Chapter 13.10 - Derivatives

In the previous section, we completed a discussion on limits. In this section, we will see a practical application of limits.

Some basics can be written in 23 steps:

1. Consider an object dropped from the top of a tall cliff.
• The distance (s) traveled by the object in time (t) is given by the formula:
s = 4.9t²
    ♦ distance should be measured in meters
    ♦ time should be measured in seconds.
2. Using this formula, we can get several ordered pairs in the form (t,s):
(i) When t = 0, the distance traveled is zero.
    ♦ Because, s = 4.9 × 0² = 0
    ♦ So we get the ordered pair (t,s) = (0,0)
(ii)  When t = 2, the distance traveled is 19.6 m.
    ♦ Because, s = 4.9 × 2² = 19.6 m
    ♦ So we get the ordered pair (t,s) = A(2,19.6)
(iii)  When t = 4, the distance traveled is 78.4 m.
    ♦ Because, s = 4.9 × 4² = 78.4 m
    ♦ So we get the ordered pair (t,s) = B(4,78.4)
(iv)  When t = 6, the distance traveled is 176.4 m.
    ♦ Because, s = 4.9 × 6² = 176.4 m
    ♦ So we get the ordered pair (t,s) = C(6,176.4)
(v)  When t = 8, the distance traveled is 313.6 m.
    ♦ Because, s = 4.9 × 2² = 313.6 m
    ♦ So we get the ordered pair (t,s) = D(8,313.6)
(vi)  When t = 10, the distance traveled is 490 m.
    ♦ Because, s = 4.9 × 10² = 490 m
    ♦ So we get the ordered pair (t,s) = E(10,490)
(vii)  When t = 12, the distance traveled is 705.6 m.
    ♦ Because, s = 4.9 × 12² = 705.6 m
    ♦ So we get the ordered pair (t,s) = F(12,705.6)
3. We can plot the above ordered pairs on a graph.
    ♦ t should be plotted along the x-axis.
    ♦ s should be plotted along the y-axis.
• Such a graph is shown in fig.13.25 below:

Fig.13.25

• The points representing the ordered pairs are connected using a smooth curve. In the above fig., the smooth curve is drawn in red color.
4. Now begins our task. The task can be explained in 3 steps:
(i) We know that, velocity of a freely falling body would be increasing continuously. It would be a uniformly increasing velocity.
(ii) The task is to find the velocity of the object when t = 8 s
(iii) That is., we want to know the velocity with which the object moves, at the instant when the reading in the stop-watch is 8 s.
5. For this task, we need to concentrate on the region near t = 8 s in the graph.
• That is., we need to concentrate on the region near point D.
• More specifically, we need to concentrate on the region between C and D.
• So we will enlarge the region between C and D. The enlarged portion is shown in fig.13.26 below:

Fig.13.26

6. Let us write two distances:
• From C we know that, when the stop-watch shows 6 s, the distance already fallen is 176.4 m.
• From D we know that, when the stop-watch shows 8 s, the distance already fallen is 313.6 m
7. So the distance fallen in the time duration of 2 s from C to D is (313.6 – 176.4) = 137.2 m
• We know that, average velocity multiplied by the " time duration of travel" will give the distance traveled.
• Let v_av(C-D) be the average velocity with which the object fell from C to D. Then we can write: 137.2 = v_av(C-D) × 2
• From this, we get: $v_{av(C-D)}~=~\frac{137.2}{2}$ = 68.6 m/s
8. Note that, average velocity is a ratio.
• Numerator of the ratio is: Distance fallen.
    ♦ It is the altitude PD of the right triangle CPD in fig.13.26
• Denominator of the ratio is: Time duration in which this distance is fallen.
    ♦ It is the base PC of the right triangle CPD in fig.13.26
◼ So we can write:
$v_{av(C-D)}~=~\frac{PD}{PC}~=~\frac{137.2}{2}$ = 68.6 m/s
9. We have the well known formula to find the velocity at any instant:
v = u + at
• Where:
    ♦ v is the velocity at time t
    ♦ u is the initial velocity
    ♦ a is the acceleration.
• In our present case,
    ♦ u is zero.
    ♦ a is the acceleration due to gravity, which is 9.8 m/s²
• So when time is 8 s, we get: v = 0 + 9.8 × 8 = 78.4 m/s
• This is very different from the average velocity obtained in (7) above.
10. So it seems that, the method of average velocity does not help us to obtain the velocity at D.
• But remember that, the time duration was taken as 2 s. The object would fall a large distance during this time. Consequently the velocity change will also be large.
• Let us try a smaller time duration.
11. Let us consider a new point C₁ between C and D. It corresponds to t = 7 s. This is shown in fig.13.27(a) below:

