1 00:00:03,350 --> 00:00:09,860 N2O5 is a reagent that was used synthesize explosives like TNT. Unfortunately, N2O5 decomposes 2 00:00:09,860 --> 00:00:14,330 relatively quickly at or above room temperature, so if you accidentally leave a bottle of it 3 00:00:14,330 --> 00:00:18,430 out on the counter, you could be in trouble! In this video, we'll approximate a solution 4 00:00:18,430 --> 00:00:22,870 to the decomposition rate equation to figure out whether a batch of N2O5 that you left 5 00:00:22,870 --> 00:00:28,400 at room temperature can still be used. We'll also determine under what conditions our approximation 6 00:00:28,400 --> 00:00:30,260 is valid. 7 00:00:30,260 --> 00:00:33,530 This video is part of the Linearity video series. 8 00:00:33,530 --> 00:00:38,640 Many complex systems are modeled or approximated linearly because of the mathematical advantages. 9 00:00:38,640 --> 00:00:44,769 Hi, my name is Ben Brubaker, and I'm a professor in the Department of Mathematics at MIT. 10 00:00:44,769 --> 00:00:49,889 Today we'll be talking about linear approximations. In mathematical terms, this is just another 11 00:00:49,889 --> 00:00:55,210 name for the tangent line to a function. But the name suggests more. Linear approximations 12 00:00:55,210 --> 00:01:01,260 can be used to simplify mathematical models that are not analytically solvable. The approximated 13 00:01:01,260 --> 00:01:06,500 model will have a solution that is only acceptable under suitable conditions. However, it can 14 00:01:06,500 --> 00:01:12,439 still illuminate the behavior of the system within a certain acceptable range. 15 00:01:12,439 --> 00:01:16,280 Before watching this video, you should know the definition of the derivative, and how 16 00:01:16,280 --> 00:01:21,780 to write the equation of a line with a given slope that passes through a given point. 17 00:01:21,780 --> 00:01:25,719 After watching this video, you should be able to recognize the linear approximation of a 18 00:01:25,719 --> 00:01:31,280 function as the tangent line to the function, apply linear approximations to solve simple 19 00:01:31,280 --> 00:01:36,710 differential equations, and explain the limitations of linear approximations both mathematically 20 00:01:36,710 --> 00:01:39,380 and graphically. 21 00:01:39,380 --> 00:01:46,009 Let's begin by defining the linear approximation. Recall that if a function is differentiable 22 00:01:46,009 --> 00:01:50,700 at a point c, then when we zoom in on the point c, the function begins to look more 23 00:01:50,700 --> 00:01:55,259 and more like a line. This only works when the function is "smooth"—it doesn't have 24 00:01:55,259 --> 00:02:02,100 any kinks, corners, or discontinuities. Given a function f(x), which is differentiable at 25 00:02:02,100 --> 00:02:07,060 the point c, we define the linear approximation to be the tangent line to the function at 26 00:02:07,060 --> 00:02:14,060 c. This is a line whose slope equals f'(c). 27 00:02:14,620 --> 00:02:21,300 As an example, let's look at the following cubic equation: f(x) = one thirtieth x times 28 00:02:21,300 --> 00:02:27,800 x minus 2 times x plus 5. This function is differentiable everywhere, and in particular 29 00:02:27,800 --> 00:02:34,520 is differentiable at the point x=3. Find the equation of the tangent line to this cubic 30 00:02:34,520 --> 00:02:41,520 equation at x=3 for yourself. 31 00:02:43,430 --> 00:02:47,700 We've also found the equation for the tangent line. To do this, we found the value of the 32 00:02:47,700 --> 00:02:53,290 function at 3, which we found to be 4/5. And then we computed the derivative and evaluated 33 00:02:53,290 --> 00:03:00,290 it at x=3, which we found to be 7/6. We'll write the tangent line as T sub 3 of (x) to 34 00:03:00,540 --> 00:03:06,220 remind us that we are finding the tangent to our function at the point 3. Then the equation 35 00:03:06,220 --> 00:03:13,220 for this line can be written as T sub 3 of x equals 4/5 + 7/6 times (x-3). 