Problem 27
Question
The specific gravity \(s\) of an object is given by the formula $$ s=\frac{A}{A-W} $$ where \(A\) is the number of pounds in the weight of the object in air and \(W\) is the number of pounds in the weight of the object in water. If the weight of an object in air is read as \(20 \mathrm{lb}\) with a possible error of \(0.01 \mathrm{lb}\) and its weight in water is read as \(12 \mathrm{lb}\) with a possible error of \(0.02 \mathrm{lb}\), find approximately the largest possible error in calculating \(s\) from these measurements. Also find the largest possible relative error.
Step-by-Step Solution
Verified Answer
The largest possible error in the specific gravity is approximately 0.004375, and the largest possible relative error is 0.175%.
1Step 1 - Write Down the Given Formula
The given formula for specific gravity is: \[ s = \frac{A}{A-W} \] where - \( A \) is the weight of the object in air - \( W \) is the weight of the object in water.
2Step 2 - Identify the Possible Errors in Measurements
Given: - Weight of the object in air \( A = 20 \text{ lb} \) with a possible error of \( \pm 0.01 \text{ lb} \). - Weight of the object in water \( W = 12 \text{ lb} \) with a possible error of \( \pm 0.02 \text{ lb} \).
3Step 3 - Use Differential to Approximate Error in Specific Gravity
The differential for the function \( s(A, W) = \frac{A}{A-W} \) can be used to estimate the error. We can write: \[ ds = \left( \frac{\partial s}{\partial A} \right) dA + \left( \frac{\partial s}{\partial W} \right) dW \] First, compute the partial derivatives.
4Step 4 - Compute the Partial Derivatives
Compute the partial derivatives of \( s \): \[ \frac{\partial s}{\partial A} = \frac{A - W + W}{(A - W)^2} = \frac{W}{(A - W)^2} \] \[ \frac{\partial s}{\partial W} = -\frac{A}{(A - W)^2} \]
5Step 5 - Substitute Values into the Partial Derivatives
Substitute \( A = 20 \), \( W = 12 \), \( dA = 0.01 \), and \( dW = 0.02 \) into the partial derivatives: \[ \frac{\partial s}{\partial A} = \frac{12}{(20 - 12)^2} = \frac{12}{64} = \frac{3}{16} = 0.1875 \] \[ \frac{\partial s}{\partial W} = -\frac{20}{(20 - 12)^2} = -\frac{20}{64} = -\frac{5}{16} = -0.3125 \]
6Step 6 - Compute the Error in Specific Gravity
Use the differential to approximate the error: \[ ds = \left( \frac{3}{16} \right)(0.01) + \left( -\frac{5}{16} \right)(0.02) \] Simplifying, we get: \[ ds = 0.1875 \times 0.01 - 0.3125 \times 0.02 \] \[ ds = 0.001875 - 0.00625 = -0.004375 \] The magnitude of the error is \( 0.004375 \).
7Step 7 - Calculate the Specific Gravity for Given Measurements
The specific gravity for \( A = 20 \) and \( W = 12 \) is: \[ s = \frac{20}{20 - 12} = \frac{20}{8} = 2.5 \]
8Step 8 - Compute the Largest Possible Relative Error
The relative error is computed as: \[ \frac{ds}{s} = \frac{0.004375}{2.5} = 0.00175 \] Therefore, the largest possible relative error is \(0.00175 \) or \( 0.175\% \).
Key Concepts
Error Propagation in MeasurementsRelative ErrorPartial Derivatives
Error Propagation in Measurements
Understanding error propagation is essential when dealing with measurements. Here, we look into how small errors in measurements of two variables affect the error in a resultant calculation. Error propagation helps us determine the uncertainty in the overall result when we know the uncertainties in individual measurements.
The formula for specific gravity is given by:
\[ s = \frac{A}{A-W} \]
where \( A \) is the weight in air and \( W \) is the weight in water. Given possible errors in these measurements, we use error propagation to find the error in \( s \).
