- #1
Johnny
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Can someone, who really knows and understands, tell me what dx (or whatever variables given) means behind the integral sign? I have seen more disagreement in Calculus books concerning this. Some authors say it's there just to "indicate the variable" your integrating with respect to, i.e., its not formally required. I had a mathematics professor tell me otherwise, namely that dx is a "real" variable and is required, multiplying at every point with f(x), defining a virtual infinite number of Reimannian rectangles as in S f(x)dx, (S means sum) and performing the normal integration.
Also, (and this ties in) if I have for example, a simple separable ODE such as:
m dv/dt = mg -kv
(where g is accelleration due to gravity, m is mass, v is velocity and k is frictional constant), and if we assume dv/dt is the standard Leibniz operator notation for a derivative, then how can one simply multiply dt through, when it's actually part of an operator? Now I have read that by "appropriately" defining dt, then defining dv as dv = v' dt, we get of course v' = dv/dt. Now, having defined dv/dt as a ratio of differentials, using the same Leibniz notation, we should NOW be able to use basic algebraic techniques with little worry. But wait, if we now integrate both sides:
S (1/(g - cv))dv = S dt where S means sum, and c = k/m,
I still don't know how to interpret S dt? Or now S (1/g-cv)dv for that matter? Do I sound confused? I am! Any help on this is greatly appreciated. I need to understand this.
Also, (and this ties in) if I have for example, a simple separable ODE such as:
m dv/dt = mg -kv
(where g is accelleration due to gravity, m is mass, v is velocity and k is frictional constant), and if we assume dv/dt is the standard Leibniz operator notation for a derivative, then how can one simply multiply dt through, when it's actually part of an operator? Now I have read that by "appropriately" defining dt, then defining dv as dv = v' dt, we get of course v' = dv/dt. Now, having defined dv/dt as a ratio of differentials, using the same Leibniz notation, we should NOW be able to use basic algebraic techniques with little worry. But wait, if we now integrate both sides:
S (1/(g - cv))dv = S dt where S means sum, and c = k/m,
I still don't know how to interpret S dt? Or now S (1/g-cv)dv for that matter? Do I sound confused? I am! Any help on this is greatly appreciated. I need to understand this.
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