In this post, we shall discuss the concept of differentiation in greater detail. In the previous post, we learned what derivatives are and why we need to learn this concept. Now let us try to find out exactly what a derivative function tells us with respect to graphs.
Rate of change = Slope of the line
For most graphs, the steepness or the slope varies from point to point. Some parts are flat, some are less steep and some are very steep (see the graph below). This pattern indicates that the rate of change of the entity we are studying is NOT the same throughout.
- A positive derivative indicates that the function is increasing.
- A negative derivative means that the function is decreasing.
- Zero derivatives indicate that the function has some special behaviour at that given point.
Consider the following graph-
We want to find out the rate of change of the value of y, from point P to point T .This is called average rate change. To find the rate of change, we need to draw a line joining points P and T as shown in the figure above. This line, that cuts a curve in two (or more) parts is called a SECANT. In the graph above, the lines PQ and PS are secants.
Thus,
Average rate change = slope of the secant line joining P&T = Δy/Δx =change in y/change in x
Later suppose we want to find out the change between point P and point R.
Average rate of change = Slope of secant line joining P & R.
However, this slope will be different from the slope of the secant line joining points P and Q , as the steepness of both lines vary. So, basically the average rate of change (i.e slope) in both cases is NOT the same.
We now try to find instantaneous rate of change (rate of change at a given instant ). This means we try to find the derivative of the function y with respect to x.
To understand the difference between average rate of change and instantaneous rate of change lets consider two cases –
- Case 1 –
| x | y |
| 0 | 5 |
| 5 | 10 |
| 10 | 15 |
| 15 | 20 |
Average rate of change for 1st and 2nd values = y2-y1 / x2-x1 = 10-5 / 5-0 = 5/5 = 1.
Average rate of change for 2nd and 3rd values = y3-y2 / x3-x2 = 15-5 / 10-5 = 5/5 = 1.
In this case, the slope remains constant.
2. Case 2-
| x | y |
| 0 | 5 |
| 5 | 15 |
| 10 | 35 |
| 15 | 0 |
Here, the average rate of change for first and second values = y2-y1/ x2-x1
= 15-5 / 5-0
= 10/5
= 2 (∴ slope = 2)
Average rate of change for next set of values= y3-y2/x3-x2
= 35-15/ 10-5
= 20/5
= 4 (∴ slope = 4)
Average rate change for next set = y4-y3/ x4- x3
= 0-35 /15-10
= -35/5
= -7 (negative slope)
In this case, the average rate of change/slope is not constant. Thus, we will NOT get a straight line on the graph. We expect to get a curve with different slopes at different points. In this case, finding the average rate of change is not very helpful. Thus, we switch to the ‘instantaneous rate of change’ concept.
The instantaneous change can be represented by a tangent line on the graph. Why a tangent line? A tangent touches the curve roughly only at one point – that is at that instant. Thus the slope of the tangent represents the instantaneous change at that given point. We say that the function is differentiable at that point.
But how does one get a tangent line from a secant line?
If we gradually reduce the time interval i.e go on reducing the length of the secant line, as shown in fig below, at one point the secant becomes tangent at that point on the graph.
This basically means we are dividing our data into small parts i.e differentiating. This helps us to go from average rate change phenomenon (which is not constant due to fluctuating data) to instantaneous rate change at a given point. We find the rate changes at every point and later integrate everything to find the desired result.
We draw tangents at every desired point on the graph.In the figure below, we have drawn tangents at points P, Q,S, and T. Let’s see how that looks on the graph (fig 1) –
We then find the slope of these tangent lines. Why do we take slope? As previously stated, the slope represents the rate of change. We plot these slopes separately as shown in fig 2. And when we do this, guess what we get?
A STRAIGHT LINE !! And we know how great it is to work with straight lines on a graph! Thus, we find a derivative to get a straight line on a graph. This helps us simplify and better understand our experimental data!
e.g. While studying Freundlich adsorption isotherm (we shall study it in detail when we learn adsorption) we have to use this concept.The isotherm is expressed as:
x/m = kp1/n
x/m ⇒ amount of gas adsorbed per unit mass of adsorbent
p ⇒ equilibrium pressure of the gas and
k, n ⇒ arbitrary constants.
Plot of this equation will not give straight line but after taking logs on both sides we get a straight line.
In the next post we shall continue our discussion on derivatives.Till then,
Be a perpetual student of life and keep learning…
Good day!
References and Further Reading –
1)http://www.maths.surrey.ac.uk/explore/vithyaspages/differential.html
2)http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/derivative/slope.html
3)http://www-math.mit.edu/~djk/calculus_beginners/chapter01/section02.html
4)https://www.math10.com/en/algebra/derivative-function.html
5) https://math.dartmouth.edu/opencalc2/cole/lecture8.pdf
7) http://www.modelab.ufes.br/xioste/papers/xioste_paper105.pdf
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