Total distance traveled with derivatives

The position of a particle moving along a number line is given by s of t is equal to 2/3 t to the third minus 6t squared plus 10t, for t is greater than or equal to 0, where t is time in seconds. The particle moves both left and right in the first 6 seconds. What is the total distance traveled by the particle for 0 is less than or equal to t is less than or equal to 6? So let's just remind ourselves what they mean by total distance. If I were to say start there, and if I were to move 3 units to the right and then I were to move 4 units to the left, and I'll say negative 4 to show that I'm moving to the left, then my total distance right over here is 7. 3 to the right and 4 to the left. Even though my position right over here is going to be negative 1. Or you could say my net distance, or you could say my displacement is negative 1. I'm 1 to the left of where I started. The total distance is 7. So now we've clarified that. I encourage you to now pause this video and try to answer the question. What is the total distance traveled by the particle in these first 6 seconds? So the easiest way I can think of addressing this is to think about, well, when is this thing moving to the right and when is it moving to the left? And it's going to be moving to the right when the velocity is positive, and it's going to be moving to the left when the velocity is negative. So this really boils down to thinking about when is the velocity positive or negative. And to think about that, let's actually graph the velocity function or make a rough sketch of it. So this is the position function. The velocity function is going to be the derivative of the position function with respect to time. So the derivative of 2/3 t to the third is going to be 2t squared. And then we have minus 12t plus 10. And so let's just try to graph this. This is going to be an upward opening parabola. This is clearly a quadratic. And the coefficient on the second degree term, on the t squared term, is a positive number, so it's going to be an upward opening parabola. It's going to look something like this. And we're assuming that it switches direction. So it's going to be positive some of the time and negative for some of the time. So it should intersect the t-axis where it's negative. The function is going to be negative in that interval, and it's going to be positive outside of that interval. So the easiest thing I could think of doing is let's try to find what the 0's are. Then we can draw this upward opening parabola. So to find its 0's, let's just set this thing equal to 0 so we get 2t squared minus 12t plus 10 is equal to 0. Divide both sides by 2 just to get rid of this 2, make this leading coefficient a 1. We get t squared minus 6t plus 5 is equal to 0. That made it a lot easier to factor. This can be factored into t minus 1 times t minus 5. Negative 1 times negative 5 is 5. Negative 1 plus negative 5 is negative 6. This is equal to 0. So this left hand side of the equation is going to be equal to 0 if either one of these things is equal to 0. Take the product of two things equaling 0, well, you get 0 if either one of them is 0. So either t is equal to 1 or t is equal to 5. So now let's graph it. So let's draw our axes. So I could say that's my velocity axis. And let me draw the-- we only care for positive values of time. So let's draw something like this. Positive time. And let's see. Let's take that 1, 2, 3, 4, 5. We could keep going. So this is t equals 1. This is t is equal to 5. This is our t-axis. And let's graph it. So it's going to be an upward opening parabola. It's going to intersect both of these points. And so its vertex is going to be when t is equal to 3 right in between those points. So it's going to look something like this. That's the only way to make an upward opening parabola that intersects the t-axis at both of these points. So it'll go like that, and it'll go like this. It'll intersect. When t equals 0, we actually can figure out. When t equals 0 our velocity is 10. So the v-intercept, we could say, is 10 right over here. So that's what it looks like. So we see that the velocity is positive for time between 0 and 1. And it's also positive for time is greater than 5 seconds. And we see that our velocity is negative, or that we're moving to the left, between 1 and 5 seconds. The velocity is below the t-axis right over here. It is negative. So let's just think about what our position is at each of these points, at time 0, at time 1, at time 5, and what we care about time 6. And then think about what the distance it would have had to travel between those times. So let's think about it. So let's make a little table here. Let's make a little table. So this is time, and this is our position at that time. So we care about time 0, time 1, time 5 seconds, and time 6 seconds. So at 0 seconds, we know that our position is 0. S of 0 is 0. At 1 second, this is going to be 2/3 minus 6 plus 10. So it's going to be 4 and 2/3. So I'll write down 4 and 2/3. At 5 seconds, let's see, it's 2/3 times-- I'm going to write this one down. So it's going to be 2/3 times 125. That's the same thing as 250 over 3, which is the same thing. Let's see, 250 over 3. That's the same. 83 times 3 is 249, so this is 83 and 1/3. That's this first term. Minus 6 times 25. So that's going to be minus 150 plus 10 times 5. So plus 50. So this is going to simplify. Minus 150 plus 50, that's going to be minus 100. 83 and 1/3 minus 100. That's going to be negative 16 and 2/3. So negative 16 and 2/3 is its position after 5 seconds. And then at the 6 seconds, it's going to be 2/3 times 6 to the third. I have to write this one down. 2/3 times 6 to the third minus 6 times 6 squared. Well, that's just going to be minus 6 to the third again. 6 times 6 squared plus 60. And let's see. How can we simplify this right over here? Well, this part right over here we can rewrite as-- we could factor out as 6 to the third. This is 6 to the third times 2/3 minus 1 plus 60. Scroll down a little bit and get some more space. So it's going to be 6 to the third times negative 1/3 plus 60. And let's see. Let's write it this way. This is going to be 6 squared times 6 times negative 1/3 plus 60. This right over here is negative 2. So it's negative 2 times 36. This is negative 72 plus 60. So this is going to be negative 12 right over there. So now we just have to think, how far did it travel? Well, it starts traveling to the right. It's going to travel to the right 4 and 2/3. So let's write this down. So we're going to have 4 and 2/3. And then it's going to travel to the left. Let's see, to go from 4 and 2/3 to negative 16 and 2/3, that means you traveled 4 and 2/3 again. You traveled 4 and 2/3 to the left, and then you traveled another 16 and 2/3 to the left. Just a reminder, we're 4 and 2/3 to the right now. We have to go 4 and 2/3 to the left back to the origin, and then we have to go 16 and 2/3 again to the left. So that's why this move from here to here is going to be 4 and 2/3 to the left followed by 16 and 2/3 to the left. Another way to think about it, the difference between these two points is what? It's going to be 4 and 2/3 plus 16 and 2/3. If you do 4 and 2/3 minus negative 16 and 2/3, you're going to have, that's the same thing as 4 and 2/3 plus 16 and 2/3. And then to go from negative 16 and 2/3 to negative 12, that means you went another 4 and 2/3 now to the right. So now this is 4 and 2/3. Now you're moving 4 and 2/3 to the right. And so we just have to add up all of these. We just have to add up all of these values. So what is this going to be? So this is going to be 2/3 times 4, so this part of it right over here, the fraction part of it. 2/3 times 4 is 8 over 3. And let's see, 4 plus 4 plus 16 plus 4 is 28. So 28 and 8/3, that's a very strange way to write it. Because 8/3 is the same thing as 2 and 2/3. So 28 plus 2 and 2/3 is 30 and 2/3. So the total distance traveled over those 6 seconds is 30 and 2/3 units.