Chapter 24.6 - Inverse Proportions - Part 2

In the previous section we completed the discussion on Direct proportions. In this section we will see Inverse proportions. We have had a discussion about the basics of inverse proportions in chapter 3.8. Here we will continue that discussion.
Fig.24.15 below shows the central angle of various regular polygons.

Fig.24.15

• For a triangle, the central angle is 120^o
• For a square, the central angle is 90^o
• For a pentagon, the central angle is 72^o
■ We can calculate the central angle for any regular polygon. The formula is:
x = ³⁶⁰⁄_n. Where x is the central angle and n is the number of sides
• Let us write the results in a tabular form:

• We can see that the product nx is a constant. The value of the constant is 360. So we can write: nx = 360
• When one quantity (n or x) increases, the other quantity decreases. 
• When one quantity decreases, the other quantity increases. 
• It is a case of inverse proportion.
■ There is another way of saying this:
• We have nx = 360. This is same as x = ³⁶⁰⁄_n ⇒ x = 360 × ¹⁄_n
• So the angle x is proportional to the reciprocal of n

Another example:
1. An object is travelling from a point 'O' at a steady speed 's' of 10 m/s. It has to reach a point P which is 100 metres away. The time 't' required for the travel is ¹⁰⁰⁄₁₀ = 10 seconds
2. If the speed is 25 m/s, then t is ¹⁰⁰⁄₂₅ = 4 seconds. 
3. If the speed is 20 m/s, then t = ¹⁰⁰⁄₂₀ = 5 seconds. Let us tabulate the results:

4. We find that st is a constant. So s and t are inversely proportional to each other.

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Many laws in physics are stated in terms of proportions. The Newtons Law of Universal Gravitation is an example. Let us see it's details:
• Any two bodies in the universe attract each other
• The force of attraction is directly proportional to 'the product of their masses'
    ♦ So, if m₁ and m₂ are the masses of the two bodies, The force of attraction F is directly proportional to m₁m₂
• The force of attraction is inversely proportional to 'the square of the distance between them'
    ♦ So, if r is the distance between the two bodies, F is inversely proportional to r². That is., F is proportional to the reciprocal of r²
• Combining the above two, we can write: F is proportional to ^m₁m₂⁄_r²
■ This is written as F = G ^m₁m₂⁄_r². Where G is the constant of proportionality.

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We will now see some solved examples:
Solved example 24.10
In rectangles of area 1 square metre, as the length of one side changes, so does the length of the other side. Write the relation between the lengths as an algebraic equation. How do we say this in the language of proportions?
Solution:
1. Let the length of the rectangle be 'l' and width be 'b'
2. Then area 'a'of the rectangle = lb
3. Area is given as a constant which is 1 sq.m
4. So we can write lb = 1
This is the algebraic equation which gives the relation between length and width of a rectangle whose area is 1 sq.m
5. Now we see if there is any proportionality between length and width:
• We have lb = 1. So if length or width increases, the other decreases
• Also, if length or width decreases, the other increases
• Their product will remain constant only if this simultaneous increase and decrease take place.
6. We can write: l = ¹⁄_b ⇒ l = 1 × ¹⁄_b
• So length is proportional to the reciprocal of the width
• That means, length and width are inversely proportional
7. The constant of proportionality is 1

Solved example 24.11
In triangles of the same area, how do we say the relation between the length of the longest side, and the length of the perpendicular from the opposite vertex? What if we take the length of the shortest side instead?
Solution:
1. We are considering triangles of the same area. That means area is a constant. Let it be 'a'
• Let the length of the longest side be 'b'
• Let the length of the perpendicular from the opposite vertex to this longest side be 'h'
2. Then we have a = ¹⁄₂ bh
• This is same as bh = 2a
• Here 2a is a constant. So if b or h increases, the other decreases
• Also, if b or h decreases, the other increases 
• Their product will remain constant only if this simultaneous increase and decrease take place.
3. We can write:
b = ^2a⁄_h ⇒ b = 2a × ¹⁄_h. So b is inversely proportional to h
4. If we change the shape of the triangle while keeping the area the same, the longest side that we considered may be come the shortest side. Then h should increase proportionately so that the area will remain the same.