36 00:03:16,010 --> 00:03:21,170 Let's take a look at the graphs of f(x) and T sub 3 of x. The function f(x) is graphed 37 00:03:21,170 --> 00:03:27,090 in red. The further we zoom into the graph at x=3, the more it begins to resemble a straight 38 00:03:27,090 --> 00:03:31,500 line. The slope of the line approaches the value of the slope of the tangent line at 39 00:03:31,500 --> 00:03:35,570 3, which is drawn in blue. 40 00:03:35,570 --> 00:03:39,390 The tangent line certainly seems to be a good approximation to our function when we are 41 00:03:39,390 --> 00:03:45,650 close to x=3. But what about when we zoom out? Is it still a good approximation? Let's 42 00:03:45,650 --> 00:03:51,420 give a mathematical justification that the tangent line is a good approximation. Our 43 00:03:51,420 --> 00:03:56,820 function is differentiable at a point, say x=3. This is equivalent to the statement that 44 00:03:56,820 --> 00:04:02,280 when x is near 3, the slope of the secant line is approximately equal to the slope of 45 00:04:02,280 --> 00:04:09,280 the tangent line. That is, for x near 3 f(x)-f(3) over x-3, is approximately equal to the value 46 00:04:12,200 --> 00:04:14,980 of the derivative at 3. 47 00:04:14,980 --> 00:04:20,339 In order to see why these statements are equivalent, we need explain what we mean by "approximately 48 00:04:20,339 --> 00:04:25,310 equal to". This means that we can make the difference between these two sides as small 49 00:04:25,310 --> 00:04:32,310 as any error bound we choose, provided that we choose x close enough to 3. 50 00:04:33,370 --> 00:04:39,940 And this is exactly the definition of differentiability at the point 3. 51 00:04:39,940 --> 00:04:46,810 To relate this to linear approximations, we use a bit of algebra. First we multiply both 52 00:04:46,810 --> 00:04:53,810 sides by (x-3), and we get f(x) -- f(3) approximately equal to f'(3)(x-3). Adding f(3) to both sides, 53 00:04:59,040 --> 00:05:06,040 we get that f(x) is approximately equal to f(3)+f'(3)(x-3). And this right hand side 54 00:05:08,210 --> 00:05:14,460 is the equation for the tangent line at 3. So indeed the tangent line closely approximates 55 00:05:14,460 --> 00:05:19,760 the function as long as x is close to 3. 56 00:05:19,760 --> 00:05:25,010 Here we see the tangent line drawn in blue, the function drawn in red. The graphs demonstrate 57 00:05:25,010 --> 00:05:31,330 our mathematical proof that the tangent line is a good approximation near x=3, but the 58 00:05:31,330 --> 00:05:35,670 further we get from x=3 the worse the approximation seems. 59 00:05:35,670 --> 00:05:42,590 Now you might see that the function and tangent line intersect at x=-9. Is the tangent line 60 00:05:42,590 --> 00:05:47,990 a good approximation for the function near x=-9? 61 00:05:47,990 --> 00:05:54,990 No, it isn't. If we move a small distance along the tangent line away from x=-9, this 62 00:05:57,020 --> 00:06:03,620 does not model the behavior of the function near x=-9. This property is extremely important 63 00:06:03,620 --> 00:06:09,010 in applications—no measurement or observed quantity is ever given exactly. 64 00:06:09,010 --> 00:06:14,550 A1 The linearization of a function that is differentiable at x=c is just another name 65 00:06:14,550 --> 00:06:17,590 for the tangent line through c. 66 00:06:17,590 --> 00:06:22,220 The A2 key properties of this tangent line are that it shares the same value as the function 67 00:06:22,220 --> 00:06:27,639 and the same first derivative of the function at x=c. And A3 we've seen through both graphical 68 00:06:27,639 --> 00:06:33,060 intuition and the definition of the derivative that this linearization closely approximates 69 00:06:33,060 --> 00:06:37,520 the function for points x sufficiently close to c. 70 00:06:37,520 --> 00:06:44,210 Let's see how we can apply a linear approximation to simplify a problem. Let's suppose that 71 00:06:44,210 --> 00:06:49,060 you are participating in undergraduate research in a chemistry lab. You have been using a 72 00:06:49,060 --> 00:06:56,060 .01 molar solution of N2O5. You bring it out of the refrigerator at 9:00 Monday morning, 73 00:06:56,270 --> 00:07:02,020 and promptly forget about it, leaving it on the counter in the lab, which is 25 degrees 74 00:07:02,020 --> 00:07:09,020 Celsius. When you realize you left the solution out, 1 hour has passed. You panic. 