if mathematically derivatives are the opposite of integrals, conceptually how is the area under a curve opposite to the slope of a tangent line?
In: 332
My physics teacher in high school drilled the “d-a-v-t-u” functions into us – distance, acceleration, velocity, time, potential enegy. In particular, distance, velocity, and acceleration are related to each other through time and *calculus* relationships.
Velocity is the derivative of the position function Z(t), where Z are three Cartesian coordinates, i.e. Z'(t). Acceleration is the derivative of the velocity function, so it is Z”(t).
Conversely, velocity is the integral of the acceleration function, and position is the integral of the velocity function.
Think of what that means if you are running. The more ground you cover, the greater Z is getting. If you’re running through a forest, you might have to go slower on uphills, and faster on downhills. The slower you run, the less ground you cover each second, right? If you look at a chart of position v time, your line will start drooping to the right. As the slope of that line gets flatter, your speed is also slowing down. But remember, the slope of the line is also the derivative of the line; that is, the slope of your position line is your speed. So as you slow down, the *slope* of the line drops, even though you continue to get farther away from the origin.
The integral of the line – the total distance travelled – will continue to grow, but not as quickly once you’ve slowed down. And if we want to know how quickly you slowed down – your acceleration – we simply have to take the derivative of the velocity line.
Now, these are inverse relationships, so they can go the other way. If I told you that you start at 0,0, accelerate instantaneously to 1/ms, then after 1 second, accelerate again to 2 m/s, and after 4 seconds, decelerate to 1 m/s, and after 5 seconds, decelerate to zero m/s. How far did you travel?
Easy right? 1 metre the first second, 6 m in the next 3, 1 m in 5th, and then you’re stopped. 8 m in 5 seconds total. And in fact, that’s the integral underneath this chart, where each X represents one second horizontally and one metre vertically.
: xxx
:xxxxx
So, once you have one of the three specified as some kind of function, you can figure out the other two. (OK, with integrals you have to account for initial conditions, e.g. what if you started at 2,4 instead of 0,0?)
Latest Answers