Solved example 24.12
A fixed volume of water is to flow into a rectangular water tank. The rate of flow can be changed using different water pipes. Write the relation between the following quantities as an algebraic equation and in terms of proportions.
(i) The rate of flow and the height of water level
(ii) The rate of flow and the time taken to fill the tank
Solution:
1. Let the rate of flow be 'r' m³/s. That means, in 1 second, 'r' m³ of water will enter the tank.
2. Let the base area of the tank be 'a' m²
• Then in the 1^st second  , that is., when t = 1, the height of water level will be:
^volume⁄_{Base area}  =  ^r⁄_a  [∵ after 1 second, the volume in the tank will be r m³]
• In the 2^nd second, that is., when t = 2, the height of water level will be:
^volume⁄_{Base area}  =  ^2r⁄_a  [∵ after 2 seconds, the volume in the tank will be 2r m³]
• In the 3^rd second, the height will be ^3r⁄_a 
3. So we can write:
• The height of water after the n^th second = h = ^nr⁄_a ⇒ h = ⁿ⁄_a × r
• n is a constant because we will put a particular value of n. We want the heigth of water at that n
• a is also a constant
• so ⁿ⁄_a  is a constant. 
4. Thus we have a relation between two quantities:
• height 'h' at the n^th second
• rate of flow 'r'
5. We can write: h = kr. This is the algebraic equation. Where K = ⁿ⁄_a 
6. From the equation, we can see that h is directly proportional to r. 
The constant of proportionality is k = ⁿ⁄_a 
part (ii):
1. We can use the same equation in (3). That is: h = ⁿ⁄_a × r 
2. In this case, h is a constant because of the following two reasons:
• A fixed volume of water is flowing into the tank
• When the tank is filled, it will have a particular value of 'h'. 
3. n is the number of seconds required to fill the tank. It will change if the rate r is increased or decreased.
4. So n and r are the variables. Let us bring them to opposite sides of the '=' sign:
• h = ⁿ⁄_a × r ⇒ r = ^ah⁄_n  ⇒ r = ah × ¹⁄_n . This is the algebraic equation.
• ah is a constant. Let it be 'k'. We can write: r = k × ¹⁄_n .
5. So r is proportional to the reciprocal of n. That means n is inversely proportional to r
• If r increases n decreases, indicating a lesser time sufficient to fill up the tank
• If r decreases, n increases, indicating a greater time required to fill up the tank

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We have completed the discussion on direct and inverse proportions. In the next section we discuss some topics in Statistics.

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Chapter 24.5 - Modifications to achieve proportions

In the previous section we saw some solved examples on proportionality between two quantities. In this section we will see a different type of proportion.

1. Consider a regular polygon. We know how to calculate the sum of all it’s interior angles. 
2. The formula is s = 180(n-2). 
• Where s is the sum 
• n is the number of sides of the regular polygon.
3. Let us use this formula for a triangle:
• For a triangle, n = 3
• So s = 180 × (3-2) = 180 × 1 = 180
4. Let us use the formula for a square:
• For a square, n = 4
• So s = 180 × (4-2) = 180 × 2 = 360
5. Let us use the formula for a pentagon:
• For a pentagon, n = 5
• So s = 180 × (5-2) = 180 × 3 = 540
6. Let us tabulate the results:

From the table we can see that s/n is not a constant. So s is not proportional to n.
7. Let us modify the formula a little:
Let s = 180 × m
• Where s is the sum
• m = (n-2) 
• n is the number of sides of the regular polygon.
Now the tabulation will be as shown below:

We can see that s is proportional to m. The constant of proportionality is 180.
8. In ordinary language, we can say this:
The sum of interior angles of a regular polygon is proportional to ‘2 less than the number of sides’.

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There are many examples where proportionality can be achieved by making modifications to one quantity. Let us see another example:
■ We have seen that the area of a square is not proportional to it’s side. Details here. The table that we saw is shown here again.

 • We know that the area of a square is not proportional to it’s side because a/s ratio is not a constant.
• But if we put p = s², a new table can be formed:

• We can see that a is proportional to p. The constant of proportionality is 1.
• In ordinary language, we can say this:
Area of a square is proportional to the square of it’s side.

