Derivatives and Integration - Modeling of Environmental Sciences | GEOG 410, Study notes of Geography

Download Derivatives and Integration - Modeling of Environmental Sciences | GEOG 410 and more Study notes Geography in PDF only on Docsity! Derivatives and Integration 1. Derivatives Example 1: Traveling from Chapel Hill to Raleigh (Distance(D)=30 miles), leaving Chapel Hill at 8:00am, arriving Raleigh City Hall at 8:45am. Therefore the speed of your travel is v=D/t = 30miles/0.75hr=40 miles/hr. This actually the average speed, you may travel at a different speed at any particular moment. If one asks what your speed is at 8:15am, we may have to figure that out in the following way: We know D=v*t, if we know the distance traveled from Chapel Hill at 8:15am, D(8:15am)=12.5miles and the distance your traveled two minute later, i.e. D(8:17am)=14.5miles, Then the distance you traveled from 8:15am to 8:17am is D(8:17am)-D(8:15am)=2miles. then the speed at 8:15am can be estimated as (D(8:17am)-D(8:15am))miles /2 min=2miles/2min=60 mph Which of the following is a more accurate estimate of speed you are driving at 8:15am? (a) 40 mph (b) 60 mph Are you sure that you are driving exactly at 60 miles/hr at 8:15am? What are the situations that you can imagine that may make your speed different from 60 mph? The actual speed at 8:15 may still be different from the above estimation, but 60 mph is a better estimation than 40 miles/hr. How can we get a more accurate estimate of your driving speed at 8:15am? The shorter the time you allow your car to travel after 8:15am, the more accurate the speed you calculate. Let t stand for time, and Δt for the time allowed for travel, the speed t for the time allowed for travel, the speed at 8:15am can be written as: Vt=8:15am ≈ t tDttD   )()( Example 2: How many of you watched a launch of a space shuttle by NASA? How fast the shuttle is traveling at the time just before it is off the launch pat? How fast the shuttle is traveling just before it reaches the orbit (18,000 miles/hour)? In order to get rid of the gravitation of Earth, an object has to travel at an accelerating speed of 7.9km/s2. If you do a plot of time and distance the shuttle is traveling, it would look like The Endeavour space shuttle was launched at 6:36pm on August 8, 2007. If I ask you how fast the shuttle was traveling at 6:37pm, how would you figure it out ? The shorter the Δt for the time allowed for travel, the speed t is, the more accurate the speed. Mathematically, t tDttD v t t     )()( lim 0 In general: If a function y=f(x) exists at x0, when x increased Δt for the time allowed for travel, the speed x at x0, i.e. x = x0+Δx, the Δt for the time allowed for travel, the speed x, the function has a corresponding increase Δt for the time allowed for travel, the speed y=f(x0+Δx, the Δt for the time allowed for travel, the speed x)-f(x0), if the limit of the ratio of Δt for the time allowed for travel, the speed y to Δt for the time allowed for travel, the speed x exists when Δt for the time allowed for travel, the speed x0, the limit is called the derivative of y=f(x) at x=x0. )( )()( 00 00 limlim0 xfx xfxxf x y y xx xx          Examples: y=f(x)=C (C=constant) miles seconds y’=(sin(u))’×u’=(sin(x2))’×(x2)’=2xcos(x2) 2) y=sin2(x) Let u=sin(x), y=sin2(x)=u2, dx du du dy dx dy    )sin(222 xuu du dy      )cos()sin( xx dx du    )cos()sin(2 xx dx du du dy dx dy  3) y=e2x+Δx, the sin(x) Let u=?, ? du dy ? dx du 4) y=e2x+Δx, the cos(2x) Integration: The inverse of derivatives: Let me ask the inverse question in Example 1 of derivatives, if I travel at 40 mph on I-40 east, where am I in 45 minutes, how far away am I from Chapel Hill? We know we traveled 30 miles in 45 min at that speed, is that sufficient to know where I am? What else do we need to know? In derivative, we can write dS/dt=40. The inverse of that is intergration, i.e.   dtdS 40 S=40t+Δx, the C Where C is a constant determined by the initial condition. e.g. (x2)’=2x, in fact (x2+Δx, the C)’=2x, where C is a constant. Cxxdx  22 Similarly we can create a table of integration: (1) Ckxkdx  (2) C n x dxx n n     1 1 , where n≠-1 (3) Cedxe xx  (4) Cxdxx  )sin()cos( (5) Cxdxx  )cos()sin( (6) Cxdx x  )ln( 1 Definite Integration: Given a function f(x) which is bounded on [a, b]. Randomly insert n points within [a, b] so that, a=x0 <x1<…<xn=b, separate the interval [a,b] into n smaller intervals, [x0, x1], [x1, x2], …, [xn-1, xn], The lengths of the invervals are, respectively, Δt for the time allowed for travel, the speed x1=x1-x0, Δt for the time allowed for travel, the speed x2=x2-x1, …, Δt for the time allowed for travel, the speed xn=xn-xn-1. Take any number εi from any interval above, calculate the product, f(εi)Δt for the time allowed for travel, the speed xi, and sum the product, i n i i xfS   )( 1  Let λ be the maximum length of the n intervals, if λ0, regardless of how [a, b] is separated, and how εi is taken from the interval [xi-1, xi], S is always approach a finite limit. The limit is the definite integration of f(x) on the interval [a, b]. Newton-Leibniz Formula: )()()( aFbFdxxf b a  , where f(x)=F’(x)  b a dxxf )( is the area under the curve from a to b. Examples: 1)  2/ 0 )sin(  dxx 2)   1 0 dxe x 3)  3 2 2dxx b y=f(x) ε i Δt for the time allowed for travel, the speed x i f(ε i ) a