WEBVTT 00:00:03.250 --> 00:00:08.490 In December 1998, the Mars climate orbiter was launched in hopes of providing detailed 00:00:08.490 --> 00:00:13.330 information about the atmosphere of Mars. The launch went well. However the two teams 00:00:13.330 --> 00:00:18.240 that collaborated on the project worked in different units--one used the metric system, 00:00:18.240 --> 00:00:23.300 and the other the English system, and the two systems of units were never properly reconciled. 00:00:23.300 --> 00:00:28.269 So while the orbiter was intended to fall into orbit around Mars, instead it crash landed 00:00:28.269 --> 00:00:33.570 on the surface. So units are pretty important. 00:00:33.570 --> 00:00:38.440 This video is part of the Problem Solving video series. Problem-solving skills, in combination 00:00:38.440 --> 00:00:43.800 with an understanding of the natural and human-made world, are critical to the design and optimization 00:00:43.800 --> 00:00:47.010 of systems and processes. 00:00:47.010 --> 00:00:52.300 Hi my name is Ken Kamrin, and I am a mechanical engineering professor at MIT. Today I want 00:00:52.300 --> 00:00:59.039 to talk to you about using unit analysis in problem solving. This might seem simple, but 00:00:59.039 --> 00:01:02.239 it is a critical tool for validating your calculations. So pay attention. 00:01:02.239 --> 00:01:07.020 To understand this example, you should be familiar with the definition of work. 00:01:07.020 --> 00:01:09.150 After watching this video, you should be able to 00:01:09.150 --> 00:01:12.920 utilize and apply the key properties of unit analysis: 00:01:12.920 --> 00:01:16.470 when two quantities are multiplied, their units also multiply. 00:01:16.470 --> 00:01:20.640 all terms added, subtracted or equated must have the same units. 00:01:20.640 --> 00:01:25.420 You should also be able to explain how derivatives and integrals affect units. 00:01:25.420 --> 00:01:29.990 Before we start the main example, let's discuss how integration and differentiation affect 00:01:29.990 --> 00:01:36.990 units. The first question is: what are the units of dx? You should think of dx as a very 00:01:39.350 --> 00:01:45.259 small change in x or a "little bit of x", Recall that a little bit of ice cream, is 00:01:45.259 --> 00:01:52.259 still ice cream. Even if it is a very very small amount. So the units of dx are precisely 00:01:59.520 --> 00:02:06.520 the units of x. Let's do an example. Let's consider the position function x(t) with units 00:02:09.780 --> 00:02:16.780 of meters, where t is time, with units of seconds. Then velocity is the time derivative 00:02:17.890 --> 00:02:24.489 of position v = dx/dt, and has units of meters per second. Let's look at how the notation 00:02:24.489 --> 00:02:30.980 is consistent with our physical interpretation: dx has units of meters, and it is divided 00:02:30.980 --> 00:02:37.980 by dt, which has units of seconds. So dx over dt should have units of meters over seconds. 00:02:40.780 --> 00:02:46.560 Now let's look at acceleration, which has units of meters per second squared. We know 00:02:46.560 --> 00:02:53.560 that acceleration is the time derivative 00:03:01.290 --> 00:03:08.290 of velocity, which is the second derivative of position with respect to time. Let's see 00:03:31.280 --> 00:03:38.280 how this plays out with notation... We can also determine how integration affects units. 00:03:41.489 --> 00:03:48.489 Let's use the example of the integral of velocity v with respect to time t. We know physically 00:03:48.930 --> 00:03:54.799 that this is position, which has units of meters. Unit analysis also makes this clear 00:03:54.799 --> 00:04:01.799 because v has units of m/s, and dt has units of s, so when we multiply, the seconds cancel, 00:04:02.840 --> 00:04:07.909 and we end up with units of seconds It might be interesting to point out that the integral 00:04:07.909 --> 00:04:14.629 symbol doesn't affect units at all. Just like the differential did not have units, it is 00:04:14.629 --> 00:04:21.160 just a symbol that represents the limit of sums. Since every term in the sum has units 00:04:21.160 --> 00:04:27.460 of v times units of t, the sums, hence the integral sign doesn't affect units at all. 00:04:27.460 --> 00:04:34.460 What we've seen: given a function f(y), differentiating with respect to y divides the units of f by 00:04:36.310 --> 00:04:43.310 the units of y. Differentiating twice divides the units of f by the units of y squared. 