Fig.13.27

• When t = 7, the distance traveled is 240.1 m.
    ♦ Because, s = 4.9 × 7² = 240.1 m
    ♦ So we get the ordered pair (t,s) = A(7,240.1)
• So the duration from C₁ to D is (8-7) = 1 s
• The distance fallen from C₁ to D is (313.6 – 240.1) = 73.5 m
• Then the average velocity can be obtained as:
$v_{av(C_1 - D)}~=~\frac{P_1 D}{P_1 C_1}~=~\frac{73.5}{1}$ = 73.5 m/s
• This is closer to the actual value of 78.4 m/s that we obtained in (9)
12. Let us decrease the time duration further.
• Consider a new point C₂ between C₁ and D. It corresponds to t = 7.5 s
• When t = 7.5, the distance traveled is 240.1 m.
    ♦ Because, s = 4.9 × 7.5² = 275.6 m
    ♦ So we get the ordered pair (t,s) = A(7.5,275.6)
• So the duration from C₂ to D is (8-7.5) = 0.5 s
• The distance fallen from C₂ to D is (313.6 – 275.6) = 38.0 m
• Then the average velocity can be obtained as:
$v_{av(C_2 - D)}~=~\frac{P_2 D}{P_2 C_2}~=~\frac{38.0}{0.5}$ = 76.0 m/s
• This is even more closer to the actual value of 78.4 m/s that we obtained in (9)
13. Let us decrease the time duration further.
• Consider a new point C₃ between C₂ and D. It corresponds to t = 7.75 s
• When t = 7.75, the distance traveled is 240.1 m.
    ♦ Because, s = 4.9 × 7.5² = 294.3 m
    ♦ So we get the ordered pair (t,s) = A(7.75,294.3)
• So the duration from C₃ to D is (8-7.75) = 0.25 s
• The distance fallen from C₃ to D is (313.6 – 294.3) = 19.3 m
• Then the average velocity can be obtained as:
$v_{av(C_3 - D)}~=~\frac{P_3 D}{P_3 C_3}~=~\frac{19.3}{0.25}$ = 77.2 m/s
• This is even more closer to the actual value of 78.4 m/s that we obtained in (9)
14. Let us write a summary of the above results. It can be written in 5 steps:
(i) The actual velocity at point D is 78.4 m/s
This is obtained from step (9)
(ii) When the time duration from C to D is 2 s, the average velocity is 68.6 m/s
This is obtained from step (7)
(iii) When the time duration from C₁ to D is 1 s, the average velocity is 73.5 m/s
This is obtained from step (11)
(iv) When the time duration from C₂ to D is 0.5 s, the average velocity is 76.0 m/s
This is obtained from step (12)
(v) When the time duration from C₃ to D is 0.25 s, the average velocity is 77.2 m/s
This is obtained from step (13)
15. So it is clear that:
As the time duration decreases, accuracy increases.
16. Note that, time duration is the length of a horizontal line segment.
• In fig.13.27(a), CP, C₁P₁, C₂P₂ and C₃P₃ are the horizontal line segments that we considered. Those line segments have decreasing lengths.
• If we decrease the "horizontal lengths" to a very small value like ‘0.00001’, then we will get a highly accurate result.
• In 0.00001, there are four zeros after the decimal point. There can be a million zeros after the decimal point. Then the length will be very close to zero.
• However, the length must not become exact zero. This is because, division by zero will give a number which is not defined.
• Also, the length cannot become -ve. This is because, the base of a triangle cannot be a -ve value.
• When the length is very close to zero, we say that, it is an infinitesimal length. The dictionary meaning can be seen here.
17. When the base CP (of the triangle CPD) is infinitesimal, we get an accurate result.
• That is:
The ratio $\frac{\text{PD corresponding to infinitesimal CP}}{\text{infinitesimal CP}}$ will give us the exact velocity at D
• The ratio $\frac{\text{PD corresponding to ordinary CP}}{\text{ordinary CP}}$ will give us only the average velocity with which the object travels from C to D.
