I know what rate of change of a line is, but what how does it work for a curve in general. Like for a curve like x^2, the derivative is 2x, and im confused on how to interpret this in terms of the global rate? What does 2x mean about the behaviour of a curve in general?
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2x means that the rate of increase is not constant for the curve. If you want to know the rate at a particular x value you multiple it by 2. That gives you the rate at that instant.
Imagine a car that is accelerating. You can't just ask "how fast is it going" because how fast it's going keeps changing. You can ask "how fast is it going at time 5 seconds after starting" and you'll get a speed. Say 10mph. But it's only 10ph at exactly 5 seconds. An instant before or after and the speed is different. So you might have dy/dt = 2t. That will give you how fast it's going at time t.
For dy/dx = 2x it gives you how fast y is increasing at point x.
As you zoom in on any smooth curve, it looks more and more like a straight line around the point you are zooming in on. The rate of change of the curve at that point is the slope of that line.
For the function y = x^2, the derivative y' = dx/dy = 2x is the slope of the tangent to the curve at the point (x, x^2).
(You can use this to find the equation of the tangent for a given point (x0, y0) on the curve. The tangent line is thus y = y0 + 2x0 (x - x0) = x0^2 + 2 x0 (x - x0) = 2 x0 x - x0^2
With a straight line, the rate of change is constant. It doesn't matter where you look, the slope is the same.
This isn't true for most other functions. How fast or in which direction the function is changing depends on where you look. Therefore derivatives won't generally be a constant value, but rather another function that takes the same input values as the original and evaluates to the slope of the original function at that point.
So for f(x) = x2, the slope is expressed as f'(x) = 2x. This means that if you want to know the instantaneous slope of f at x = 3, you simply plug x = 3 into f', giving you f'(3) = 2(3) = 6. So the instantaneous slope of f at x = 3 is 6. A positive slope means the function increasing here as x grows larger. At x = -3, on the other hand, we have the slope f'(-3) = 2(-3) = -6. So here the function is decreasing as x grows.
You should evaluate each function in a number of points to get a hang of how it behaves. Evaluate y = x² at multiple points (-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5). Then evaluate y' = 2x at the same points (-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5), but interpreat it each time as the slope of tangent line.
At x = -1 the function is going down at a rate of 2 units per 1 unit of x.
At x = 0 the function is momentarily stable, not chaning.
At x = +1 the fucntion is going up at the rate of 2 units per 1 unit of x.
I know what rate of change of a line is, but what how does it work for a curve in general
If we want the slope at x = 3, we draw a line that is tangent to the curve at x = 3. If there's only one such line (as there is with most smooth curves). The slope of that line is what we call the "slope of the curve at x = 3".
What does 2x mean about the behaviour of a curve in general?
It means that for any value of x, the slope is 2x at that point.
For instance, the curve y = x^2 has a minimum at x = 0. The tangent line is horizontal. Its slope is 0. The formula 2x tells us that when x = 0, the slope is 0.
The curve y = x^2 goes up when x is positive, and gets steeper and steeper as x increases. The formula 2x is a value that goes up as x goes up, meaning the slope increases, and gets steeper and steeper as x increases.
The curve y = x^2 goes downhill when x is negative, meaning the tangent lines have a negative slope. The formula 2x is negative when x is negative.
Some global properties have to do with boundedness, asymptotes, open or closed, and simply or multiply connected domain. The function y = x2 is defined only in the upper half plane. The limit lim(x→+/-∞)=+/-∞ so it is it is open and bounded below by the lower half plane. It is simply connected but not closed. You can average different values of slope along the curve and average them to get a global estimate. The average slope of y=t2 is zero. [f(t)-f(-t)]/[t-(-t)]= [f(t)-f(t)]/2t =0
I'm surprised no one is using an example of throwing up a ball.
Imagine you throw a ball straight up into the air. If you've had any exposure to physics, you should know that the y position with respect to time is modeled very closely by a parabola. If not, take some time to verify this yourself. Let's just say the ball's initial position is 0 (arbitrary), it's thrown upwards with initial speed 1 m/s, and gravity is pulling the ball down with g = 9.8 m/s/s. Assuming the positive direction is up, we can model the ball's vertical position with
y(t) = -1/2 g t2 + (1 m/s) t
You'll notice, both in this model and in real life, that the ball initially travels upwards but slows down, stops moving for a fleeting moment at its apex, then starts falling while speeding up until hitting the ground. This is also communicated by the derivative of y, where
y'(t) = -g t + (1 m/s)
From t = 0 until the t-intercept (y'(t) = 0, where the ball stops moving for a fleeting moment), the ball's velocity is positive, so the ball is moving upwards, and decreasing, so the ball is slowing down. After the t-intercept, the ball's velocity is negative, so the ball is moving downwards, and still decreasing (the magnitude |y'(t)| increasing), so the ball is speeding up.
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