00:04:45.470 --> 00:04:52.470 Similarly, integrating with respect to y multiplies the units off by the units of y 00:04:52.800 --> 00:04:57.659 In this example, we'll see how unit analysis can help us check our calculation of the work 00:04:57.659 --> 00:05:04.660 done by an applied force. A machine in a factory is programmed to apply a force to a 3.0kg 00:05:05.720 --> 00:05:11.970 object to move it back and forth in the horizontal direction. The position of the object as a 00:05:11.970 --> 00:05:18.970 function of time is given by the equation x equals 3.0t minus 4.0 t squared plus 1.0 00:05:19.930 --> 00:05:25.680 t cubed, where x is measured in units of meters, and t in seconds. Find the work done on the 00:05:25.680 --> 00:05:32.680 object by the force from t=0 to t=2. Note that work is done when a force is applied 00:05:34.419 --> 00:05:41.419 over a distance on an object. Let x0 be the position of the object at time t = 0. Let 00:05:42.419 --> 00:05:49.419 x2 be the position at time t = 2. Then the amount of work done on the object is computed 00:05:49.990 --> 00:05:56.990 by the equation W = integral x2 F ยท dx. Let's start by figuring out the units of work. We 00:06:00.910 --> 00:06:06.710 do this by analyzing our equation, which tell us that the units of work are the units of 00:06:06.710 --> 00:06:13.710 force times the units of dx. If we forget what the units of force are, we can use unit 00:06:14.639 --> 00:06:21.639 analysis to figure it out. We know that F = ma, and acceleration is in units of meters 00:06:23.610 --> 00:06:27.110 per second squared, and mass has units of kilograms, So the unit of force, also known 00:06:27.110 --> 00:06:29.740 as the Newton, is equivalently written as kilogram-meters per second squared. So the 00:06:29.740 --> 00:06:35.830 unit of force, also known as the Newton, is equivalently written as kilogram-meters per 00:06:35.830 --> 00:06:42.830 second squared. Recall that the units of dx are the units of x, which is meters. Thus 00:06:43.039 --> 00:06:49.930 the unit of work is a Newton-meter. So now we know what units our answer needs to be 00:06:49.930 --> 00:06:55.669 in. So let's start solving our problem. In order to solve our problem, we need to know 00:06:55.669 --> 00:07:02.669 the force acting on our object. We are given the mass of the object, the position function 00:07:02.949 --> 00:07:09.780 for the object, and F=ma. Because we can determine acceleration as the second derivative of position 00:07:09.780 --> 00:07:16.780 with respect to time, we can use our given information to determine the force. Let's 00:07:16.930 --> 00:07:22.060 go ahead and solve this problem, perhaps incorrectly. See if you can catch my mistake. Alright, 00:07:22.060 --> 00:07:27.050 so here we have position as a function of time. So let's differentiate to get the velocity. 00:07:27.050 --> 00:07:33.530 Where I've left off the decimal points. And we'll go ahead and differential the velocity 00:07:33.530 --> 00:07:40.530 to get the acceleration. And finally to get the force, we go ahead with F=ma. So there 00:07:44.430 --> 00:07:51.430 is my m. And according to this, our expression for a is -8 + 6t. So I look at this and I 00:07:55.259 --> 00:07:59.940 think to myself, uhoh, this can't be right, because this doesn't have units of force. 00:07:59.940 --> 00:08:03.870 We know that force should have units of Newtons, but it doesn't look like this has the correct 00:08:03.870 --> 00:08:10.620 units. So this entices us to go back into our problem to figure out what went wrong. 00:08:10.620 --> 00:08:15.190 Looking back up at the first equation for position, I see immediately that something 00:08:15.190 --> 00:08:22.190 has gone wrong, because position is in units of meters, and each of these terms don't have 00:08:22.190 --> 00:08:25.250 units of meters. In fact they don't even agree with each other. 00:08:25.250 --> 00:08:31.800 This tells me there had to be units for each of these constants that somehow got swept 00:08:31.800 --> 00:08:38.799 under the rug. So let's be rigorous, and put those units back in. Here we need units of 00:08:39.219 --> 00:08:46.219 meters per second to cancel with the seconds from time, and leave us with units of meters. 00:08:47.959 --> 00:08:54.959 Note that this means something physical, it is the initial velocity. Let's go back into 00:08:59.399 --> 00:09:04.570 this term here, and see that I have a t squared, and I need something with units of meters 00:09:04.570 --> 00:09:09.490 per seconds squared if I am going to get from this term something with units of meters. 