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We will now see some solved examples:
Solved example 24.8
For circles, is the area proportional to the square of the radius? If so, what is the constant of proportionality?
Solution:
1. We have seen that the area of a circle is not proportional to it’s radius. Details here. 
But the area may be proportional to the 'square of the radius'. Let us try:
2. We know that area a = πr²
• Put r² = q
• Then a = πq
    ♦ π is a constant
    ♦ when q increases a increases
    ♦ When q decreases a decreases
3. So a is proportional to q. That means, a is proportional to the square of the radius.
4. The constant of proportionality is π.

Solved example 24.9
For equilateral triangles, is the area proportional to the square of the side? If so, what is the constant of proportionality?
Solution:
1. We know that area of an equilateral triangle is given by a = (^√3⁄₄)s² Details here.
Where s is the length of side.
• Put s² = q
• Then a = (^√3⁄₄)q
    ♦ ^√3⁄₄ is a constant
    ♦ when q increases a increases
    ♦ When q decreases a decreases
2. So a is proportional to q. That means, a is proportional to the square of the side.
3. The constant of proportionality is ^√3⁄₄.

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So we find that, in some cases, proportionality can be achieved by modifying one quantity. In the next section we will see Inverse proportions.

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Chapter 24.4 - More solved examples on Proportionality

In the previous section we saw some solved examples on proportionality between two quantities. In this section we will see a few more solved examples.

Solved example 24.5
In the angle shown in fig.24.10 below, an object moves along the slanting line. As the distance of the object from the vertex changes, it's height from the horizontal line also changes.

Fig.24.10

(i) Prove that height is proportional to the distance
(ii) Calculate the constant of proportionality for 30^o, 35^o and 60^o angles
Solution:
1. The given angle in fig.24.10 is reproduced in fig.24.11(a) below. Some modifications are also made:
• The blue object which moves along the slanting line is marked as Q
• A perpendicular is dropped from Q to the horizontal leg of the given angle. The foot of the perpendicular is marked as P

Fig.24.11

• So we get a right triangle APQ
2. There is also another right triangle ABC. This is our base triangle. That is., we are going to do the calculations based on ⊿ABC:
• We assume that BC is fixed at it's position.
• Also we assume that all sides of ⊿ABC are known.
3. But QP is not fixed. Because, Q can be at any point along the slanting line.
4. Now we find the relation between ⊿ABC and ⊿APQ:
• Angle at A is denoted as x. It is same for both the triangles
• Angle at P and B are both 90
• Angle at C and Q will both be equal to [180 – (90+x)] = (90-x)
• So the two triangles are have the same angles. They are similar triangles. We can apply theorem 19.2:
5. ^{Side opposite angle x in ABC}⁄_{Side opposite angle x in APQ} =
^{Side opposite angle (90-x) in ABC}⁄_{Side opposite angle (90-x) in APQ} =
^{Side opposite angle 90 in ABC}⁄_{Side opposite angle 90 in APQ} 
6. This is same as:
^BC⁄_PQ  = ^AB⁄_AP = ^AC⁄_AQ 
7. From the above, we will take the first and last. Because they connect the distance and height
^BC⁄_PQ  = ^AC⁄_AQ 
8. From this we get:
• AQ = ^AC⁄_BC × PQ
^{• AC}⁄_BC is a constant because ⊿ABC is fixed. We can calculate ^AC⁄_BC. So we can write:
• AQ = A constant × PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
Part (ii)
1. In this part we explore the cases when the angle x at vertex A is 30^o, 60^o and 45^o
2. First we will take 30^o. Consider the fig.24.11(b). A base triangle ABC is drawn in it. How is it drawn?
• First mark a point C such that AC = 2 cm. Next drop perpendicular BC onto the horizontal leg. Then:
    ♦ BC will be 1 cm
    ♦ AB will be √3 cm
• The above two will naturally occur because: √[(√3)²+1²] = √[3+1] = √[4] = 2
3. So we established a base triangle. As we saw in fig.(a), here also, ⊿ABC and ⊿APQ are similar.
4. So we get:
^BC⁄_PQ  = ^AB⁄_AP = ^AC⁄_AQ 
5. From the above, we will take the first and last. Because they connect the distance and height
^BC⁄_PQ  = ^AC⁄_AQ 
8. From this we get:
• AQ = ^AC⁄_BC × PQ ⇒ AQ = ²⁄₁ × PQ ⇒ AQ = 2 PQ
9. In the above result, '2' is a constant. So AQ is proportional to PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
• And the constant of proportionality is '2'