(We use the word “corresponding: because:
For each length of the base CP, there will be a corresponding length for the altitude PD. Even when the base is infinitesimal, there will be a corresponding altitude)  
18. In the above steps, we considered the portion between C and D.
• We gradually decreased the horizontal distance between C and D. We saw that, by decreasing the horizontal distance to an infinitesimal length, we will get the exact velocity at D
• Just like the portion between C and D, we can consider the portion between D and E also. This is shown in fig.13.27(b) above.
    ♦ First take the points D and E. (time at E is 10 s)
    ♦ Next take D and E₁. (time at E₁ must be 9 s)
    ♦ Next take D and E₂. (time at E₂ must be 8.5 s)
    ♦ Next take D and E₃. (time at E₃ must be 8.25 s)
• In this way, we will be decreasing the horizontal distance between D and E. When the length become infinitesimal, we will get the velocity at D.
• The reader may write all the steps related to fig.13.27(b) and become convinced about this fact.
19. We see that, we can move from both sides.
    ♦ We can move from left towards D.
    ♦ We can move from right towards D.
• Both will give the same value for the velocity at D.
20. The velocity at D is called instantaneous velocity at D.
• It is called "instantaneous" because:
The velocity at D is valid only at the instant when the reading in the stop-watch is 8 s
    ♦ At any instant before 8 s, the velocity will be different.
    ♦ At any instant after 8 s, the velocity will be different.
21. At this stage, we can note an interesting fact. It can be written in steps:
(i) We calculated the velocity as a ratio:
$\frac{\rm{Vertical~distance~between~C~and~D}}{\rm{Horizontal~distance~between~C~and~D}}$
(ii) The vertical distance between C and D is the change in distance. (distance fallen up to D minus distance fallen up to C)
(iii) Horizontal distance between C and D is a time duration.
(iv) That means, we are dividing change in distance by time. So it is the rate of change of displacement.
• Indeed, we know that, velocity is the rate of change of displacement.
• When the denominator becomes infinitesimal, we get the instantaneous velocity.
(v) We can write:
When the denominator is infinitesimal, we are getting the instantaneous rate of change.
22. Remember that, the graph we plotted, is the graph of the function f(t) = s = 4.9t²
• We took a ratio in which:
    ♦ Denominator is infinitesimal.
    ♦ Numerator corresponds to that infinitesimal denominator.
• Such a ratio is called the derivative.
• We can say:
    ♦ The derivative of f(t)
    ♦ at t = 8 s
    ♦ is 78.4 m/s
23. Similarly, t can be any instant shown by the stop-watch.
• At that instant, by taking infinitesimal denominator, we will be able to find the derivative at that instant.
• In our present case, the derivative is the instantaneous velocity.
• In some other case, it will be the amount of water flowing into a reservoir at that instant.
• In some other case, it will be the pressure (at that instant), experienced by a balloon which is being filled with gas.
• Derivatives are very important in many science and engineering topics. In the next few sections, we will see easy methods to find derivatives.

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In the next section, we will see another application of derivatives. 

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