75 00:07:09,169 --> 00:07:16,169 The problem is that N205 decomposes into NO2 and O2 at room temperature. If the molarity 76 00:07:16,510 --> 00:07:22,419 of the solution has changed significantly, it might ruin your experiments. Experiments 77 00:07:22,419 --> 00:07:27,940 have shown that the rate of decomposition follows first order kinetics. This means that 78 00:07:27,940 --> 00:07:34,190 the instantaneous rate of change in concentration of N205 is proportional to the concentration 79 00:07:34,190 --> 00:07:41,190 of N205. The constant of proportionality, k, has been found experimentally to be 1.72 80 00:07:43,010 --> 00:07:48,540 times ten to the negative 5 inverse seconds. 81 00:07:48,540 --> 00:07:52,990 Because you are panicking, you are having a hard time solving this differential equation. 82 00:07:52,990 --> 00:07:59,990 Instead, find the _approximate_ decomposition in the N205 solution after 1 hour using a 83 00:08:00,340 --> 00:08:07,150 linear approximation at time t=0. Note that because the reaction constant has units of 84 00:08:07,150 --> 00:08:14,150 inverse seconds, the variable t must have units of seconds. 85 00:08:16,110 --> 00:08:22,770 In order to produce a linear approximation at t=0, we need two ingredients: the concentration 86 00:08:22,770 --> 00:08:28,760 at time 0, and the value of the derivative at time 0. 87 00:08:28,760 --> 00:08:34,870 The initial concentration was .01 molar. To find the value of the derivative, we use our 88 00:08:34,870 --> 00:08:40,719 differential equation, and we find that the value of the derivative at time t=0 is just 89 00:08:40,719 --> 00:08:47,719 minus k times the initial concentration at t=0, or negative k times .01 molar. Remember 90 00:08:50,540 --> 00:08:56,019 that k has units of inverse seconds. 91 00:08:56,019 --> 00:09:00,470 Putting this together into the formula for the tangent line at zero, we see that the 92 00:09:00,470 --> 00:09:07,470 linear approximation, is given by T0(t)= (.01 molar) times 1 minus $kt$. 93 00:09:12,420 --> 00:09:19,420 Now we are interested in approximating the value at 1 hour, or 3600 seconds. Plugging 94 00:09:19,649 --> 00:09:26,649 in the value for k, we find that to two significant figures T0(3600)=.0094 Molar. So the molarity 95 00:09:30,730 --> 00:09:33,860 is not so different. Phew!! 96 00:09:33,860 --> 00:09:38,220 Now you are feeling relieved, so when you look back at the reaction rate, you realize 97 00:09:38,220 --> 00:09:43,170 that in fact, a first order reaction rate is telling you that the concentration is some 98 00:09:43,170 --> 00:09:49,529 function of time whose derivative is proportional to itself. 99 00:09:49,529 --> 00:09:54,860 The function with this property is the exponential function! So the concentration function is 100 00:09:54,860 --> 00:09:59,670 equal to some constant times e to the negative kt. 101 00:09:59,670 --> 00:10:06,670 The constant must be the initial concentration, which is .01 Molar, so [N2O5]=(.01)e^{-kt}. 102 00:10:14,079 --> 00:10:20,889 Evaluating this expression at time t=3600, we find that the exact concentration is also 103 00:10:20,889 --> 00:10:27,889 .0094 Molar to two significant figures. To be more precise, we can look at the error, 104 00:10:28,920 --> 00:10:34,790 which is the absolute value of the exact solution minus the approximate solution all divided 105 00:10:34,790 --> 00:10:37,899 by the exact solution. 