I was wondering what the second derivative means about the first derivative....in physics that 's the "impulse". But I had trouble remembering if the point of intersection with the function had any relationship...(the graphs intersections_) .I think if I remember right, that there is none. I used to get this stuff down pat. I still do but it was extremely easy with Mr Woo's help! What a joy Eddie Woo is in the classes he put in his channel.
When the first derivative is near zero, the second derivative approximately indicates the curvature of the function. When the first derivative is much larger, a greater slope is gives the second derivative a harder job to do, to make the function curve. For instance, a parabola has a uniform second derivative, but it doesn't have uniform curvature, otherwise it would look like a circle. Locally near the vertex, it approximately looks like a circle, and this is the point where its curvature is maximum. Curvature is a complicated combination of the first and second derivative, but for a first derivative near zero, it approaches being equal to the second derivative. An application of this geometric interpretation, is Euler's beam theory. The theory of the distribution of stress and the way a linear-elastic material responds to stress, tells us that the curvature is directly determined by the bending moment in the beam. Integrating the curvature once, we get the slope. Integrating the slope, we get the deflection, and the equation of the shape of the deformed beam.
In calculus, students are usually taught "the second derivative test for concavity". The second derivative gives you information about the point of inflection. Namely, whenever the second derivate is positive, the first derivative is concave up; and when it is negative, the first derivative is concave up. (When it's zero the first derivative is horizontal)
The content and lecture delivery was perfect... the sound quality ruined everything.... I can't use headphones because it gives me headache at least a full day ... so no other option left for me....
He proved that the derivative of sin is cos by using some particular points but he has to prove that it is true for all points ... that’s how things work in maths in general
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One of the best math instructions in my entire life. Thank you Mr. Eddie Woo
This is how these trig functions should be introduced.
He’s an underrated teacher and this channel too
The quality of video can be improved by jus placing the can bit more closer to board…
Superb explanation ۔ This is what I was looking for
Love your enthusiasm!
I watch math classes from MIT you tube and the professor write really big, so people in the back can see. This would be great here.
Hey edd...I appreciate your videos man. .thanks
I was wondering what the second derivative means about the first derivative....in physics that 's the "impulse". But I had trouble remembering if the point of intersection with the function had any relationship...(the graphs intersections_) .I think if I remember right, that there is none. I used to get this stuff down pat. I still do but it was extremely easy with Mr Woo's help! What a joy Eddie Woo is in the classes he put in his channel.
When the first derivative is near zero, the second derivative approximately indicates the curvature of the function. When the first derivative is much larger, a greater slope is gives the second derivative a harder job to do, to make the function curve.
For instance, a parabola has a uniform second derivative, but it doesn't have uniform curvature, otherwise it would look like a circle. Locally near the vertex, it approximately looks like a circle, and this is the point where its curvature is maximum. Curvature is a complicated combination of the first and second derivative, but for a first derivative near zero, it approaches being equal to the second derivative.
An application of this geometric interpretation, is Euler's beam theory. The theory of the distribution of stress and the way a linear-elastic material responds to stress, tells us that the curvature is directly determined by the bending moment in the beam. Integrating the curvature once, we get the slope. Integrating the slope, we get the deflection, and the equation of the shape of the deformed beam.
In calculus, students are usually taught "the second derivative test for concavity". The second derivative gives you information about the point of inflection. Namely, whenever the second derivate is positive, the first derivative is concave up; and when it is negative, the first derivative is concave up. (When it's zero the first derivative is horizontal)
thank you eddie you are a life saver
Everything is fine, but how have you start with 1 and then to zero. Fro sinx and then for dy/dx of sinx=cost. Positive & xome down to zero,?
Why does my teacher not show us this stuff
I don't know dude. I don't know why teachers don't do this stuff
I show these stuffs to my students but they say I m not a good teacher
M making things complex and dont go straight to the point
@@alimensah1153 they are not students they are parrots
@@rhuturajmirashi7369 😂😂 probably
They want to be exercises solver machine
@@alimensah1153 Dont worry you are explaining the things right way, your students are bad not you :D
Could you please write larger, very diff to see.
You play video on 1080p
You're so bubbly! I like it!
Amazing lecture.but could you write a little big.
What an amazing video
Thanks :) no one explained this before!
Lost me on that one. First time.
Tutions in whics country
Very nice
School in Australia
The content and lecture delivery was perfect... the sound quality ruined everything.... I can't use headphones because it gives me headache at least a full day ... so no other option left for me....
He proved that the derivative of sin is cos by using some particular points but he has to prove that it is true for all points ... that’s how things work in maths in general
A bit confusing cause the local minimum of the sine curve should level up with the local minimum of the cosine curve. He didn't draw it correctly.
Wait wym, the local minimum is a stationary point?