<p>It is estimated tat at the current rate of consumption rgallons per year, the oil supply of the earth will last 200 years. However, the rate of consumption, R(t) is increasing at the rate of 5% per year; that is dR/dt = .05R.</p>
<p>a.) In terms of r, how many gallons of oil are currently available?</p>
<p>b.) Use the diffential equation to find R(t).</p>
<p>c.) If no additional oil is discovered, how long, to the nearest year, will the current oil supply actually last? Show how you arrived at your solution.</p>
<p>eeh i'm lazy so i'll just post the thought process</p>
<p>a)
r*200=200r gallons</p>
<p>b)
integrate dR/dt and u get R=Ce^t/.05, then evaluate at (0, 200r)</p>
<p>but this will produce an equation in terms of t and r...is that wut the question is askin for?</p>
<p>c)</p>
<p>find zero of the equation found in part b</p>
<p>any correction?</p>
<p>I dont know but this is an actual AP question and this is how it was worded. This is what i did</p>
<p>dR/dt = .05R
ln|.05R| = t + C
C * e ^ T = .05R
R(t) = Ce^t /.05
R(t) = Ce ^ t (Can the C obsorb that dividing by .05... I think so but im not sure.)</p>
<p>This equation gives the rateat any given time so wouldnt we have to integrate again for part c to find out the oil at any given time. Please help</p>
<p>a) yeah let x be the current supply, then x-200r(0)=0 x=200r(0)</p>
<p>b)dR/dt=.05R
dR/R=.05dt
ln|R|=.05t+c
R=Ke^(.05t)</p>
<p>we know that x = 200R(0) so R(0) = x/200. Plug these in to solve for k.</p>
<p>x/200 = Ke^(0)
x/200=k</p>
<p>So R=(x/200)e^(.05t)</p>
<p>c) {integral from 0 to t of [(x/200)e^(.05a) da]} = x</p>
<p>Remember (x/200) is just a constant, and i picked "a" randomly as the axis of integration.
Solve using fundamental theorem of calculus. hope this helped..(and is right)</p>
<p>^^ i don't get what u did in part c</p>
<p>the integral gives u the amount consumed from 0 to t years, i don;t think that's wut the question;s askin tho</p>
<p>we're solving for the value of t that makes the amount consumed from 0 to t years equal to x.(Remember the amount consumed is the integral of R(t)) If you evaluate that integral the x's will cancel I believe and you can solve for the value of t.</p>
<p>ah that makes sense now...i didn't realize u were solvin for t instead of x :)</p>