Back To CourseMath 104: Calculus
16 chapters | 135 lessons | 11 flashcard sets
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I often have this dream where I'm back in high school, and I've just learned how to drive. My father's there, and he doesn't exactly trust me in the car, so he puts a GPS tracker on my car. He was going to sit at his computer and watch exactly where I was at any given point in time. Let's say that this GPS tracker measured the distance I was, at any given point in time, from home. So for example, one day he's sitting at his computer, and he sees this graph of my location as a function of time. Over here, he sees that I left home, I hit a traffic jam, I went to the mall and then, eventually, I came home - a little bit late, he was sure to remind me.
Other than my location as a function of time, he was also interested in how fast I was going - my velocity. So he can use the fact that my velocity is the derivative of my position as a function of time to determine, approximately, how fast I was going at any given point in time. So dx/dt.
Right here, I've just left home, and to make it look good, I pull out of the driveway very, very slowly. My derivative, which is the slope of the tangent, is very small. It's positive because I'm leaving home, but it's small. As I get out of the house, I speed up. My position changes a lot faster as a function of time. That's true until I hit this traffic jam, which slows me down again. So up here, my derivative is smaller. When I get to the mall, I stop the car. When I park the car, its position is not changing as a function of time. So here, my derivative is flattened out. It looks like it's zero. When I'm done shopping, I start to head home. Now here the slope of the tangent is negative, because I'm on my way back home. On this particular day, I forgot something at the mall, so I had to turn around and go back to the mall. By that time I was really late, so I sped home. My derivative is very steep here; the slope of this line is very steep. I'm driving very quickly because I'm very late. So the derivative here is a very large negative number.
So let's say my father wants to know exactly how fast I'm going, and let's say he zooms in really close to one part of this map. To find out exactly how fast I was going, he draws a tangent on this line, and he finds the slope of this tangent. In this case, my slope is dx/dt or the change in x divided by the change in time. Now as x went from 348 to 349, the change is 1. In this case, it is 1 foot. The change in time was 1 second, going from t=1 to t=2. So at this point in time, I was traveling at a whopping 1 foot per second, which is 60 feet per minute, which is not all that fast.
So let's get rid of all of the numbers here, and just look at one example of my position as a function of time. And let's see what my father can glean from this single graph. This graph is pretty simple. I leave home, I get to the mall, I turn around and I come back home. So let's focus on this area where I'm leaving home. I'm going away from home. Because I'm leaving home, I know that my derivative is going to be positive. It's going to be greater than zero for this entire time where I'm leaving home. So this is where x is increasing as t increases, but let's be a little more specific. Let's say in that first minute, I sped off very quickly. I've gone a long way in just one minute. In this case, the slope of the tangent is pretty large and positive. Once I get further from home, I slow down a little bit. I'm not moving - I'm not getting quite as far - as a function of time. In one minute, maybe I go half the distance that I went before. Here, the slope of the tangent isn't quite as steep as it was when I first left home; I'm not moving quite as fast. When I get even further from home, I slow down even more. So here in one minute, I'm hardly moving. My derivative is still positive, but it's pretty close to zero because x is not changing very much as a function of time.
So coming home, it's like the same thing in reverse, like I'm driving backward. Now my derivative is going to be negative. How fast I'm going will determine how negative it is. Is the value small but negative, or is it large but negative? If it's small but negative, I'm going backward slowly. And if it's large and negative, I'm going backward very quickly. When I first leave the mall, my position is not changing very much as a function of time. My derivative is negative, but it's close to zero. As I get closer to home, I speed up some. So my derivative is still negative, it's just slightly larger. It's further away from the x-axis. As I get even closer to home, I realize that I'm very late, and I've sped up a lot. And so the magnitude of my velocity is very large, but it's still negative because I'm still driving backward to get home.
Let's put all of this together, and let's take our position as a function of time and use it to graph our velocity, our x`, as a function of time. So here when we're leaving home, we started off with a really fast velocity, and positive because we were going to the mall forward. As we got to the mall, we slowed down. Our derivative decreased. When we got to the mall, we actually stopped for a second. Our derivative, right here, is actually equal to zero. The slope of the tangent here is zero. Then we started to come back from the mall; we started driving backward. At first we were driving slowly. Our slope was very shallow, so our derivative was negative but close to zero. Then we started to speed up, getting more and more and more negative. By the time we got home, we were going very fast, but we were still going backward. So our derivative is down here. We can connect all of these points and come up with an approximation for x`. And it looks like this straight line that goes through zero when we were stopped at the mall.
Let's take a second to make this a little more formal. Let's look at x as a function of time. When our function x is getting larger, we're leaving home. We have a positive derivative. The derivative x` is going to be greater than zero. If the function is getting smaller (we're driving backwards here), we have a negative derivative. If our function isn't changing, the derivative will be zero. This is like that instant when we were at the mall.
