Problem 55
Question
The diameter of a sphere is measured as \(100 \pm 1 \mathrm{cm}\) and the volume is calculated from this measurement. Estimate the percent- age error in the volume calculation.
Step-by-Step Solution
Verified Answer
The percentage error in the volume calculation is approximately 3%.
1Step 1: Understand the Formula for Volume
The volume of a sphere is calculated using the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere. To find the radius given the diameter, we use \( r = \frac{d}{2} \).
2Step 2: Calculate the Radius
The given diameter of the sphere is \( 100 \pm 1 \) cm. Therefore, the radius \( r = \frac{100}{2} = 50 \) cm. The uncertainty in the radius is half of that in the diameter, so it is \( \pm 0.5 \) cm.
3Step 3: Derive the Formula for Volume Error
The error in the volume \( \Delta V \) can be estimated using propagation of error rules for multiplication and powers: \( \Delta V = 3 \times \frac{\Delta r}{r} \times V \). Since all are directly proportional, this leads to \( \frac{\Delta V}{V} \approx 3 \times \frac{\Delta r}{r} \).
4Step 4: Calculate the Percentage Error in Radius
The percentage error in the radius is \( \frac{\Delta r}{r} \times 100\), which is \( \frac{0.5}{50} \times 100\% = 1\% \).
5Step 5: Calculate the Percentage Error in Volume
Using the formula from Step 3, the percentage error in volume is \( 3 \times 1\% = 3\% \).
6Step 6: Conclusion
The percentage error in the volume calculation due to the diameter measurement error is approximately 3%.
Key Concepts
Volume of a SphereDiameter and Radius MeasurementPercentage Error CalculationError Propagation Rules
Volume of a Sphere
The volume of a sphere is a measurement of how much space it occupies. To calculate the volume of a sphere, we use the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Here, \( V \) stands for the volume, and \( r \) is the radius of the sphere. This formula shows that the volume is directly proportional to the cube of the radius. Thus, small changes in the radius can result in larger changes in the calculated volume.
Let's think about why the radius is cubed. Imagine inflating a balloon. As you blow more air, the balloon's surface stretches in three dimensions — forward, sideways, and upwards. Therefore, volumetric changes scale with the cube of the radius.
\[ V = \frac{4}{3} \pi r^3 \]
Here, \( V \) stands for the volume, and \( r \) is the radius of the sphere. This formula shows that the volume is directly proportional to the cube of the radius. Thus, small changes in the radius can result in larger changes in the calculated volume.
Let's think about why the radius is cubed. Imagine inflating a balloon. As you blow more air, the balloon's surface stretches in three dimensions — forward, sideways, and upwards. Therefore, volumetric changes scale with the cube of the radius.
Diameter and Radius Measurement
Understanding the relationship between the diameter and radius is crucial when dealing with spheres. The diameter of a sphere is the length of a straight line going from one side to the other, passing through the center. The radius is simply half of this distance. So, given a sphere's diameter \( d \), you can find the radius \( r \) using the formula:
\[ r = \frac{d}{2} \]
In our exercise, the diameter is given as \( 100 \pm 1 \) cm. This means the measured diameter might actually be anywhere between 99 cm and 101 cm.
\[ r = \frac{d}{2} \]
In our exercise, the diameter is given as \( 100 \pm 1 \) cm. This means the measured diameter might actually be anywhere between 99 cm and 101 cm.
- The calculated radius is \( 50 \pm 0.5 \) cm, reflecting that measurement errors in the diameter directly affect the radius.
Percentage Error Calculation
Percentage error helps us understand how significant an error is relative to the measurement size. It is calculated by dividing the absolute error by the actual measurement and then multiplying by 100% to get a percentage.
Such percentage error calculations are crucial for evaluating the accuracy and reliability of measurements, allowing us to quantify uncertainty in an understandable way.
- For the radius error \( \Delta r \), we have \( \Delta r = 0.5 \) cm.
- The percentage error in the radius is calculated as: \[ \text{Percentage Error in Radius} = \frac{\Delta r}{r} \times 100\% = \frac{0.5}{50} \times 100\% = 1\% \]
Such percentage error calculations are crucial for evaluating the accuracy and reliability of measurements, allowing us to quantify uncertainty in an understandable way.
Error Propagation Rules
Error propagation refers to how measurement errors affect a calculated value. When calculating a sphere's volume, any error in measuring the radius propagates through the volume formula, affecting the final result.
For our case:
For our case:
- Given the volume formula \( V = \frac{4}{3} \pi r^3 \), error propagation uses derivatives to find how each variable's uncertainty affects the result.
- For power functions (like \( r^3 \)), the rule states: \( \frac{\Delta V}{V} \approx 3 \times \frac{\Delta r}{r} \).
- We calculate the percentage error in the volume: \( 3 \times 1\% = 3\% \).
Other exercises in this chapter
Problem 54
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In Exercises \(53-58,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) . $$ f(u)=\cot \frac{\pi u}{10}, \quad u=g(x)=5 \sqrt{x}, \quad x
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