1. Next we will take 60^o. Consider the fig.24.11(c) below. 

Fig.24.12

A base triangle ABC is drawn in it. How is it drawn?
• First mark a point C such that AC = 2 cm. Next drop perpendicular BC onto the horizontal leg. Then:
    ♦ BC will be √3 cm
    ♦ AB will be 1 cm
• The above two will naturally occur because: √[(√3)²+1²] = √[3+1] = √[4] = 2
3. So we established a base triangle. As we saw in figs.24.11(a) and (b), here also, ⊿ABC and ⊿APQ are similar.
4. So we get:
^BC⁄_PQ  = ^AB⁄_AP = ^AC⁄_AQ 
5. From the above, we will take the first and last. Because they connect the distance and height
^BC⁄_PQ  = ^AC⁄_AQ 
8. From this we get:
• AQ = ^AC⁄_BC × PQ ⇒ AQ = ²⁄_√3 × PQ
9. In the above result, ²⁄_√3 is a constant. So AQ is proportional to PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
• And the constant of proportionality is ²⁄_√3

1. Finally we will take 45^o. Consider the fig.24.11(d) above.
A base triangle ABC is drawn in it. How is it drawn?
• First mark a point B on the horizontal line such that AB = 1 cm. Next erect perpendicular BC upto the slanting leg. Then:
    ♦ BC will be 1 cm
    ♦ AC will be √2 cm
• The above two will naturally occur because: √[1²+1²] = √[1+1] = √2
3. So we established a base triangle. As we saw in figs.24.11(a),(b) and (c), here also, ⊿ABC and ⊿APQ are similar.
4. So we get:
^BC⁄_PQ  = ^AB⁄_AP = ^AC⁄_AQ 
5. From the above, we will take the first and last. Because they connect the distance and height
^BC⁄_PQ  = ^AC⁄_AQ 
8. From this we get:
• AQ = ^AC⁄_BC × PQ ⇒ AQ = ^√2⁄₁ × PQ = √2PQ
9. In the above result, √2 is a constant. So AQ is proportional to PQ
• That means, the distance of Q from the vertex A is proportional to the height of Q from the horizontal line.
• And the constant of proportionality is √2

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Based on the above problem, we can have a discussion on a similar problem that is commonly encountered in science and engineering.
1. Consider fig.24.13 below. BC is a vertical line. 'A' is another point to the left of BC. This point is joined to B and C. Thus we get a triangle ABC. This is our base triangle. 

Fig.24.13

2. PQ is a line parallel to BC. This line can move towards the left or towards the right
3. • If it moves towards the left, it's distance from vertex A decreases. 
    • If it moves towards the right, it's distance from vertex A increases.
4. • Also, if it moves towards the left, it's own length PQ decreases  
    • If it moves towards the right, it's own length PQ increases
5. We have to check whether there is any proportionality between the two quantities:
(i) Distance of the line PQ from A
(ii) Length of the line PQ
6. For that, we add some details to the given fig.24.13(a). The modified fig. is 24.13(b). 
• A line AR is drawn through A, perpendicular to BC
• AR will be perpendicular to PQ also
• AR meets BC at D
7. Based on the calculations that we did in the solved example 24.5 above, we can make the following conclusions:
(i) ⊿ADC and ⊿ARQ are similar.
(ii) ⊿ADB and ⊿ARP are similar
8. From 7(i) we get:
^CD⁄_QR  = ^AD⁄_AR = ^AC⁄_AQ 
• From the above, we will take the first and second. Because they connect the distance from A and length of QR
^CD⁄_QR  = ^AD⁄_AR ⇒ QR = ^AR⁄_AD × CD
9. From 7(ii) we get:
^BD⁄_PR  = ^AD⁄_AR = ^AB⁄_AP 
• From the above, we will take the first and second. Because they connect the 'distance from A' and 'length of PR'
^BD⁄_PR  =  ^AD⁄_AR ⇒ PR = ^AR⁄_AD × BD
10. Now we add the results in (8) and (9):
QR + PR = (^AR⁄_AD × CD) + (^AR⁄_AD × BD)
⇒ QR + PR = ^AR⁄_AD × (CD +BD)
⇒ PQ = ^AR⁄_AD × BC [∵ (QR + PR) = PQ AND (CD +BD) = BC]
⇒ PQ = ^BC⁄_AD × AR
11. In the above result, ^BC⁄_AD is a constant because ΔABC is the base triangle, whose dimensions are known.
12. So PQ is proportional to AR 
• When AR, which is the distance from vertex A increases, the length PQ increases 
• When AR decreases, the length PQ decreases