106 00:10:37,899 --> 00:10:44,899 The error at time t=3600 seconds is .002 or .2%. 107 00:10:46,329 --> 00:10:51,829 This is really quite good. Generally speaking, we might consider an error of less than 5% 108 00:10:51,829 --> 00:10:55,110 to be acceptable. 109 00:10:55,110 --> 00:11:02,110 So our linear approximation was definitely within the acceptable range 1 hour later. 110 00:11:02,199 --> 00:11:08,100 But now it's Tuesday. Today you leave the .01 Molar N2O5 on the counter for a full 10 111 00:11:08,100 --> 00:11:14,550 hours. Oops. We know that the exact solution is an exponential 112 00:11:14,550 --> 00:11:17,139 decay. 113 00:11:17,139 --> 00:11:22,199 So do you think that the linear approximation will be an over estimate, an under estimate, 114 00:11:22,199 --> 00:11:29,199 or equal to the exact solution after 10 hours? [pause] 115 00:11:31,670 --> 00:11:38,670 The linear approximation of the concentration after 10 hours at room temperature is T0(36000)= 116 00:11:39,889 --> 00:11:46,889 .0038 Molar. The exact solution of the concentration after 117 00:11:47,709 --> 00:11:54,709 10 hours is .0054 Molar. Comparing these two solutions, we see that linear approximation 118 00:11:56,160 --> 00:12:03,160 is an under estimate as we would expect, with error is .29 or 29%. So a linear approximation 119 00:12:06,529 --> 00:12:12,379 is definitely not acceptable for this time range. 120 00:12:12,379 --> 00:12:18,290 But now you begin to wonder, for what times is the linear approximation at t=0 within 121 00:12:18,290 --> 00:12:25,290 5% error of the exact solution? [pause] Because we are only interested in positive time values, 122 00:12:30,529 --> 00:12:36,509 you should have found that the approximation is acceptable for times 0 less than t less 123 00:12:36,509 --> 00:12:43,209 than 16, 700 seconds, which is about 4.5 hours. Now that you've carefully examined this problem 124 00:12:43,209 --> 00:12:48,149 using linear approximation, it is probably time to tell your graduate student adviser 125 00:12:48,149 --> 00:12:55,149 that you need to order a new batch of N_2 O_5 solution. 126 00:12:55,920 --> 00:13:00,910 Let's review. A1 The linear approximation to a function at a point c is the tangent 127 00:13:00,910 --> 00:13:07,040 line of a function at c. A2 This linear approximation only accurately models the function for points 128 00:13:07,040 --> 00:13:13,100 sufficiently close to c. A3 It is easy to read off the derivative at a point from a 129 00:13:13,100 --> 00:13:19,449 differential equation, and thus give a linear approximation to the solution. In our example 130 00:13:19,449 --> 00:13:25,449 with the concentration of N205, we could also solve the differential equation exactly and 131 00:13:25,449 --> 00:13:32,239 demonstrate that the approximation was acceptable near the point of approximation t=0, but not 132 00:13:32,239 --> 00:13:34,879 so for larger time intervals. 133 00:13:34,879 --> 00:13:39,389 Notice that we used this simple example to illustrate linear approximation because we 134 00:13:39,389 --> 00:13:45,959 could also solve it exactly and use the exact solution to determine the error. This is somewhat 135 00:13:45,959 --> 00:13:51,049 misleading, since we often want to use linear approximation precisely when the differential 136 00:13:51,049 --> 00:13:57,399 equation that describes our physical situation does not have an analytic solution. 137 00:13:57,399 --> 00:14:02,170 So you may wonder, is there a way to estimate the error in a linear approximation without 138 00:14:02,170 --> 00:14:08,089 knowing the exact solution. The answer is yes! However, it involves Taylor's Remainder 139 00:14:08,089 --> 00:14:13,309 theorem, which finds an acceptable range in terms of higher order derivatives, and is 140 00:14:13,309 --> 00:14:19,980 beyond the scope this video.