In the previous section, we saw the details about absolute error and relative error. In this section, we will see combination of errors.
We will write the basics in steps:
1. Consider an experiment to determine the density of the material of an object
• We first find the mass of that object
• For that, we weigh that object in a balance
• We make several observations so as to get the most accurate mass
2. Then we find the volume of that object
• For finding the volume, we may have to measure it’s length, width and thickness
Or, we may have to measure the diameter
• Here also, we make several observations so as to get the most accurate dimension/dimensions
3. Even after taking all the necessary precautions, there will be some error in the measured mass
4. Also there will be some error in the measured dimensions
• So there will be some error in the calculated volume
5. When we apply the equation density = mass⁄volume, we are combining two quantities (mass and volume) which have errors in them
■ So we want to know 'how much error will be there in the density'
6. To make such estimates, we must learn how errors combine when we do the mathematical operations
• We can derive rules for each of the mathematical operations: addition/subtraction, multiplication/division
1. In an experiment, lengths of two rods are to be determined
■ The total length of the two rods is to be reported
2. Length of first rod is denoted as A
• Length of second rod is denoted as B
• Total length is denoted as Z
• So Z = A + B
3. But while measuring the lengths, errors (even though small), are unavoidable
• The length of the first rod can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the length of rod
• The length of the second rod can be written only as: (B ± 𝚫B) cm
♦ Where 𝚫B is the absolute error in the measurement of the length of second rod
4. So the total length Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
5. We have: Z ± 𝚫Z = (A ± 𝚫A) + (B ± 𝚫B)
= A + B + (± 𝚫A ± 𝚫B)
6. So the maximum possible value that (Z ± 𝚫Z) can have is: A + B + (+ 𝚫A + 𝚫B)
= A + B + ( 𝚫A + 𝚫B)
• The least possible value that (Z ± 𝚫Z) can have is: A + B + (- 𝚫A - 𝚫B)
= A + B - ( 𝚫A + 𝚫B)
7. So the total length has the probability to be anywhere between
[A + B + ( 𝚫A + 𝚫B)] and [A + B - ( 𝚫A + 𝚫B)]
• This can be represented as: [A+ B ± (𝚫A + 𝚫B)]
■ So the error 𝚫Z in Z is (𝚫A + 𝚫B)
1. In an experiment, lengths of two rods are to be determined
■ The difference in the lengths of the two rods is to be reported
2. Length of first rod is denoted as A
• Length of second rod is denoted as B
• Difference in length is denoted as Z'
• So Z' = A - B
3. But while measuring the lengths, errors (even though small), are unavoidable
• The length of the first rod can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the length of rod
• The length of the second rod can be written only as: (B ± 𝚫B) cm
♦ Where 𝚫B is the absolute error in the measurement of the length of second rod
4. So the difference Z' will contain some error. Let us denote it as 𝚫Z'
• We want to find this 𝚫Z'
5. We have: Z' ± 𝚫Z' = (A ± 𝚫A) - (B ± 𝚫B)
= A - B + (± 𝚫A) - (± 𝚫B)
= A - B + (± 𝚫A) + (± 𝚫B)
= A - B + (± 𝚫A ± 𝚫B)
6. So the maximum possible value that (Z' ± 𝚫Z') can have is: A - B + (+ 𝚫A + 𝚫B)
= A - B + ( 𝚫A + 𝚫B)
• The least possible value that (Z' ± 𝚫Z') can have is: A - B + (- 𝚫A - 𝚫B)
= A - B - ( 𝚫A + 𝚫B)
7. So the difference has the probability to be any where between
[A - B + ( 𝚫A + 𝚫B)] and [A - B - ( 𝚫A + 𝚫B)]
• This can be represented as: [A- B ± (𝚫A + 𝚫B)]
■ So the error 𝚫Z' in Z' is (𝚫A + 𝚫B)
• This is the same expression that we obtained for 𝚫Z
When two quantities are added or subtracted, the absolute error in the final result is the sum of the absolute errors in the individual quantities
The volumes of two bodies are (8.4 ± 0.03) cm3 and (3.7 ± 0.01) cm3. Find the sum and difference in volumes with error limits
Solution:
1. We have: V1 = (8.4 ± 0.03) cm3 and V2 = (3.7 ± 0.01) cm3
