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# Lecture1 - M Calculus II for Engineering Lecture Notes...

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Unformatted text preview: M Calculus II for Engineering Lecture Notes David Hamsworth Department of Applied Mathematics University of Waterloo 2009 Chapter 1: Approximation Methods §1.l: Introduction In Math 117 you were introduced to the basic tools of calculus, and you have already begun to see that there are some problems which we simply cannot solve exactly (recall that the integral J‘sin(x2 )dx has no antideiivative in terms of the functions we commonly use). Unfortunately, these problems are not unusual, and so we often find that the only option available to us is to seek an approximate solution. We have two basic strategies for finding these approximate solutions: numerical methods and analytical methods. Numerical methods generally use our mathematical definitions to compute If approximations by “brute force”; for example to approximate Isin(x2ylx we could refer to the 0 definition of the definite integral, and calculate the area of n rectangles of width Z and heights n determined by the function sin(x2). Analytical methods, on the other hand, use the theory of calculus (also known as analysis) to recognize reasonable approximations for the functions involved. As an example, we could argue that if x is small, then sinx z x (and sin(x2) z x2 ), so Isin(x2)dx = Ixzdx, which we can evaluate easily. Each of the two approaches has its own strengths and weaknesses. With the advances in computer technology over the last several decades the numerical approach has become extremely powerful, but analytical methods still have their place: 0 Analytical methods allow us to find approximations for functions, without specifying the numerical values involved. In a numerical approach, we must usually assign values to all of the parameters involved, whether we know what they should be or not. 0 Analytical methods can often be used to check that our numerical methods are giving realistic results. In these notes you will find a brief introduction to some common numerical methods, but the course will concentrate on the analytic approach. Specifically, we’ll be looking at how polynomials can be used as approximations for more complicated functions. §1.2 Our Simplest Option: the Linear Approximation (aka. the Tangent Line Approximation, or Linearization) Recall the definition of the derivative at a point from Math 117: f ’(a)is defined as ”man—m 3““: x—a x=a: . Graphically, we interpret this as the slope of the tangent line to y = f (x) at For values of xnear a, t e tangent line gives a reasonable approximation to f (x) : Ll ad 51th) What is L(x) ? Recall: the line through (0,3)) with slope 100 has equation (y —-b) = f’(a)(x— a) , i.e. HI) = y = fta)+ f’(a)(x-a). In lip-‘0. main Lint] m ““30 . This can be helpful when the function f (x) is easy to evaluate at x = a but difficult to work with at points nearby. mm Consider f(x)=J§.We know f(4)=2 ,but what’s f(3.98)? - It’s hard to evaluate 113.98 exactly (without a calculator), but the equation of the tangent line is simpIer: Le) = f(4)+f’(4)(X'4) = 2 + Eu - 4) -> 1.0.98) = 2 + 11190.02) = 2 — 0.005 = 1.995 We conclude that J3.98 = 1.995 . This should look familiar. We introduced the same concept in Math 117 with the notation of differentials: f(a+Ax) = f (a) +Af , where Af = f ’(a)Ax: In our example we identify a = 4 . Ax = «0.02 , and then Af =%(—0,02) = —0.005, and f(3.98) 9-: 2 — 0.005 = 1.995. Example: A particularly common example of a linear approximation is the approximation sin x = x when x is near zero. f(x)=sinx f(0)=0 :L(x)=0+1(x—0) l f’(x)=con f’(0)=1 =x Exam The “simple" pendulum problem. Consider the motion of a mass m suspended by a rod of length I from a fixed point. In developing a mathematical model of a physical problem, we’ll need some preliminary approximations, before we I I even get to the mathematics! Here we’ll assume the mass to be '9 concentrated at a point, the rod to be 1 massless and rigid, and we'll ignore I both friction and air resistance; we’ll hﬁ consider gravity as the only force. 5 If we let 30) be the displacement along the are (positive when 6‘ is positive), then Newton's Second Law of Motion (F = ma) gives us d2: _ m— = —m sm 6'. at:2 g _ dzs r1219 . We have too many vanables here, but we know that s = l6, so F =! d1: , which 2 means that ml Elm—:9 = —mg sin 6', 2 i.e. “1—3: —£sina. dr 1 If we specify our initial conditions (say 6(0) = 60 and 9(0) = 0) then our model is complete. Unfortunately, we can’t solve it... not even with Math 211 techniques! At this point we need an approximation. We know that if the amplitude of the oscillations remains small 0.6. as long as 19 is close to zero), then the equation above is approximately equivalent to the equation 2 d 9 = _i ‘9’ an2 t which is so much simpler that we can guess the solution for the initial conditions we’ve speciﬁed! (we‘re looking for a function whose second derivative is just 4 i times itelf ...... this must be a combination of cosines or sines of 1’%I, and to 1 match our initial conditions it must just be cos( g/l )). Example: The resistivity of a wire of a certain metal depends on the temperature according to the equation p(t) = pzoe“(“2°’ Ohm-meters, where ,920 is the resistivity of the metal at 20°C and a is a constant (the “temperature coefficient”). This is often replaced with its linear approximation at t = 20 : arc—20) ’ so €00) : p20 )0 (20) = apzo This is easier to calculate, and gives excellent approximations for temperatures << 1 a For example, for copper wire, a = 0.0039 /°C (so a 2 256°C), and p20 2 1.7 X10"8 0- m. At 40°C, then, p z 1.7x10'8 +(0.0039)(1.7><10'8)(20) =1.8x10‘8 Q—m, which matches the precise value to our two significant digits. p’(t) = apzoe }=> L(t) 2 p20 + apzo (t — 20) . ...
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Lecture1 - M Calculus II for Engineering Lecture Notes...

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