I got into a little bit of trouble for my last exploit. My father didn't let me out of the house for a while, but when he did, I again went to the mall. And he plotted my position as a function of time, and it looked like this. Now here when I left the house, I kept a pretty constant velocity. So all along this time, the slope of the tangent, everywhere along here, was the same because this is a straight line. I got to the mall and I stayed there for a while, and then I came home. And again, I had a straight line between the mall and home.
What does my velocity look like as a function of time? Well, on my way to the mall, my velocity was constant, thus x` was constant, and it was positive because I was going away from the house. When I got to the mall, I parked the car, and the car wasn't moving. So the entire time I was at the mall, the velocity of the car was zero. When I came back home, driving backward, my velocity was negative but still constant, because again I have a straight line. Now what happens at these intersections here? Right when I parked and right when I left the mall? Well, on the left-hand side of these points, when I'm looking at x as a function of t, I have a velocity, a tangent here, that is positive. Just on the right side of that point, I have a tangent that is equal to zero. So the tangent at that point really depends on what side you're looking at - the left side or the right side. Because those two values are different, there is no tangent at that point in particular. There is no derivative. The derivative is discontinuous. So on my graph of the derivative as a function of time, I'm going to put a hole. The same thing happens when I'm leaving the mall. On the left-hand side, I have zero velocity; my car isn't moving. On the right-hand side, I'm speeding home, so my derivative is negative. But at that point exactly, there is no tangent; the derivative is discontinuous.
One more point with this graph. On the way to the mall, I was driving relatively reasonably. I wasn't driving too fast; I wasn't driving too slow. The slope here might be something like 1. On the way home from the mall, the slope is much steeper and negative. I'm going to reflect this on the derivative as a function of t graph by writing my velocity on the way to the mall as being closer to this t-axis than when I was leaving the mall. Since my velocity was much faster, it was further away from the axis.
You can use these principles for other functions, like y=f(x). Let's consider my rules, and try to write out y` as a function of x. So first, I'm going to note where the function is increasing - where I'm going to have a positive derivative - and I'm going to shade those regions in on y`. Then, I'm going to note where the function is decreasing - where my derivative is going to be negative - and I'm going to shade those in on my graph y`. These couple points where the function is constant just for an instant, I'm going to mark on my graph y`, as red points. So in these cases, I have a zero derivative. Finally, I'm going to make a special note about this last part here, where my function is decreasing, but it's decreasing in a straight line. So I know that my derivative is going to be constant, but negative.
So what does y` look like in these regions? Well let's look at this first region on the left side. Here my slope starts out pretty steep, and as I move toward this red point, the slope actually decreases; it gets closer to zero. I'm going to mimic that on my y` graph where my derivative starts out very large, gets smaller and approaches zero. In this middle region, the function is decreasing the entire time, but it starts out decreasing at a very shallow rate; the slope here is very shallow. Then it gets steep, and then it actually becomes more shallow again. So I'm going to try to put that on my graph. It starts out close to zero, very shallow, gets big and then gets shallow again, but all negative.
All right, then I've got my zero derivative where my function is constant. Then I've got this region where the function is increasing. So at the left-hand side, the slope is very shallow; it's very close to zero. Then it increases, and by the time I get to this point, the slope is pretty steep. Let's put that on our y` graph. Now at this point here, the derivative on the left-hand side of the point and the right-hand side of the point are not the same. At this point right here, the function is discontinuous, but so is the derivative, y`. So I'm going to put a circle in my graph. Now my function is a straight line for this part of the graph, and it's decreasing. So the derivative is negative. I know that my derivative is going to be negative, but it's going to have a constant value. Let's put it down here. It might not be exactly right, but it gets the idea across. And let's make sure to put the circle, the open circle, at the left-hand side because I know that there is no derivative at this point where I had my discontinuity.
Let's review. You can find d/dt of x, from any x. This is especially true if you just have the graph and not the function itself. If you have the graph, you might not be able to get the numbers correct, but at least you can get an idea of what is going on. To do that, you want to keep in mind three things: 1.) If the function is increasing, the derivative is going to be positive. This is like when I was driving to the mall. 2.) If the function is decreasing, the derivative is going to be negative. Like when I was driving backward from the mall, back home. 3.) If the function is constant, the derivative is going to be zero. When my car was parked, I wasn't moving; my velocity was zero.
The other two things that you might want to keep in mind are that straight lines give you constant derivatives. When you're going 60 miles per hour for exactly 10 minutes, your derivative will be 60 for those 10 minutes. Finally, the derivative is not always continuous. If the slope on the left-hand side of the point and the slope on the right-hand side of the point are not the same, then your derivative doesn't exist.
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Back To CourseMath 104: Calculus
16 chapters | 135 lessons | 11 flashcard sets