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.
• For a triangle, the central angle is 120o
• For a square, the central angle is 90o
• For a pentagon, the central angle is 72o
■ 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.
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.
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 1st 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 2nd 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 3rd second, the height will be 3r⁄a
3. So we can write:
• The height of water after the nth 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 nth 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
We have completed the discussion on direct and inverse proportions. In the next section we discuss some topics in Statistics.
Fig.24.15 below shows the central angle of various regular polygons.
 |
| Fig.24.15 |
• For a square, the central angle is 90o
• For a pentagon, the central angle is 72o
■ 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.
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.
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 1st 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 2nd 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 3rd second, the height will be 3r⁄a
3. So we can write:
• The height of water after the nth 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 nth 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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