The oil consumption rate satisfies the equation C(t) = C0e^rt, where C0 is the consumption rate at t=0 (the number of barrels per year) and r is a constant. If the consumption rate is C0 = 2.5x10^10 barrels per year in 1976 and r= 0.06 how long will it take before 2x10^12 barrels are used up?

b) As the fuel is almost used up, the prices will probably skyrocket and other sources of energy will be turned to. Let S(t) be the supply left at time t. Assume that S= ds/dt = - (alpha)S, where (alpha) is a constant. Find S(t).

2. Originally Posted by ewander
Can anyone help me with this problem?

a) The oil consumption rate satisfies the equation C(t) = C0e^rt, where C0 is the consumption rate at t=0 (the number of barrels per year) and r is a constant. If the consumption rate is C0 = 2.5x10^10 barrels per year in 1976 and r= 0.06 how long will it take before 2x10^12 barrels are used up?
$C(t) = C_0e^{rt}$

from the problem $C_0 = 2.5\times 10^{10}$ and $r= 0.06$ so

$C(t) = 2.5\times 10^{10}\times e^{0.06t}$

We need to find t for when $C(t) = 2\times 10^{12}$ so we sub this into what we have and solve for t.

$2\times 10^{12}= 2.5\times 10^{10}\times e^{0.06t}$

$\frac{2\times 10^{12}}{2.5\times 10^{10}}= e^{0.06t}$

$ln\left(\frac{2\times 10^{12}}{2.5\times 10^{10}}\right) = 0.06t$

Can you finish it from here?

3. Originally Posted by ewander

b) As the fuel is almost used up, the prices will probably skyrocket and other sources of energy will be turned to. Let S(t) be the supply left at time t. Assume that S= ds/dt = - (alpha)S, where (alpha) is a constant. Find S(t).
$\frac{ds}{dt}=-\alpha \times s$

First of all separate the variables

$\frac{ds}{s}= -\alpha dt$

Now integrate both sides

$\ln(s)=-\alpha t+c$

You now have no more derivatives, can you get $s$ on its own?

4. a) ln(100) = 0.06t
t=ln(100)/0.06 = 77
Thanks!!

but b) ?????

5. Originally Posted by ewander

but b) ?????
$\frac{ds}{dt}=-\alpha \times s$

First of all separate the variables

$\frac{ds}{s}= -\alpha dt$

Now integrate both sides

$\ln(s)=-\alpha t+c$

$s(t)=e^{-\alpha t+c}$

any clearer...?