We apply the following formula for error propagation in a function \( s = f(A, W) \):
\[ ds = \left( \frac{\partial s}{\partial A} \right) dA + \left( \frac{\partial s}{\partial W} \right) dW \]
This expression combines the partial derivatives of \( s \) with respect to \( A \) and \( W \), multiplied by their respective measurement errors, \( dA \) and \( dW \).
Errors from each source combine to affect the final uncertainty in the result. Knowing this helps us understand the impact of measurement precision and improve it wherever possible.
The formula for specific gravity is given by:
\[ s = \frac{A}{A-W} \]
where \( A \) is the weight in air and \( W \) is the weight in water. Given possible errors in these measurements, we use error propagation to find the error in \( s \).
We apply the following formula for error propagation in a function \( s = f(A, W) \):
\[ ds = \left( \frac{\partial s}{\partial A} \right) dA + \left( \frac{\partial s}{\partial W} \right) dW \]
This expression combines the partial derivatives of \( s \) with respect to \( A \) and \( W \), multiplied by their respective measurement errors, \( dA \) and \( dW \).
Errors from each source combine to affect the final uncertainty in the result. Knowing this helps us understand the impact of measurement precision and improve it wherever possible.
Relative Error
Relative error provides a dimensionless number that expresses the uncertainty of a measurement relative to the size of the measurement. It is an important concept because it helps us compare the precision of different measurements irrespective of their scales.
For the specific gravity calculation, the relative error formula is:
\[ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Measured Value}} \]
In this problem, the measured value of specific gravity is \( 2.5 \) and the absolute error is \( 0.004375 \). Therefore, the relative error is:
\[ \frac{0.004375}{2.5} = 0.00175 \text{ or } 0.175\% \]
This tells us that the error in the specific gravity measurement is 0.175% of its true value.
Relative error is useful because it represents the accuracy of the measurement in a normalized form, allowing for easy comparison between different measurements.
For the specific gravity calculation, the relative error formula is:
\[ \text{Relative Error} = \frac{\text{Absolute Error}}{\text{Measured Value}} \]
In this problem, the measured value of specific gravity is \( 2.5 \) and the absolute error is \( 0.004375 \). Therefore, the relative error is:
\[ \frac{0.004375}{2.5} = 0.00175 \text{ or } 0.175\% \]
This tells us that the error in the specific gravity measurement is 0.175% of its true value.
Relative error is useful because it represents the accuracy of the measurement in a normalized form, allowing for easy comparison between different measurements.
Partial Derivatives
Partial derivatives measure how a function changes as its input variables change, holding other variables constant. For multivariable functions, they are critical in understanding how each variable individually affects the output.
In our specific gravity problem, the function \( s \) depends on both \( A \) and \( W \). To find the error propagation, we found partial derivatives:
\[ \frac{\partial s}{\partial A} = \frac{W}{(A - W)^2} \text{ and } \frac{\partial s}{\partial W} = -\frac{A}{(A - W)^2} \]
Partial derivatives break down how changes in \( A \) and \( W \) impact \( s \):
These are then used to estimate the error in \( s \).
Partial derivatives are very handy in multi-variable calculus and error analysis, helping us break down complex dependencies into manageable pieces.
In our specific gravity problem, the function \( s \) depends on both \( A \) and \( W \). To find the error propagation, we found partial derivatives:
\[ \frac{\partial s}{\partial A} = \frac{W}{(A - W)^2} \text{ and } \frac{\partial s}{\partial W} = -\frac{A}{(A - W)^2} \]
Partial derivatives break down how changes in \( A \) and \( W \) impact \( s \):
- \( \frac{\partial s}{\partial A} \) reveals the sensitivity of \( s \) to changes in \( A \)
- \( \frac{\partial s}{\partial W} \) reveals the sensitivity of \( s \) to changes in \( W \)
These are then used to estimate the error in \( s \).
Partial derivatives are very handy in multi-variable calculus and error analysis, helping us break down complex dependencies into manageable pieces.
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