Higher Derivatives and Their Applications 

Distance incremental fan know this shiet in 1030163 BC

@@thiennhanvo2591 Derivatives of functions imply that the functions means something in the real world. Just because you can apply a derivative formula to a function like X^7 and get answers does not mean that the function has a meaning in the real world, and that your rote derivatives are meaningful. Math is more than rote iteration of a rule - it should stand for some aspect of reality. To take the fourth derivative of a function and end up with a position, you first have to have a meaningful higher function - one that signifies motion in reality. Math is not just repetitive fun and the following of rules, it makes a contact with reality.

@@charleshudson5330 i aint reading allat

i agreeeeee!!!!!

@@ProfessorDaveExplains you got a million subs dude. good job.

@@ProfessorDaveExplains much more than one mil now 😂

same. the intro got me goin too 🤣 ... you ever find anything about snap, crackle and pop?

@@devmehta4144 bro my course is no longer doing these things I cant even remember what the fuck they were lmao. reading back my comment I sound like an alien like what the fuck are derivatives lmaooo 1 year makes u forget things u dont use beyond college exams

You're actually pretty spot on with both speculations. If the object was a car then its position graph says that it was (1) stationary, then (2) sped up away from the origin, (3) maintained a speed, (4) slowed to a stop, (5) stayed stationary for some time, (6) slowly sped "up" to reverse towards the origin, (7) maintained that speed, and finally (8) abruptly slowed down to a stop at approximately the origin. For reference, let each of the points above refer to successive vertices on the second graph. Imagine slicing the three graphs vertically at each of these points: The second graph describes the car's velocity (or speed) in the same time frame as the position graph. Simply put, if the car moving forward then its velocity hovers above the x-axis (points 2 to 5) and if it is reversing, then its velocity hovers below the x-axis (points 6 to 8). If there is no change in velocity (intervals where the velocity graph is flat) then the car is either stationary (velocity is zero - points 0 to 1, 5 to 6, 8 to 9) or it is moving at a constant speed (where the velocity graph is flat but not at zero - points 2 to 3, 6 to 7). The gradual rise/fall on the velocity graph refers to a change in velocity. Let's try and develop an intuition for why this is true but before we can do that, let's first review mathematics. Recall the definition of a slope: rise/run. When we talk about "the rate of change" of a function, we are referring to its slope within some interval. In calculus, this is referred to as the derivative of the function - its slope over an infinitely small interval. The second graph demonstrates the rate of change in position over time and the third graph demonstrates the rate of change in velocity over time; thus, we say that *velocity is the first derivative of position* and *acceleration is its second derivative.* But what does all this mathematical jargon actually mean? We all know from experience that a car cannot just go from rest and magically travel at 60mph at the snap of a finger - it has to *speed up* from rest and traverse through 1mph, 2mph, 3mph and so on until it reaches 60mph over some time. This gradual change in velocity is just the car's acceleration ("speeding up/down"). Notice the first rising curve in the first graph (points 2 to 3)? Well that section corresponds to: (1) a positive change in velocity (the first rising slope in the second graph) and (2) the first positive "step" in the acceleration graph. By interpreting the acceleration vs time graph, we find that the car is speeding up at a constant rate in this interval. The steepness of this slope describes how fast the car is accelerating (speeding up) or decelerating (slowing down) - where a steeper slope = greater acceleration (analyze the first two slopes in the second graph and compare it with the magnitude of acceleration in the third graph). In English, we would describe a steep positive slope in the velocity graph as "speeding up quickly," and conversely, a steep negative slope as "slowing down quickly." At the extreme end, we would describe a nearly vertical negative slope as "abruptly stopping." Notice how language agrees with math - [speeding up] [quickly] = [velocity/time] per [time] = [distance/time] per [time] = [meters/second] per [second] = [m/s]/s = *m/s^2.* Voila - we have derived the SI units for acceleration using just English! When the car is speeding up, we should expect: (1) a positive "step" in the acceleration graph, (2) a rising slope in the velocity graph, and (3) a rising curve in the position graph. Conversely, if the car is slowing down we should expect: (1) a negative "step" in the acceleration graph, (2) a downward slope in the velocity graph, and (3) a flattening curve in the position graph. It is also important to keep in mind that *while a negative velocity implies that the car is in reverse, a negative acceleration does not imply the same* - a negative acceleration just means the magnitude of its velocity (speed) is decreasing relative to the direction of motion - a car can still be moving forward even if its acceleration is negative. Given enough time, a car moving in the positive direction with a negative acceleration will eventually come to a stop. This is evident in points 3 to 4. Moreover, a negative acceleration also does not imply that the car is slowing down. Take points 5 to 6 for example. The car is speeding up but in reverse! Bonus: The area under the curve of a velocity graph is the total displacement (vector counterpart of distance) of the car - in this case, by our analysis of the position graph, we can extrapolate this value to be zero which means that the car returned to the origin. By analyzing the geometry of the velocity graph we see that the area (in blue) above the x-axis is approximately the same area below the x-axis. A positive area "I" plus a negative area "I" equals zero, thus the net displacement of the car is zero. This is yet another relationship between the position and velocity graphs. The same relationship applies to velocity and acceleration. In calculus, displacement (position) is known as the integral of velocity (speed) and velocity is known as the integral of acceleration. In essence, integrals are anti-derivatives *wink*, vice versa. Professor dave actually goes over this in this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-bqf8m7xNvLg.html

@@Sooper35 Holy moly you were bored

@@aleksszukovskis2074 just a little cracked out on caffeine :) drank too much coffee for a midterm hehe.

It's also a tangential slope but of that derived function. In this case of graph of velocities at given time and tangential slop is acceleration.

For slopes close to zero, the second derivative is proportional to the curvature. Actual curvature (reciprocal of radius of a circle that matches the function in question) is a combination of both first and second derivative.

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Short answer: history, that's how Leibnitz wrote derivative notations. It is implied that dx^2 really means (dx)^2.

no just a science communicator!

Professor Dave Explains ok boss

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Yeah! It's its own derivative.

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