• The constant of proportionality is ^BC⁄_AD

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Now we will consider a similar problem in circles:
1. Consider fig.24.14(a) below. ABC is a semi circle with centre at "O'. AOC is the diameter. OB is drawn perpendicular to the diameter. OB meets the semicircle at 'B'.    

Fig.24.14

2. Several chords are drawn on the left side of AOC. 
Chords nearer to B are shorter.
Chords away from B are longer
3. We have to check whether there is any proportionality between the two quantities:
(i) Distance of the chord from B
(ii) Length of the chord
4. For that, we modify the given fig.24.14(a). The modified fig. is 24.13(b). 
• A single chord EF is considered
• It meets OB at D
• EF is parallel to AC. So EF is perpendicular to OB
• OA is the radius 'r'. OE is also equal to 'r'
5. Now we can begin the calculations: Apply Pythagoras theorem to the right triangle OED.
We get: ED = √[OE² - OD²] = √[r² - OD²] = √[r² - (OB-BD)²] = √[r² - (r-BD)²]
 = √[r² - (r² - 2rBD + BD²)] = √[r² - r² + 2rBD - BD²] = √[2rBD - BD²]  = √[BD(2r - BD)].
6. The perpendicular OB from the centre will bisect the chord EF. Details here
So EF = 2ED = 2√[BD(2r - BD)]
7. So we get a relation between two quantities:
(i) The distance BD of the chord from B
(ii) The length of the chord EF
The relation is: EF = 2√[BD(2r - BD)]
8. But on the right side, we are getting square root of BD. So they are not proportional.

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Solved example 24.6

In calcium carbonate, the masses of calcium, carbon and oxygen are in the ratio 10:3:12. When 150 grams of a compound was analysed, it was found to contain 60 grams of calcium, 20 grams of carbon and 70 grams of oxygen. Is that compound calcium carbonate?
Solution:
1. The ratio is 10:3:12. So, if we divide a sample of calcium carbonate into (10+3+12 =)  25 equal parts, then:
• 10 such equal parts will be calcium 
• 3 such equal parts will be carbon
• 12 such equal parts will be oxygen
2. 150 grams of an unknown compound was analysed. Let us divide it into 25 equal parts. Then each part will be 150/25 = 6 grams
• 10 such equal parts = 10 × 6 = 60 grams. Analysis result also shows 60 grams 
• 3 such equal parts = 3 × 6 = 18 grams. Analysis result shows 20 grams. So the unknown compound is not calcium carbonate. 
• 12 such equal parts = 12 × 6 = 72 grams. Analysis result shows 70 grams. So the unknown compound is not calcium carbonate.

Solved example 24.7
A person invests Rs. 10000 and Rs. 15000 in two different schemes. After one year, he got Rs. 900 as interest for the first amount and Rs 1500 as interest for the second amount.
(i) Are the interests proportional to the investments?
(ii) What is the ratio of the interest to the amount invested in the first scheme? What about the second?
(iii) What is the annual rate of interest for the two schemes?
Solution:
1. We know that the interest (i) obtained is proportional to the principal amount (p)
• When p increases, i increases
• When p decreases, i decreases
• The constant of proportionality is the rate of interest (r)
2. So we get: i = pr. From this we get: r = i/p
• Thus, r for first scheme = 900/10000 = 9/100
    ♦ This is usually expressed as a percent. So r = 9/100 x 100 = 9%
• r for the second scheme = 1200/15000 = 12/150 = 4/50
    ♦ In percentage, it is 4/50 x 100 = 8%
3. So the annual rate of interest are:
First scheme - 9%
Second scheme - 8%
This is the solution for part (iii)
4. Since the rates are different, they are not proportional. This is the solution for part (i)
5. Part (ii):
• We need this ratio:
Interest : Amount invested
• Scheme 1:
900 : 10000 = 9 : 100
• Scheme 2:
1200 : 15000 = 12 :150 = 4 : 50 = 8 : 100

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In the next section we will see a different type of proportion.