2. Sum of the absolute errors = (0.03 + 0.01) = 0.04
3. Sum of the volumes (V) = (8.4 + 3.7) ± 0.04 cm3 = 12.1 ± 0.04 cm3
4. Difference of the volumes (V') = (8.4 - 3.7) ± 0.04 cm3 = 4.7 ± 0.04 cm3
■ The product of the length and width is to be reported as the area
2. Length is denoted as A
• Width is denoted as B
• Area is denoted as Z
• So Z = AB
3. But while measuring the lengths, errors (even though small), are unavoidable
• The length can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the length
• The width can be written only as: (B ± 𝚫B) cm
♦ Where 𝚫B is the absolute error in the measurement of the width
4. So the area Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
5. We have: Z ± 𝚫Z = (A ± 𝚫A) × (B ± 𝚫B) = AB ± B𝚫A ± A𝚫B ± 𝚫A𝚫B
⇒ Z ± 𝚫Z = AB ± B𝚫A ± A𝚫B ± 𝚫A𝚫B
6. Let us divide both sides by Z
• But from (2), we have: Z = AB
• So we will divide left side by Z and right side by AB. We get:
$\mathbf\small{1 \pm \frac{\Delta Z}{Z}=1 \pm \frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm \frac{\Delta A \Delta B}{AB}}$
$\mathbf\small{\Rightarrow \frac{\Delta Z}{Z}=\frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm \frac{\Delta A \Delta B}{AB}}$
• 𝚫A and 𝚫B are small. So their product (𝚫A𝚫B) will be very small
♦ Example: (0.01× 0.02) = 0.0002
♦ We can ignore the last term containing (𝚫A𝚫B)
7. We get: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A} \pm \frac{\Delta B}{B}}$
• The error 𝚫Z will be maximum when we add the two terms on the right
• So we can discard the '-' sign from '±'
• We can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
• Note that, each of the three terms are 'relative errors'
8. What if we want to multiply 3 measurements?
An example:
If (U ± 𝚫U) = (A ± 𝚫A) × (B ± 𝚫B) × (C ± 𝚫C), find 𝚫U
Solution:
(i) (A ± 𝚫A) × (B ± 𝚫B) × (C ± 𝚫C)
= [(A ± 𝚫A) × (B ± 𝚫B)] × (C ± 𝚫C)
= [(Z ± 𝚫Z)] × (C ± 𝚫C)
⇒ (U ± 𝚫U) = (Z ± 𝚫Z) × (C ± 𝚫C)
(ii) Applying the relation in (7), we can write:
$\mathbf\small{\frac{\Delta U}{U}=\frac{\Delta Z}{Z}+\frac{\Delta C}{C}}$
(iii) But $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
• So we get 𝚫U from the relation:
$\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{\Delta C}{C}}$
• In this way, we can multiply any number of measurements we want
• This method can be used to find the volume of a rectangular prism, where we have to multiply length, width and height together
■ The quotient mass⁄volume is to be reported as the density
2. Mass is denoted as A
• Volume is denoted as B
• Density is denoted as Z
• So Z = A⁄B
3. But while measuring the mass, errors (even though small), are unavoidable
• The mass can be written only as: (A ± 𝚫A) kg
♦ Where 𝚫A is the absolute error in the measurement of the mass
4. The volume is calculated as the product of length, width and height
• All those 3 measurements will contain errors. So the volume will contain errors
• We have seen how to report the product of three 'measurements with errors'
• The volume that we have in our present experiment is B
• It will be written as: (B ± 𝚫B) cm3
♦ Where 𝚫B is the absolute error in the calculation of volume
5. So the density Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
6. We have: $\mathbf\small{Z\pm \Delta Z=\frac{(A \pm\Delta A)}{(B \pm\Delta B)}}$
$\mathbf\small{\Rightarrow Z\left(1\pm\frac{\Delta Z}{Z} \right)=\frac{A\left(1 \pm\frac{\Delta A}{A} \right)}{B\left(1 \pm\frac{\Delta B}{B} \right)}}$
7. But from (2), we have:
The Z on the left side is equal to A⁄B on the right side
• So they will cancel each other
• We will get:
$\mathbf\small{\left(1\pm\frac{\Delta Z}{Z} \right)=\frac{\left(1 \pm\frac{\Delta A}{A} \right)}{\left(1 \pm\frac{\Delta B}{B} \right)}}$
$\mathbf\small{\Rightarrow \left(1 \pm\frac{\Delta Z}{Z} \right)=\left(1 \pm\frac{\Delta A}{A} \right)\left(1 \pm \frac{\Delta B}{B} \right)^{-1}}$
8. The term $\mathbf\small{\left(1 \pm \frac{\Delta B}{B} \right)^{-1}}$ can be expanded by applying binomial theorem
• After simplification, the result in (7) will become:
