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Anonymous Tue 28 Jul 21:50:34 2020 No.11947665 View Reply Original Report

Quoted By: >>11947719 >>11947769 >>11947808 >>11947908

Can someone explain to me what does it mean when someone in calculus says how fast a function is changing, what is the meaning of that, and if you could relate that to physics? I've watched Prof. Leonard's Calc 1 and 2 videos and I've got a pretty good understanding, but I feel I'm missing something, I still can't connect it to physics. I'm missing something.

Anonymous

Anonymous Tue 28 Jul 2020 22:04:50 No.11947719 Report

Quoted By:

>>11947665
f(t) = position at time t
how fast f changes = velocity

Anonymous

Anonymous Tue 28 Jul 2020 22:19:53 No.11947769 Report

Quoted By:

>>11947665
Imaging driving down a highway, you cross a sign labeled 10km to destination. lets use that as position, -10 km. Now look at your speedometer, it reads 100 km / h. and start a stopwatch

Just by sitting in your car we can get an intuition that your function is your position relative to time. and the rate at which that function is change is yes, your speedometer! velocity is how fast you are changing with time.

But wait there is more, imagine that there is traffic and you have to slow down, you press the breake, watching the speedometer fall, and a slight force into the seat belt. The rate at which your velocity changes is your acceleration.

Anonymous

Anonymous Tue 28 Jul 2020 22:29:50 No.11947808 Report

Quoted By:

>>11947665
I had no trouble understanding the theory, but in truth it doesn't help and having a solid algebraic footing is 90% of the battle ; most things in calculus are just pattern matching

Anonymous

Anonymous Tue 28 Jul 2020 23:11:50 No.11947908 Report

Quoted By:

>>11947665
let x be the x-coordinate of the position of a particle.
suppose that particle's position depends on time, then
then the particle's velocity (in the x-coordinate) is dx/dt
and the particle's acceleration (again, in the x-coordinate) is dv/dt = d^2 x / dt^2
we relate this to physics, by noting that F=ma relates the acceleration of a particle to the force on the particle, thus we have
F = m d^2x /dt^2

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