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
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
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
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
• 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.
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
■ 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
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.
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
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:
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
In the above discussion, we were considering squares. We considered:
• Perimeter of squares • Diagonal of squaresNow 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
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:
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
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.
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:
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
• 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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