$\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A} \pm \frac{\Delta B}{B}}$
9. The error 𝚫Z will be maximum when we add the two terms on the right
• So we can discard the '-' sign from '±'
• We can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
• Note that, each of the three terms are 'relative errors'
• This is the same equation that we obtained for multiplication. In both cases, the relative errors are being added
• Rule for multiplication:
When two quantities are multiplied, the relative error in the product is the sum of the relative errors in the multipliers
• Rule for division:
When a quantity (dividend) is divided by another quantity (divisor), the relative error in the quotient is the sum of the relative errors in the dividend and divisor
1. In an experiment, side of a square is to be determined
■ The square (raising to the power 2) of the side is to be reported as the area
2. Side is denoted as A
• Area is denoted as Z
• So Z = A2
3. But while measuring the side, errors (even though small), are unavoidable
• The side can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the side
4. So the area Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
5. We have: Z ± 𝚫Z = (A ± 𝚫A)2
⇒ Z ± 𝚫Z = (A ± 𝚫A) × (A ± 𝚫A)
6. So now we have a product of two measurements. We know how to find 𝚫Z in such a case
• Applying the rule for multiplication, we can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta A}{A}}$
• That means:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=2 \left(\frac{\Delta A}{A} \right)}$
7. What if the exponent is '3'?
We will see an example:
• In an experiment, side of a cube is to be determined
■ The cube (raising to the power 3) of the side is to be reported as the volume
8. Side is denoted as A
• Volume is denoted as Z
• So Z = A3
9. But while measuring the side, errors (even though small), are unavoidable
• The side can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the side
10. So the volume Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
11. We have: Z ± 𝚫Z = (A ± 𝚫A)3
⇒ Z ± 𝚫Z = (A ± 𝚫A) × (A ± 𝚫A) × (A ± 𝚫A)
12. So now we have a product of three measurements. We know how to find 𝚫Z in such a case
• Applying the rule for multiplication, we can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta A}{A}+\frac{\Delta A}{A}}$
• That means:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=3 \left(\frac{\Delta A}{A} \right)}$
13. In general, we can write:
• If Z = Ap, then $\mathbf\small{\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)}$
We will write the basics in steps:
1. Consider an experiment to determine the density of the material of an object
• We first find the mass of that object
• For that, we weigh that object in a balance
• We make several observations so as to get the most accurate mass
2. Then we find the volume of that object
• For finding the volume, we may have to measure it’s length, width and thickness
Or, we may have to measure the diameter
• Here also, we make several observations so as to get the most accurate dimension/dimensions
3. Even after taking all the necessary precautions, there will be some error in the measured mass
4. Also there will be some error in the measured dimensions
• So there will be some error in the calculated volume
5. When we apply the equation density = mass⁄volume, we are combining two quantities (mass and volume) which have errors in them
■ So we want to know 'how much error will be there in the density'
6. To make such estimates, we must learn how errors combine when we do the mathematical operations
• We can derive rules for each of the mathematical operations: addition/subtraction, multiplication/division
Error of a sum
We will write the derivation by demonstrating an example in steps:1. In an experiment, lengths of two rods are to be determined
■ The total length of the two rods is to be reported
2. Length of first rod is denoted as A
• Length of second rod is denoted as B
• Total length is denoted as Z
• So Z = A + B