00:09:09.490 --> 00:09:16.000 And lastly, we have this term over here that has a t cubed. And likewise the missing unit 00:09:16.000 --> 00:09:23.000 must have units of meters per second cubed, which physically is the unit of jerk, or the 00:09:23.850 --> 00:09:24.730 rate of change of acceleration. 00:09:24.730 --> 00:09:27.920 Now every term has units of meters, and so this equation makes sense. You may want to 00:09:27.920 --> 00:09:33.060 pause the video here and find the units for the constants in the formulas for velocity 00:09:33.060 --> 00:09:39.610 and acceleration. Show that these units agree with the units you find by differentiating 00:09:39.610 --> 00:09:46.610 x(t) Now that you have checked the units of acceleration, let's put it back in the force 00:09:50.330 --> 00:09:55.990 formula, and see that everything works out ok. So we got F=ma. M as before equals 3 kilograms. 00:09:55.990 --> 00:10:00.310 And the acceleration written with correct units is negative 8 point 0 meters per second 00:10:00.310 --> 00:10:07.310 squared plus 6 point 0 t meters per second cubed. To see that this has the correct units, 00:10:08.019 --> 00:10:15.019 lets expand this formula. -24 kg m per s squared yep, that's a Newton. + 18t kg m per s cubed. 00:10:22.260 --> 00:10:29.260 Just to check, note that while this doesn't have units of N, time has units of seconds, 00:10:29.440 --> 00:10:36.440 so this term is entirely in units of newtons as well. Now that we've verified the of Force 00:10:43.930 --> 00:10:50.930 is in units of Newtons. we can compute the work integral. That means we want to evaluate 00:10:56.279 --> 00:11:03.279 the integral: x0 to x2 of F dx. We have an expression for F, so let's go ahead and put 00:11:04.269 --> 00:11:09.010 that in. I'm going to go ahead and leave off the units for now. We leave it as an exercise 00:11:09.010 --> 00:11:16.010 for later that you will do that. What do we get? Integral x0tox2 minus 24 plus 18t dx. 00:11:17.890 --> 00:11:22.050 So we look at this and we realize that we actually have a slight problem. Because the 00:11:22.050 --> 00:11:27.760 force is written in terms of time, and we are integrating with respect to x. Fortunately 00:11:27.760 --> 00:11:34.760 this is easy to resolve. Just remember that a small change in x, dx is equivalently velocity 00:11:36.850 --> 00:11:43.380 times a small change in time. This is easy to see if you express velocity as dx dt, because 00:11:43.380 --> 00:11:49.200 you multiply by dt, and it is clear that this gives you a small change in x. So in your 00:11:49.200 --> 00:11:54.389 calculus class this little maneuver is called a change of variable. So let's go ahead and 00:11:54.389 --> 00:12:00.880 do that. This means our integral can be written as an integral over time, from time 0 to time 00:12:00.880 --> 00:12:07.880 t=2 of force times velocity times dt. We write F as ma, and we recall that acceleration is 00:12:11.730 --> 00:12:18.730 the time derivative of velocity. Now we see how to integrate this, and we find that the 00:12:19.120 --> 00:12:26.120 integral is m times velocity squared divided by 2 evaluated at the two time end points. 00:12:29.139 --> 00:12:32.980 Plugging in our formula for the velocity, and evaluating, we get that the answer is 00:12:32.980 --> 00:12:39.329 negative 12. I intentionally left off the units while doing this computation. Now I 00:12:39.329 --> 00:12:44.680 leave it as an exercise for you to use unit analysis to check that this integral does 00:12:44.680 --> 00:12:51.680 in fact give you 12 Newton-meters or 12 Joules, the units for work. In the example, we saw 00:12:58.899 --> 00:13:04.290 that when you are adding numbers with units, it is important for those numbers to have 00:13:04.290 --> 00:13:10.389 the same units if you want your quantity to have physical meaning. We also learned that 00:13:10.389 --> 00:13:17.389 the differential quantity dx has the same units as x. We also saw that it is helpful 00:13:17.660 --> 00:13:21.920 to use unit analysis to check your work at points during and after the computation. You 00:13:21.920 --> 00:13:27.149 can see now that unit analysis can be a useful part of your problem solving strategy. It 00:13:27.149 --> 00:13:33.459 is important to understand the properties of units, and how they are affected by mathematical 00:13:33.459 --> 00:13:40.459 operations. Checking units at the end of a computation useful to see if solution is reasonable. 00:13:40.769 --> 00:13:46.720 Sometimes, unit analysis can suggest a formula for an unknown quantity in terms of given 00:13:46.720 --> 00:13:53.720 information. But physical knowledge is necessary to validate solution completely.