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Homework Help: Separation of Variables

  1. Jan 16, 2008 #1
    I do not understand the process of separating variables such as in derivatives. I thought that dy/dx is just the rate of change of y with respect to the independent variable x. Why can you physically move dx (like multiply it on both sides)?? What would "dy" be reffered to as then? Simply the change in y? I also to not get why dx, dy, dt, etc. are used in integrals when it is just (with respect to). They seem to be variables now instead of simply definitions. Please help me understand why we can separate variables and just everything related to it in general. (I'm only in High School btw, so please don't use abstruse terms or concepts) Thanks.
  2. jcsd
  3. Jan 16, 2008 #2

    Gib Z

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    dy is called the differential. It is defined to be equal to the expression [itex] f'(x) dx[/itex]. This leads to the conclusion that the derivative is the quotient of two differentials, though this is no special result but merely a tautology: it is just a verbal restatement of the definition of the derivative.

    The easiest way to think of this is as thus: For derivatives, we choose two sets of unknown values of a function. Representing them as co-ordinates, [itex](x, f(x) ), (x_1, f(x_1))[/itex]. The difference between the two y values is just [itex] f(x_1) - f(x)[/itex], and for the x values, [itex] x_1 - x[/itex].

    We often shorten that expression with the upper case delta, denoting "change in". So we can write the differences in the y and x values between the two points as [itex]\Delta y[/itex] and [itex]\Delta x[/itex].

    Now, we can take the limiting process of letting x_1 become arbitrarily close to x. That is the same as letting x_1 - x approach 0, or [itex] \Delta x \to 0[/itex]. The same is done for y.

    However, it comes up so often in calculus that we need to take this limit, writing [itex] \lim_{x_1 \to x} x_1 - x[/itex] that we have shortened this to the symbol dx. This is in agreement with the derivative being df/dx, as by the definition of our symbols df and dx, we get;

    [tex]\frac{dy}{dx} = \lim_{x_1 \to x} \lim_{f(x_1) \to f(x)} \frac{ f(x_1) - f(x)}{x_1 - x}[/tex]. We can omit one of the limit operators - the fact that we are assuming the derivative exists means we are also assuming continuity. Since f is continuous, the statement x_1 approaches x implies that f(x_1) approaches f(x). They are the same statement. Hence we have the same, normal definition.

    For integrals, it has exactly the same meaning. The Riemann definition of the integral defines the set of symbols [itex]\int^b_a f(x) dx[/itex] to mean [itex]\lim_{n\to \infty} \sum_{v=0}^{n} f(t_v) ( x_v - x_{v-1} ) [/itex], where the interval of integration, a= x_0, b= x_v, and t_v is some value satisfying [itex] x_v \leq t_v \leq x_{v-1}[/itex].

    Now, a nice geometric way to interpret this definition of the integral is as such;
    We graph the function f(x) over the closed interval [a,b]. Between these points, we construct (n-1) ordinates, basically divided the region into n slices. These slices need not be equal width. Each slice now has the general sub-interval, [x_p, x_(p+1)], p is some number between (inclusive) 1 and n-1. Then we have in the summation f(t_v). t_p is any value in the general closed sub-interval i just mentioned.

    So for each product in the summation, ie f(t_1) ( x_1 - x_0), we interpret as a rectangle with height f(t_1) and width ( x_1 - x_0). So now the summation part of our definition is reduced to the sum of n rectangles, which give the area under the function quite well, but with some error. Taking limits as n goes to infinity, there error goes to zero, whilst the interval [a,b] remains the same 'length' implies that x_v - x_(v-1) goes to zero. But from our earlier definition, that means we can put in the symbol dx for that. Thats why the dx appears in integrals, alone.
    Last edited: Jan 16, 2008
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