# Instantaneous Rate of Change at a Point

Derivatives #1b The Definition of the Derivative

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Darshan Senthil

## Key Questions

• The average rate of change of a function $f \left(x\right)$ on an interval $\left[a , b\right]$ is the slope of the secant line, which can be found by
$\frac{f \left(b\right) - f \left(a\right)}{b - a}$,
and the instantaneous rate of change of $f \left(x\right)$ at $x = a$ is the slope of the tangent line, which can be found by
$f ' \left(a\right)$.

• Yes, it is possible for the instantaneous rate of change to be 0.

For a specific example, imagine the function $f \left(x\right) = 3$. This is a horizontal line parallel to the x-axis at the value y=3. This function is unchanging for any value of x, therefore its rate of change is zero.

For a physics example, if I stand still in a spot for 5 minutes, then my rate of change of position after those 5 minutes is 0. Another physics example, if I walked for 5 minutes at a constant velocity of 2.5 miles per hour, then my rate of change of velocity would be zero.

However, a function need not be constant throughout to have the instantaneous rate of change at a given point be 0. For example, suppose that for five seconds I walk forward, and after those five seconds I turn and walk back the way I came for five seconds, ending up at my original position. When I started walking back, assuming that I was walking in the 'positive' direction initially, I began walking in the negative direction.

Thus, my initial velocity was positive (recall that velocity is not just speed, but speed and direction combined), but for the second half, my velocity was negative. If my velocity function is continuous, then at some point between when I was walking forward and when I walked back i must have had a velocity of 0; in other words, I must have stopped and turned around. At that point, my instantaneous rate of change of position was zero.

• Most certainly! When the instantaneous rate of change of a function at a given point is negative, it simply means that the function is decreasing at that point. As an example, given a function of the form $y = m x + b$, when m is positive, the function is increasing, but when m is negative, the function is decreasing. For a line, the rate of change at any given point is simply m.

This can also be seen in physics. In physics, when your velocity, or your rate of change of position, is positive, that means that you are moving in the 'positive' direction (such as towards the right on a number line). When your velocity is negative, it means that you are moving in the 'negative' direction (such as towards the left on the number line). Further, if your acceleration, your rate of change of velocity, is positive, it means that your velocity is increasing, and if your acceleration is negative, then your velocity is decreasing.

• For an estimation of the instantaneous rate of change of a function at a point, draw a line between two points ("reference points") very close to your desired point, and determine the slope of that line. You can improve the accuracy of your estimate by choosing reference points closer to your desired point.

Note In this explanation, I assume the reader is aware of and familiar with the calculus concept of limits. For those who are not, a link to a website with what I consider a good explanation of the concept is below.

http://www.mathsisfun.com/calculus/limits.html

In addition, here's a video I made several years ago that tries to give a clear and concise explanation of limits and derivatives in layman's terms. There's a brief volume spike at the 0:05 second mark that might startle viewers.

If you would prefer a written explanation, feel free to read on.
End Note

When looking at a graph of a function $f \left(x\right)$, one can graph the rate of change over a given interval of $x$. For example, if I state that I ran a distance of 3 miles over the course of 30 minutes, since we know my average velocity will equal my distance traveled divided by the time it took me, we calculate that my average velocity is 0.1 miles per minute, or 6 miles per hour. In this case we used the formula

${v}_{a , b} = \left[\frac{f \left(b\right) - f \left(a\right)}{b - a}\right]$

to calculate the average velocity $v$ of my distance function with $a = 0 , f \left(a\right) = 0 , b = 30 , f \left(b\right) = 3.$

On a mathematical graph, if my distance as a function of time were a curve, this value would be the slope of a secant line which intersects the curve at our two reference points (my starting distance of 0 at 0 minutes, and my final one of 3 miles at 30 minutes). If my actual velocity was constant at 0.1 miles per minute throughout the run, then this secant line is identical to my distance function.

However, if my velocity fluctuated at all, then the distance will be a curve as opposed to a line, and the secant will not be identical to my distance function, though it will at least still intersect at my two reference points (the beginning and end of my run in this case). In this case, I can get a better idea of my exact velocity at that point by choosing a secant line with reference points closer to my target point.

As an exercise, suppose that I tell you that the instantaneous rate of change of the function $f \left(x\right) = {x}^{3} - 1$ at the point $x = 2$ is 12. You can estimate this by using intervals of different size. In this case, try first the interval $\left[- 1 , 5\right]$ followed by the interval $\left[0 , 4\right]$ and then the interval $\left[1 , 3\right]$. You should notice that as your interval gets smaller, your estimate approaches the actual value of 12!

Citations:

Pierce, Rod. "Limits (An Introduction)" Math Is Fun. Ed. Rod Pierce. 13 Jan 2014. 11 Aug 2014 http://www.mathsisfun.com/calculus/limits.html

## Questions

• · 1 week ago
• · 4 months ago
• · 4 months ago
• · 4 months ago
• · 4 months ago
• · 5 months ago
• · 6 months ago
• · 8 months ago
• · 8 months ago
• · 8 months ago
• · 9 months ago
• · 9 months ago
• · 9 months ago
• · 10 months ago
• · 11 months ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago
• · 1 year ago