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Chapter 24.2 - Examples of Proportionality

In the previous section we saw proportionality in the case of A0, A1, A2 series of writing papers. In this section we will see Proportionality in some more cases.

Let us write a summary of what we have discussed so far in this chapter:
■ The ‘3:2 rectangle family’ has all the rectangles whose length is always ³⁄₂ times the width
• That means length is always proportional to the width
   ♦ ³⁄₂ is the constant of proportionality
■ The ‘16:9 rectangle family’ has all the rectangles whose length is always ¹⁶⁄₉ times the width
• That means length is always proportional to the width
    ♦ ¹⁶⁄₉ is the constant of proportionality
■ The ‘√2:1 rectangle family’ has all the rectangles whose length is always √2 times the width
• That means length is always proportional to the width
    ♦ √2 is the constant of proportionality

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Now we will continue our discussion:
1. Consider a square of side 's' = 1 cm. It’s perimeter 'p' will be 4×1 = 4 cm
2. Let us change the side and see how it affects the perimeter:
• let s be 2 cm. Now p becomes 4×2 = 8 cm 
• let s be 2.5 cm. Now p becomes 4×2.5 = 10 cm
• let s be 0.5 cm. Now p becomes 4×0.5 = 2 cm
3. We will write the above results in a tabular form, and calculate the ^p⁄_s ratio in each case:

4. From the table it is clear that, ^p⁄_s is a constant. It’s value is 4
5. So ^p⁄_s = 4 ⇒ p = 4s
• Perimeter is always a constant times the side
• That means perimeter is proportional to the side.
• And 4 is the constant of proportionality
We can use the result in (5) as an equation to find p for any value of s

Another case:
1. Consider a square of side ‘s’= 1 cm. The length of it’s diagonal ‘d’ will be √2. See fig.16.5
[It is a simple application of the Pythagoras theorem. In this way we can find the diagonal of any given square] 
2. Let us change the side and see how it affects the diagonal:
• let s be 2 cm. Now d becomes  = √[2²+2²] = √[2(2²)] = 2√2 cm 
• let s be 2.5 cm. Now p becomes  = √[2.5²+2.5²] = √[2(2.5²)] = 2.5√2 cm 
• let s be 3.25 cm. Now p becomes  = √[3.25²+3.25²] = √[2(3.25²)] = 3.25√2 cm 
• let s be 0.3 cm. Now p becomes  = √[0.3²+0.3²] = √[2(0.3²)] = 0.3√2 cm 
3. We will write the above results in a tabular form, and calculate the d/s ratio in each case:

4. From the table it is clear that, ^d⁄_s is a constant. It’s value is √2
5. So ^d⁄_s = √2 ⇒ d = √2s
• Diagonal is always a constant times the side
• That means diagonal is proportional to the side.
• And √2 is the constant of proportionality
We can use the result in (5) as an equation to find d for any value of s

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In the above discussion, we were considering squares. We considered:

• Perimeter of squares • Diagonal of squares
Now we consider: • Area of squares
1. Consider a square of side ‘s’= 1 cm. It’s area ‘a’ will be 1² = 1 cm²
2. Let us change the side and see how it affects the area:
• let s be 2 cm. Now a becomes 2² = 4 cm²
• let s be 2.25 cm. Now a becomes 2.25² = 5.0625 cm²
• let s be 3 cm. Now a becomes 3² = 9 cm²
• let s be 0.4 cm. Now a becomes 0.4² = 0.16 cm²
3. We will write the above results in a tabular form, and calculate the a/s ratio in each case:

4. From the table, we can see that ^a⁄_s is not a constant.
We will not get a by multiplying s by a fixed number. So a is not proportional to s

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Now we will consider some examples in physics.