3. But while measuring the lengths, errors (even though small), are unavoidable
• The length of the first rod can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the length of rod
• The length of the second rod can be written only as: (B ± 𝚫B) cm
♦ Where 𝚫B is the absolute error in the measurement of the length of second rod
4. So the total length Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
5. We have: Z ± 𝚫Z = (A ± 𝚫A) + (B ± 𝚫B)
= A + B + (± 𝚫A ± 𝚫B)
6. So the maximum possible value that (Z ± 𝚫Z) can have is: A + B + (+ 𝚫A + 𝚫B)
= A + B + ( 𝚫A + 𝚫B)
• The least possible value that (Z ± 𝚫Z) can have is: A + B + (- 𝚫A - 𝚫B)
= A + B - ( 𝚫A + 𝚫B)
7. So the total length has the probability to be anywhere between
[A + B + ( 𝚫A + 𝚫B)] and [A + B - ( 𝚫A + 𝚫B)]
• This can be represented as: [A+ B ± (𝚫A + 𝚫B)]
■ So the error 𝚫Z in Z is (𝚫A + 𝚫B)
Error of a difference
1. In an experiment, lengths of two rods are to be determined
■ The difference in the lengths of the two rods is to be reported
2. Length of first rod is denoted as A
• Length of second rod is denoted as B
• Difference in length is denoted as Z'
• So Z' = A - B
3. But while measuring the lengths, errors (even though small), are unavoidable
• The length of the first rod can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the length of rod
• The length of the second rod can be written only as: (B ± 𝚫B) cm
♦ Where 𝚫B is the absolute error in the measurement of the length of second rod
4. So the difference Z' will contain some error. Let us denote it as 𝚫Z'
• We want to find this 𝚫Z'
5. We have: Z' ± 𝚫Z' = (A ± 𝚫A) - (B ± 𝚫B)
= A - B + (± 𝚫A) - (± 𝚫B)
= A - B + (± 𝚫A) + (± 𝚫B)
= A - B + (± 𝚫A ± 𝚫B)
6. So the maximum possible value that (Z' ± 𝚫Z') can have is: A - B + (+ 𝚫A + 𝚫B)
= A - B + ( 𝚫A + 𝚫B)
• The least possible value that (Z' ± 𝚫Z') can have is: A - B + (- 𝚫A - 𝚫B)
= A - B - ( 𝚫A + 𝚫B)
7. So the difference has the probability to be any where between
[A - B + ( 𝚫A + 𝚫B)] and [A - B - ( 𝚫A + 𝚫B)]
• This can be represented as: [A- B ± (𝚫A + 𝚫B)]
■ So the error 𝚫Z' in Z' is (𝚫A + 𝚫B)
• This is the same expression that we obtained for 𝚫Z
■ So we can write the rule for addition/subtraction:
Solved example 2.21
Solution:
1. We have: V1 = (8.4 ± 0.03) cm3 and V2 = (3.7 ± 0.01) cm3
2. Sum of the absolute errors = (0.03 + 0.01) = 0.04
3. Sum of the volumes (V) = (8.4 + 3.7) ± 0.04 cm3 = 12.1 ± 0.04 cm3
4. Difference of the volumes (V') = (8.4 - 3.7) ± 0.04 cm3 = 4.7 ± 0.04 cm3
Error of a product
1. In an experiment, length and width of a rectangle is to be determined■ The product of the length and width is to be reported as the area
2. Length is denoted as A
• Width is denoted as B
• Area is denoted as Z
• So Z = AB
3. But while measuring the lengths, errors (even though small), are unavoidable
• The length can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the length
• The width can be written only as: (B ± 𝚫B) cm
♦ Where 𝚫B is the absolute error in the measurement of the width
4. So the area Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
5. We have: Z ± 𝚫Z = (A ± 𝚫A) × (B ± 𝚫B) = AB ± B𝚫A ± A𝚫B ± 𝚫A𝚫B
⇒ Z ± 𝚫Z = AB ± B𝚫A ± A𝚫B ± 𝚫A𝚫B
6. Let us divide both sides by Z
• But from (2), we have: Z = AB
• So we will divide left side by Z and right side by AB. We get:
$\mathbf\small{1 \pm \frac{\Delta Z}{Z}=1 \pm \frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm \frac{\Delta A \Delta B}{AB}}$
$\mathbf\small{\Rightarrow \frac{\Delta Z}{Z}=\frac{\Delta A}{A} \pm \frac{\Delta B}{B} \pm \frac{\Delta A \Delta B}{AB}}$
• 𝚫A and 𝚫B are small. So their product (𝚫A𝚫B) will be very small
♦ Example: (0.01× 0.02) = 0.0002
♦ We can ignore the last term containing (𝚫A𝚫B)
7. We get: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A} \pm \frac{\Delta B}{B}}$
• The error 𝚫Z will be maximum when we add the two terms on the right
• So we can discard the '-' sign from '±'
• We can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
• Note that, each of the three terms are 'relative errors'
