Posted: Mon Jul 31, 2006 1:29 pm Post subject: second derivative
i was looking over a math book, and i can not comprehend the idea of a second derivative
could someon explain it to me?
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rdrake
Posted: Mon Jul 31, 2006 2:44 pm Post subject: (No subject)
So, you've derived your equation and you now have F'(x). Congrats, you have the first derivative, which is slope/distance. Now, what if you wanted to derive F''(x) from F'(x), what would that give you? Why the velocity of course. When you decide to get F'''(x), you finally have the acceleration.
Your typical equations would go something like so:
Quote:
s(t) = x^2 - x + 5
v(t) = 2x - 1
a(t) = 2
I'm sure somebody else can explain it better though, this is just a simple explaination .
Oh, and a little tip when solving questions involving derivatives. Derive then set equal to 0, then solve for x/t/the variable.
Last Edit (I promise): Star pointed out to me that this is at an instant in time. Forgot to go mention that little bit.
1of42
Posted: Mon Jul 31, 2006 4:39 pm Post subject: (No subject)
The velocity example is good.
Velocity is how much distance is changing relative to time.
Acceleration is how much velocity is changing relative to time.
Thus, if the relationship between distance and time is F(t), the velocity is F'(t), and the acceleration is F''(t).
Cervantes
Posted: Mon Jul 31, 2006 4:42 pm Post subject: (No subject)
More edits!
Except these won't come in edit format.
Given your function f(x), f'(x) represents the slope of f(x) at your given point x. So far so good.
f''(x) is the slope of f'(x) at a given point x.
So, what does the slope of f'(x) represent when it comes to f(x)? If f''(x) is the slope of f'(x), and f'(x) is the slope of f(x), then f''(x) is the slope of the slope of f(x).
Now think physics. If f(x) represents the displacement of an object, then f'(x) is the rate of change (slope) of displacement, which is also referred to as velocity. f''(x) is the rate of change (slope) of velocity, which is also referred to as acceleration.
So the second derivative represents the "acceleration" of the graph f(x).
You could do this once again. Get the third derivative. It's the rate of change of acceleration. This is referred to as "jerk". The fourth derivative is the rate of change of jerk. It's referred to as "jerk^2". And so on.
Please note that cartoon_shark and I have said two different things. I've said f'(x) is like velocity and f''(x) is like acceleration, while he has said f''(x) is velocity and f'''(x) is acceleration. He's on crack.
rdrake
Posted: Mon Jul 31, 2006 4:45 pm Post subject: (No subject)
Cervantes wrote:
Please note that cartoon_shark and I have said two different things. I've said f'(x) is like velocity and f''(x) is like acceleration, while he has said f''(x) is velocity and f'''(x) is acceleration. He's on crack.
Pft, that's all that calculus is though. It's just physics on crack .
Tony
Posted: Mon Jul 31, 2006 7:19 pm Post subject: (No subject)
cartoon_shark wrote:
Pft, that's all that calculus is though. It's just physics on crack .
Actually Calculus is just a tool that is highly used in the field of Physics. Before Newton, one had to essentially write out limits for everything. Without calculators.. Ouch.