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Mathematics in the Earth Sciences

by: Jillian Moore

Mathematics in the Earth Sciences EART 111

Marketplace > University of California - Santa Cruz > Earth Sciences > EART 111 > Mathematics in the Earth Sciences
Jillian Moore
GPA 3.78


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This 2 page Class Notes was uploaded by Jillian Moore on Monday September 7, 2015. The Class Notes belongs to EART 111 at University of California - Santa Cruz taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/182200/eart-111-university-of-california-santa-cruz in Earth Sciences at University of California - Santa Cruz.


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Date Created: 09/07/15
ES 111 Mathematical Methods in the Earth Sciences Lecture Outline 15 Mon 17th Nov 708 First Order Differential Equations Part I Integrating Factor Here is a powerful technique which will work only for linear rst order ordinary differential equations If a rst order equation is not separable see Lecture 14 then this technique is the next one to try Any such equation can be written in the so called Standard Form 11 New W The trick to solving this equation is to multiply both sides by an integrating factor wr where w 6fpltmgtdm Doing so allows us to rewrite the Standard Form as i Wham 1mg paw dx which we can always solve This looks more formidable than it really is see example below Its a very powerful tool7 but beware it only works on linear rst order ODEs Example 1 Let7s solve our ooids problem from last time dr Eaibr Example 2 Now lets solve a more complicated version of the ooid problem Note that you cant solve this example just by collecting all the r7s on one side and all the t7s on the other d CT ae b Z 7 btr Does the answer make physical sense Example 3 Solve CL 2 dx 4zy 2x Example The deposition of sand grains in a uid Newton7s second law relates force F to the rate of change of momentum F m where m4 and t are mass velocity and time respectively and the second equality is obtained by assuming that mass of a sand grain is constant There are two forces acting on a particle settling out of suspension the buoyancy force B which acts downwards and the uid drag force D which acts upwards The buoyancy force is just Amg where Am is the excess mass of the particle and g is the acceleration due to gravity note that this quantity is constant The uid drag force however is proportional to and in the opposite direction to the velocity of the falling particle D 71w So combining these equations we get d1 m7 Am 7 k1 dt 9 We can rewrite this in Standard Form k Am 1 in 79 m m and solve using the method outlined above In this case the integrating factor w is given by and the general solution is vt ce ktm Assuming that the particle was initially motionless an initial condition of 00 0 we obtain the particular solution 11t lt1 7 Kidm As useful check on physical problems like this is to make sure that the units are consistent What does this solution mean physically If the drag force were not important the velocity would increase continually due to the force of gravity However the drag force becomes more important and at some point the particle stops


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