Thread: What, exactly, is a rate of change?

1. What, exactly, is a rate of change?

Hello, this will sound stupid but I can't grasp what exactly a rate of change is. I know it's the rate at which something is increasing or decreasing etc, but what does that mean? Does rate in this context mean like the speed of change? Like the rate of change of y with respect to x would be the speed of change of y relative to the change of x?

For some reason I just can't wrap my mind around it and I've been doing calc and derivatives all year. Any help in a more elaborate meaning would be helpful.

2. Rate of Change is the slope of the Tangent line:

Which the first derivative is velocity, the norm of velocity is speed, and second derivative is acceleration.

3. Yeah, I know it's the slope of the tangent line. But for some reason I still have trouble interpreting like velocity when it is stated as "the rate of change of displacement with respect to time."

I have no problem with the harder calc problems but interpreting them in plain english for some reason is difficult for me. Thanks for the help though.

4. Originally Posted by JDabs
Yeah, I know it's the slope of the tangent line. But for some reason I still have trouble interpreting like velocity when it is stated as "the rate of change of displacement with respect to time."

I have no problem with the harder calc problems but interpreting them in plain english for some reason is difficult for me. Thanks for the help though.
This is good mate. Hard problems come from first principles, so a firm grasp of the basics is key!

What is a slope? Well a slope is simply rise over run. Think of it as elevation. So say we have a function and we want to know the instantaneous speed at a specific point. This means that we want the slope at that specific point.

Again, this is simply Rise over Run. However, with the derivation of derivatives we know this as the rate of change in the rise over the rate of change in the run. So essentially, we have the rate of change of the rise with respect to the run.
So it can then be said, we want to know how fast we are climbing (rise) with respect to how fast we are moving horizontally (run)
Does that clear it up at all?

Also visit: http://www.mathhelpforum.com/math-he...-tutorial.html

The link above us! It gives a great intro into calc/derivatives.

5. Originally Posted by AllanCuz
This is good mate. Hard problems come from first principles, so a firm grasp of the basics is key!

What is a slope? Well a slope is simply rise over run. Think of it as elevation. So say we have a function and we want to know the instantaneous speed at a specific point. This means that we want the slope at that specific point.

Again, this is simply Rise over Run. However, with the derivation of derivatives we know this as the rate of change in the rise over the rate of change in the run. So essentially, we have the rate of change of the rise with respect to the run.
So it can then be said, we want to know how fast we are climbing (rise) with respect to how fast we are moving horizontally (run)
Does that clear it up at all?
I think I understand what you're saying. So by rate in this context you mean like the speed/pace of change? Like, the rate of change of y with respect to x would be how fast y changes as x changes?

Thanks ton for the help so far.

6. Originally Posted by JDabs
I think I understand what you're saying. So by rate in this context you mean like the speed/pace of change? Like, the rate of change of y with respect to x would be how fast y changes as x changes?

Thanks ton for the help so far.
Absolutely correct.

If you see the form

$dy/dx$

This is slope of the tangent to the curve. Otherwise known as the rate of change.

So let us define our curve and go through an example.

$y=x^2$

Thus,

$y=dy/dx=2x$

So now we have a curve expressed as a function of x and the first derivative of that curve with respect to x. What does this mean?

Well if we plug in any point to our y that will be the rate of change of y with respect to x. In other words, how much Y is moving as X moves at that particular point.

So let us say

$x=2$

Then

$y`= 4$

Which means, for every unit we move in the x direction, we move 4 units in the y direction!

Does this make sense?

Also, the tutorial at the top of this forum is excellent!

http://www.mathhelpforum.com/math-he...-tutorial.html

7. Alright, that makes a lot of sense. I grasp it intuitively I just get lost in the wording sometimes haha. But hey, thanks again for the help. I really appreciate it.