Anonymous

# if dy/dt means the derivative of y with respect to t, then what does simply dy mean? (or just dt for that?

matter?)

Update:

if you have an equation dy/dt = 3y+4, then the LHS is telling you that the RHS is a derivative of some function.

If that is true then how can you separate dy from dt?? I mean they are not a value or anything???

Relevance

dy/dt means more than "the derivative of y with respect to t". You must understand what it means to take a derivative.

The derivative is most commonly thought of as the slope at a given point. Let's say that you have a line:

y = 3x + 1

The slope of the line is 3, and you can use the derivative to find that the slope of the line tangent to each point on that line is also 3 (which makes sense).

What is the definition of slope?

(change in y) / (change in x)

What is the definition of a derivative?

(change in y) / (change in x) as the y - y0 -> 0 and x - x0 -> 0

The infinitesimal (very, very small) change between (x0, y0) and (x, y) is what dx and dy represent, respectively. Since dy, dx (and dt, dA, etc.) both represent just very small differences, you can separate them based on the same rules that you would use to separate macroscopic (y - y0, x - x0) differences.

So for:

dy/dt = 3y + 4

You can just do this:

(dy/dt) * dt = (3y + 4) * dt

dy = (3y + 4) * dt

Source(s): B.S. degrees in Physics and Mathematics
• They are called "differentials" and are the differences between two values as those values become arbitrarily close. For instance "dy" is the difference between two values of y, y1-y0, as that difference approaches zero.

dy/dt is called "the derivative of y with respect to t", but in fact it is the ratio of two differentials, both of which approach zero. So, in a sense, dy/dt = 0/0. And all this time, you've been taught that 0/0 is undefined. Well, in differential calculus, that ain't always so.

your example, dy/dt = 3y + 4, is telling you that the ratio of the differential y to the differential t is 3y+4 or that dy = (3y + 4)dt.

You'll get a lot of opportunity to "separate dy from dt" when you get into integral calculus and later on when you hit differential equations.

• dy is a measure of a very small change in y, and dt is the measure of very small change in t. dy/dt is the change in y with respect to t. Remember that the derivative is an equation for the slope of the curve at any value of t. So dy and dt are definite values, but more useful than either dy or dt is the ratio dy/dt. Just like when graphing a line, the knowing slope is more important than just knowing the rise.

• if f(t) is a differentiable function the df is termed the " differential of f " and is given by df = [ f"(t)] dt , and if g(t) = t then dg = 1 dt..{note that g is the identity function so that it moves in a 1 to 1 fashion with the domain movement }...dy / dt = 3y + 4 means find a function whose derivative is 3 times the function plus 4.....the only function which has a derivative being a multiple of itself is an exponential...so y(t) = e^(3t ) + K..but then if dy / dt = 3y + 4 we see that 3[ e^3t + K ] +4 = 3e^3t means that K = -4/3....solution is y = e^3t - 4/3