1. Consider an object moving at a steady speed of 10 m/s.
2. Since the speed is steady, it will travel a distance ‘d’ of 10 m in a time ‘t’ of one second
• In a time t of 2 seconds it will travel a distance ‘d’ of 2×10 = 20 m
• In a time t of 2.5 seconds it will travel a distance ‘d’ of 2.5×10 = 25 m
• In a time t of 4.25 seconds it will travel a distance ‘d’ of 4.25×10 = 42.5 m
• In a time t of 3.6 seconds it will travel a distance ‘d’ of 3.6×10 = 36 m
3. We will write the above results in a tabular form, and calculate the ^d⁄_t ratio in each case:

4. From the table it is clear that, ^d⁄_t is a constant. It’s value is 10
5. So ^d⁄_t = 10 ⇒ d = 10t
• Distance is always 'a constant × t'
• That means distance is proportional to the time.
• And 10 is the constant of proportionality. Note that, this 'constant of proportionality' is the 'steady speed'
We can use the result in (5) as an equation to find d for any value of t

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In the above example we saw an object which is moving at a steady speed. Now we will consider an object which is moving at a varying speed. 

1. An 'object dropped from a height' gives an example for such a motion. It’s speed will go on increasing. In physics classes we have derived the formula to calculate the distance travelled by such a object. Let us see the details:
• At the instance when the object is dropped, the distance travelled by it is zero
• After a time of t seconds, the distance travelled by it (from the spot where it is dropped) is given by 4.9t² metres
2. So, after 1 second, it will be at a distance of 4.9×1² = 4.9 m from the spot where it is dropped
• After 2 seconds, it will be at a distance of 4.9×2² = 19.6 m from the spot where it is dropped   
• After 2.5 seconds, it will be at a distance of 4.9×2.5² = 30.625 m from the spot where it is dropped
• After 4 seconds, it will be at a distance of 4.9×4² = 78.4 m from the spot where it is dropped
• After 4.2 seconds, it will be at a distance of 4.9×4.2² = 86.436 m from the spot where it is dropped
We will write the above results in a tabular form, and calculate the ^d⁄_t ratio in each case:

3. From the table, we can see that ^d⁄_t is not a constant.
We will not get d by multiplying t by a fixed number. So d is not proportional to t
4. In the earlier case we got ^d⁄_t as a constant. This is because, in that case, the object was moving with a steady speed. But in the present case, the speed is increasing.

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Another example from physics:

1. The density of a material is 12 kg/m3.
2. The mass 'm' of  volume 'v' 1 m³ of that material will be 1×12 = 12 kg
• The mass 'm' of volume 'v' 2 m³ of that material will be 2×12 = 24 kg
• The mass 'm' of  volume 'v' 15 m³ of that material will be 1.5×12 = 18 kg
• The mass 'm' of  volume 'v' 2.25 m³ of that material will be 2.25×12 = 27 kg
3. We will write the above results in a tabular form, and calculate the ^m⁄_v ratio in each case:

4. From the table it is clear that, ^m⁄_v is a constant. It’s value is 12
5. So ^m⁄_v = 12 ⇒ m = 12v
• Mass is always 'a constant × volume'
• That means mass is proportional to the volume.
• And 12 is the constant of proportionality. Note that, this 'constant of proportionality' is the 'density of the material'
We can use the result in (5) as an equation to find m for any value of v

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■ So we have considered a number of cases. In all those cases, there are two quantities.
• In some cases one quantity is proportional to the other. Some examples that we saw in this category are:
    ♦ In the ‘16:9 rectangle family’, the the length is always 16/9 times the width. The constant of proportionality is 16/9
    ♦ In the ‘family of squares’, the diagonal is always 2 times the side. The constant of proportionality is 2. (Note that, all squares belong to the ‘1:1 rectangle family’)
    ♦ The distance travelled by an object moving at a steady speed is always: the 'steady speed' times the time. The constant of proportionality is the ‘steady speed’ 
• In the rest of the cases, neither quantity is proportional to the other. Some examples that we saw in this category are:
    ♦ In the family of squares, the area is not proportional to the side
    ♦ For a freely falling body, the distance travelled is not proportional to the time

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In this chapter we consider those cases where one quantity is proportional to the other. In the next section we will see some solved examples.

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