Hello and greetings everyone. 1. A quantity of oil is dropped into water. When the oil hits the water it spreads out as a circle. The radius of the circle is r cm after t seconds and when t = 3 the radius of the circle is increasing at the rate of 0.5 centimetres per second One observer believes that the radius increases at a rate which is proportional to [tex]\frac{1}{t+1}[/tex] i) Write down the a differential equation for this situation, using k as a constant of proportionality. ii) Show that k = 2 iii) Calculate the radius of the circle after 10 seconds according to this model Another observer belives that the rate of increase of the the radius of the circle is proportional to [tex]\frac{1}{(t+1)(t+2)}[/tex] iv) Write down a new differential equation for this new situation. Using the same initial conditions as before, find the the new value for the constant v) Hence solve the differential equation vi) Calculate the radius of the circle after 10 seconds according to this model. 2. ? 3. The attempt at a solution I'm pretty sure I can do the first two, i am getting stuck on part iii) however. I may beable to do the rest by myself after this is sorted out, but thought i would post the full Question just incase. i) [tex]\frac{dr}{dt}=\frac{k}{t+1}[/tex] ii) 0.5 = k/(3+1) so, k = 2 iii) [tex]\frac{dr}{dt}=\frac{2}{t+1}[/tex] so.... [tex]\frac{dr}{dt}=\frac{2}{11}[/tex] ?? Then i really dont have a clue what to do. I dont know how to separate the r from the dr/dt to find the radius form this relationship. Thanks in advance. Adam.
in order to solve for the radius you need to solve the differential equation. try separation of variables.
Good! No, that's dr/dt when t= 10. As the problem says, you need to solve the differential equation. If dr/dt= 2/(t+1) then dr= (2/(t+1))dt. It's that easy!
Thanks. I had a mental block remembering that to solve differential equations you have to integrate (kind of fundamental, i know!) so... dr= (2/(t+1))dt => r = 2 ln (t+1) + c t=0, r=0 => c=0 When t = 10, r = 2ln11 = (approx) 4.796 That agrees with the answer in the book. cheers.