8. What if we want to multiply 3 measurements?
An example:
If (U ± 𝚫U) = (A ± 𝚫A) × (B ± 𝚫B) × (C ± 𝚫C), find 𝚫U
Solution:
(i) (A ± 𝚫A) × (B ± 𝚫B) × (C ± 𝚫C)
= [(A ± 𝚫A) × (B ± 𝚫B)] × (C ± 𝚫C)
= [(Z ± 𝚫Z)] × (C ± 𝚫C)
⇒ (U ± 𝚫U) = (Z ± 𝚫Z) × (C ± 𝚫C)
(ii) Applying the relation in (7), we can write:
$\mathbf\small{\frac{\Delta U}{U}=\frac{\Delta Z}{Z}+\frac{\Delta C}{C}}$
(iii) But $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
• So we get 𝚫U from the relation:
$\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}+\frac{\Delta C}{C}}$
• In this way, we can multiply any number of measurements we want
• This method can be used to find the volume of a rectangular prism, where we have to multiply length, width and height together
Error of a quotient
1. In an experiment, mass and volume of a rectangular prism is to be determined■ The quotient mass⁄volume is to be reported as the density
2. Mass is denoted as A
• Volume is denoted as B
• Density is denoted as Z
• So Z = A⁄B
3. But while measuring the mass, errors (even though small), are unavoidable
• The mass can be written only as: (A ± 𝚫A) kg
♦ Where 𝚫A is the absolute error in the measurement of the mass
4. The volume is calculated as the product of length, width and height
• All those 3 measurements will contain errors. So the volume will contain errors
• We have seen how to report the product of three 'measurements with errors'
• The volume that we have in our present experiment is B
• It will be written as: (B ± 𝚫B) cm3
♦ Where 𝚫B is the absolute error in the calculation of volume
5. So the density Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
6. We have: $\mathbf\small{Z\pm \Delta Z=\frac{(A \pm\Delta A)}{(B \pm\Delta B)}}$
$\mathbf\small{\Rightarrow Z\left(1\pm\frac{\Delta Z}{Z} \right)=\frac{A\left(1 \pm\frac{\Delta A}{A} \right)}{B\left(1 \pm\frac{\Delta B}{B} \right)}}$
7. But from (2), we have:
The Z on the left side is equal to A⁄B on the right side
• So they will cancel each other
• We will get:
$\mathbf\small{\left(1\pm\frac{\Delta Z}{Z} \right)=\frac{\left(1 \pm\frac{\Delta A}{A} \right)}{\left(1 \pm\frac{\Delta B}{B} \right)}}$
$\mathbf\small{\Rightarrow \left(1 \pm\frac{\Delta Z}{Z} \right)=\left(1 \pm\frac{\Delta A}{A} \right)\left(1 \pm \frac{\Delta B}{B} \right)^{-1}}$
8. The term $\mathbf\small{\left(1 \pm \frac{\Delta B}{B} \right)^{-1}}$ can be expanded by applying binomial theorem
• After simplification, the result in (7) will become:
$\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A} \pm \frac{\Delta B}{B}}$
9. The error 𝚫Z will be maximum when we add the two terms on the right
• So we can discard the '-' sign from '±'
• We can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta B}{B}}$
• Note that, each of the three terms are 'relative errors'
• This is the same equation that we obtained for multiplication. In both cases, the relative errors are being added
■ So we can write the rules:
When two quantities are multiplied, the relative error in the product is the sum of the relative errors in the multipliers
• Rule for division:
When a quantity (dividend) is divided by another quantity (divisor), the relative error in the quotient is the sum of the relative errors in the dividend and divisor
Error in the case of a measured quantity raised to a power
1. In an experiment, side of a square is to be determined
2. Side is denoted as A
• Area is denoted as Z
• So Z = A2
3. But while measuring the side, errors (even though small), are unavoidable
• The side can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the side
4. So the area Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
5. We have: Z ± 𝚫Z = (A ± 𝚫A)2
⇒ Z ± 𝚫Z = (A ± 𝚫A) × (A ± 𝚫A)
6. So now we have a product of two measurements. We know how to find 𝚫Z in such a case
• Applying the rule for multiplication, we can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta A}{A}}$
• That means:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=2 \left(\frac{\Delta A}{A} \right)}$
