this is not going to be a long video we're just going to go over one problem here in the last video we talked about the tangent problem and finding the equation of the tangent line and we did two examples so what we're going to do now is look at an example of a velocity problem and the main point here is I want you to notice as we're going through the problem even though there's different variable names and you know the function is different notice that the steps were going through so that the process we're going through is really mathematically identical to what we were doing with the velocity problem even though it's stated a little bit differently so recall that the instantaneous velocity is the instantaneous rate of change of an object's position and I mentioned in the overview of calculus that there's a little bit of a problem here and the problem is that it is peers like it doesn't really make any sense to calculate a velocity at a single instance of time single instant of time and this is the paradox of the derivative that we will get into talking about at different times during the beginning of the semester but at first and I remember when I first heard this one you know I took calculus class I was like way to say that that doesn't make any sense at all how can you say something has an instantaneous velocity because in order to measure velocity you need two points just like it's the same thing with the secant line in order to find the slope of a line you need two points and so it doesn't make any sense to say that there's a velocity at an instant of time however we can use the same kind of approximation procedure that we were looking at in the previous section so there we had the idea of taking these secant lines that are very close to the tangent line and seeing if the slopes of those seem to be settling in on some value and that's basically what we're going to do here with the velocity problem although we cannot calculate the velocity right at an instant we can calculate it over extreme small time intervals and so we're going to do is look at some of those time intervals and see if we can see those getting closer and closer to a specific value so here's the problem we're going to look at a ball is thrown straight up from the top of a 64 foot tall building initial speed of 48 feet per second then we can write the height as this function H of T over here so negative 16t squared plus 48 T plus 64 so the general picture just so just to give you of what's going on so the general picture of what's going on is let's say here's their building okay there's a building here's the ground and the height of the building is 64 and there's somebody here on top it's my best representation of a person and they throw an object straight up so think about what's going to happen with the object it's going to start off here it's gonna go up and then eventually of course everything has to come down so that's going to come back down so that's the general path of course it's not moving left right it's just going straight up or downs but I had to move a little bit to show you the emotion of it so it starts here goes up a little bit comes to a maximum height and then starts going back down so what we're gonna try and do is figure out what's the instantaneous velocity two seconds after it is thrown so T is the time here since the ball is thrown so it's thrown at time T equals zero so what we're going to do is look at some average velocities over time intervals very small time interval starting at two so to do this I'm going to make a kind of a table like we did before you so let's say that we're looking over the time interval from 2 to T and the idea here is that T is just a little bit bigger than 2 so that's our time interval and we want to find average velocities over these time intervals so I'm going to make a table that's going to have T in the left-hand column and then the average velocity over that interval in the right-hand column so let's pick say two point one to point zero one two point zero zero one and two point zero zero one so should look very familiar to what we were doing in the last video now the average velocity over here since this is our function H of T our average velocity is going to be H of t minus H of two all over t minus two oops so the H of t minus H of two that's the change in height over the time interval from 2 to T t minus 2 is the time elapsed over that little time interval now if we simplify this we get -16 t-squared plus 48 T plus 64 let's see what is H of to pause the video ok so you should have gotten H of 2 is 96 so we're going to subtract 96 and this is all over t minus 2 so this is -16 t-squared Plus 48c and what is the difference of those two 32 so - 32 all over t minus 2 okay so let's go ahead and use Wolfram Alpha again to do this so I'm going to enter in this expression here and ask it to evaluate it at t equals 2 point 1 to point 0 1 and so on okay so we're asking it again to evaluate this expression that we just got at t equals two point one two point zero one two point zero zero one two point zero zero zero one so let's go ahead and do that you okay and there we go so minus seventeen point six minus sixteen point one six - sixteen point zero one six - sixteen point zero zero one six okay so let's go and copy those into the table okay so here are these values here and remember these are these are actually physical values it's not just like you know numbers abstract numbers the unit's here are feet per second so what we're saying is the average velocity over the time interval of this would be one tenth of a second after time two is negative seventeen point six feet per second over one one hundredth of a second after that instant is minus sixteen point one six feet per second after when one thousandth of a second one ten thousandth of a second so you see that we get this pattern here and you can probably guess what the next pattern is after this but it's pretty clear that these are going closer and closer to just negative sixteen so based on that guess we're going to say the instantaneous velocity at time two is exactly negative 16 feet per second so we say that since the average velocity seemed to be approaching negative 16 feet per second we we guess that the instantaneous velocity at time 2 seconds right at that instant is exactly negative 16 feet per second so a couple things here let's see if we can pardon this fireworks going off in the background I don't know if you can hear let's see if we can figure out what's going on here with regard to the picture so at t equals 2 we already got that H of 2 was 96 so we know 96 is somewhere up here at T equals to H equals 96 because it's I I asked you to not just find the instantaneous velocity find and describe so let's describe it in terms of the actual problem so two seconds after you throw it up the height is 96 so it's still above the building's still above the person that threw it but since its velocity is negative 16 that means it's on it's on its way down so it's already reached the maximum height and it's started to fall back down toward earth so that's the point that we're at in its path and just one last comment to make here and then we're gonna go on to developing all the machinery with limits that we need when we say that this instantaneous velocity at T equals 2 is negative 16 feet per second it's important to know what we're saying the instantaneous velocity is exactly negative 16 feet per second even though it's a limiting process that we use a limiting process of approximations to get that value of negative 16 we don't say that the instantaneous velocity is approaching 16 but we don't say that the instantaneous velocity is getting closer and closer to negative 16 the instantaneous velocity is exactly negative 16 it's not not a little bit next to it or close to it it's exactly negative 16 so hopefully you see how even though this is a physical problem especially if you look at the the calculations we made the table we set up the whole process we went through is exactly identical to what we were doing with the slopes of the secant line in the equation of the tangent line so what we're going to do over the next several sections the next big topic is to develop all the machinery of limits we need to make this idea this whole this whole thing here about you know getting closer and closer you know what's really going on here so we need to make this idea of this limiting process more precise and that's what we're going to do when we develop the machinery of limits so we're going define and in we're going to give an informal definition of limits and then we're going to give a more formal definition but we're not really going to work with the formal definition we're going to get some basic limit properties and then develop everything we need to know about limits from that and then the limit stuff is the things that we need to really do calculus when we get to derivatives later so it's kind of some technical stuff and it's going to seem like it doesn't it's not really related to the problems that we were talking about but it's very important to develop it so that we can do what we need to do