He said at the beginning that all turning points are stationary points, then proceeds to give an example of the absolute value of x that shows otherwise.
If the graph in the upper right corner of the whiteboard is for the function "|x|" then it has two derivatives, right? One is for the negative part of the x-axis and that is "-1", the second is for the positive part of the x-axis and that is "1". If we imagine that this is a displacement function then the derivatives show the velocity, right? The object was approaching to the origin with the constant speed of -1, stopped for a moment and went back with the constant speed of 1. The moment when they stopped is "x = 0". In that position, the function value is also "0". So the object is 0 units away from the origin meaning stopping for a moment, right? As none of these derivatives have a value of 0 then the speed of the object is never 0, right? Does it mean that without stopping, the object just bounces back with the same speed? How come then in the case of a parabola we call that bounce "stopping for the moment" as it's smooth and for that sharp-edged graph, we don't do it as it's not smooth?
The absolute value graph is not differentiable. A kink or a cusp in a graph, means that the derivative is undefined at that point. With a Fourier transform, it will converge to the midpoint of the jump discontinuity, averaging the derivatives on both sides of the kink or cusp, but that is a topic for another day. The absolute value graph cannot possibly describe how any object's position vs time, assuming it has non-zero mass. Because speed cannot suddenly change, since that would require infinite acceleration, which isn't possible. Jump discontinuities in speed are only possible if the acceleration duration is negligible, and you are simply ignoring it for the purposes of your problem. In reality, it will take a non-zero amount of time to change speed, so any graph describing position vs time, will have to be continuous and differentiable. It could not contain kinks, cusps, or jump discontinuities
Even for a quick rebound of a bouncy ball, if you film it in slow motion, you will find that it takes a non-zero amount of time to slow down to rest, and speed back up for the rebound. If you are interested in determining the force of impact, you might want to know what this time is, as the shorter the rebound time, the greater the stopping and restoring force needs to be.
Turning point is the point on a graph where the gradient changes from positive to negative, or vice versa. Stationary point is the point where the gradient is zero. Very similar, but the only difference between the two is that turning points only applies to minimum and maximum points, but stationary points includes the point of inflection as well as maximum and minimum points.
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Thank you so much Eddie Woo! I go to online school, and it's difficult to focus when all the content is boring, but you make it fun to learn again.
Awesome video. I seriously rather learn things from you Eddie woo than with my teacher. Cheers for Eddie woo!
I’d totally be failing AP calc without this dude
He said at the beginning that all turning points are stationary points, then proceeds to give an example of the absolute value of x that shows otherwise.
Wow very good analogy with the water. Helps me understand.
If the graph in the upper right corner of the whiteboard is for the function "|x|" then it has two derivatives, right? One is for the negative part of the x-axis and that is "-1", the second is for the positive part of the x-axis and that is "1". If we imagine that this is a displacement function then the derivatives show the velocity, right? The object was approaching to the origin with the constant speed of -1, stopped for a moment and went back with the constant speed of 1. The moment when they stopped is "x = 0". In that position, the function value is also "0". So the object is 0 units away from the origin meaning stopping for a moment, right? As none of these derivatives have a value of 0 then the speed of the object is never 0, right? Does it mean that without stopping, the object just bounces back with the same speed? How come then in the case of a parabola we call that bounce "stopping for the moment" as it's smooth and for that sharp-edged graph, we don't do it as it's not smooth?
The absolute value graph is not differentiable. A kink or a cusp in a graph, means that the derivative is undefined at that point. With a Fourier transform, it will converge to the midpoint of the jump discontinuity, averaging the derivatives on both sides of the kink or cusp, but that is a topic for another day.
The absolute value graph cannot possibly describe how any object's position vs time, assuming it has non-zero mass. Because speed cannot suddenly change, since that would require infinite acceleration, which isn't possible. Jump discontinuities in speed are only possible if the acceleration duration is negligible, and you are simply ignoring it for the purposes of your problem. In reality, it will take a non-zero amount of time to change speed, so any graph describing position vs time, will have to be continuous and differentiable. It could not contain kinks, cusps, or jump discontinuities
Even for a quick rebound of a bouncy ball, if you film it in slow motion, you will find that it takes a non-zero amount of time to slow down to rest, and speed back up for the rebound. If you are interested in determining the force of impact, you might want to know what this time is, as the shorter the rebound time, the greater the stopping and restoring force needs to be.
Some one coughing at about 2:17
Covid ¿¿
I don't think so mate. This video was uploaded in 2015.
@@jackyye0416 COVID-14?
Whats the difference between a turning point and a stationary point?
Turning point is the point on a graph where the gradient changes from positive to negative, or vice versa. Stationary point is the point where the gradient is zero. Very similar, but the only difference between the two is that turning points only applies to minimum and maximum points, but stationary points includes the point of inflection as well as maximum and minimum points.
@@jackyye0416 Thank you for your answer.
I think, basically, Turning Points are Global maxima or Global Minima.
@@ynat957 yes, local maxima and local minima.
@@jackyye0416 Thanks bro. Now its clear for me.
@@ynat957 Taken from this thread, Jacky Ye is a better teacher than Eddie Woo?
5 years since the video was released
still 0 dislikes