7. What if the exponent is '3'?
We will see an example:
• In an experiment, side of a cube is to be determined
■ The cube (raising to the power 3) of the side is to be reported as the volume
8. Side is denoted as A
• Volume is denoted as Z
• So Z = A3
9. But while measuring the side, errors (even though small), are unavoidable
• The side can be written only as: (A ± 𝚫A) cm
♦ Where 𝚫A is the absolute error in the measurement of the side
10. So the volume Z will contain some error. Let us denote it as 𝚫Z
• We want to find this 𝚫Z
11. We have: Z ± 𝚫Z = (A ± 𝚫A)3
⇒ Z ± 𝚫Z = (A ± 𝚫A) × (A ± 𝚫A) × (A ± 𝚫A)
12. So now we have a product of three measurements. We know how to find 𝚫Z in such a case
• Applying the rule for multiplication, we can write:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=\frac{\Delta A}{A}+\frac{\Delta A}{A}+\frac{\Delta A}{A}}$
• That means:
Maximum 𝚫Z is given by the relation: $\mathbf\small{\frac{\Delta Z}{Z}=3 \left(\frac{\Delta A}{A} \right)}$
13. In general, we can write:
• If Z = Ap, then $\mathbf\small{\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)}$
14. Based on the above, we can derive two more results:
(A) If Z = (Ap × Bq), then $\mathbf\small{\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)+q \left(\frac{\Delta B}{B} \right)}$
Proof:
(i) We have:
Z ± 𝚫Z = [(A ± 𝚫A) × (A ± 𝚫A) × . . . p terms] × [(B ± 𝚫B) × (B ± 𝚫B) × . . . q terms]
(ii) Then:
$\mathbf\small{\frac{\Delta Z}{Z}=\left(\frac{\Delta A}{A}+\frac{\Delta A}{A}+\;.\;.\;.\;\text{p terms} \right)+\left(\frac{\Delta B}{B}+\frac{\Delta B}{B}+\;.\;.\;.\;\text{q terms} \right)}$
$\mathbf\small{\Rightarrow\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)+q \left(\frac{\Delta B}{B} \right)}$
(B) If $\mathbf\small{Z=\frac{A^p B^q}{C^r}}$, then $\mathbf\small{\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)+q \left(\frac{\Delta B}{B} \right)+r \left(\frac{\Delta C}{C} \right)}$
• Proof can be written using binomial theorem and the quotient rule
The relative error in a physical quantity raised to the power k is k times the relative error in the individual quantity
(A) If Z = (Ap × Bq), then $\mathbf\small{\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)+q \left(\frac{\Delta B}{B} \right)}$
Proof:
(i) We have:
Z ± 𝚫Z = [(A ± 𝚫A) × (A ± 𝚫A) × . . . p terms] × [(B ± 𝚫B) × (B ± 𝚫B) × . . . q terms]
(ii) Then:
$\mathbf\small{\frac{\Delta Z}{Z}=\left(\frac{\Delta A}{A}+\frac{\Delta A}{A}+\;.\;.\;.\;\text{p terms} \right)+\left(\frac{\Delta B}{B}+\frac{\Delta B}{B}+\;.\;.\;.\;\text{q terms} \right)}$
$\mathbf\small{\Rightarrow\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)+q \left(\frac{\Delta B}{B} \right)}$
(B) If $\mathbf\small{Z=\frac{A^p B^q}{C^r}}$, then $\mathbf\small{\frac{\Delta Z}{Z}=p \left(\frac{\Delta A}{A} \right)+q \left(\frac{\Delta B}{B} \right)+r \left(\frac{\Delta C}{C} \right)}$
• Proof can be written using binomial theorem and the quotient rule
■ So we can write the rule:
In the next section, we will see some solved examples
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