Problem 201 - Calculus 3

Question

The radius \(r\) and height \(h\) of a right circular cylinder are measured with possible errors of 4\(\%\) and 5\(\%,\) respectively. Approximate the maximum possible percentage error in measuring the volume (Recall that the percentage error is the ratio of the amount of error over the original amount. So, in this case, the percentage error in \(V\) is given by \(\frac{d V}{V} . )\)

Step-by-Step Solution

Answer

The maximum possible percentage error in measuring the volume is 13\%.

Step 1: Recall the Volume Formula

The first step is to remember the formula for the volume of a right circular cylinder, which is given by \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height.

Step 2: Derive the Expression for Error in Volume

To find the error in volume \( dV \), we need to differentiate \( V \) with respect to both \( r \) and \( h \). Using partial derivatives, we get: \( dV = \frac{\partial V}{\partial r} dr + \frac{\partial V}{\partial h} dh \). This leads to \( dV = 2\pi rh \, dr + \pi r^2 \, dh \).

Step 3: Calculate Percentage Errors for \( r \) and \( h \)

The initial step to calculate \( dr \) and \( dh \) is to express them in terms of the original measurements as percentages: \( dr = 0.04r \) and \( dh = 0.05h \).

Step 4: Substitute Error Values into Error Expression

Substitute the expressions for \( dr \) and \( dh \) from the percentage errors into the error expression: \( dV = 2\pi rh(0.04r) + \pi r^2(0.05h) \). Simplifying gives \( dV = 0.08\pi r^2 h + 0.05\pi r^2 h \).

Step 5: Simplify and Calculate the Maximum Percentage Error

Combine terms to simplify \( dV = 0.13\pi r^2 h \). Divide by the original volume \( V = \pi r^2 h \) to find \( \frac{dV}{V} = \frac{0.13\pi r^2 h}{\pi r^2 h} = 0.13 \). Hence, the maximum possible percentage error in volume is 13\%.

Key Concepts

Volume of a Cylinder

The volume of a cylinder is an essential formula in geometry and calculus. Calculating the volume helps in understanding how much space is inside the cylindrical shape. For a right circular cylinder, the volume formula is expressed as \( V = \pi r^2 h \). Here, \( \pi \) is a constant (approximately 3.14159), \( r \) is the radius of the cylinder's base, and \( h \) is the height of the cylinder.
To calculate the volume:

• Square the radius (\(r^2\)).
• Multiply by \( \pi \).
• Multiply by the height (\(h\)).

This formula gives the capacity of the cylinder in cubic units. Understanding this calculation is crucial, especially in fields like engineering and manufacturing, where precise volume calculations are required for design purposes.

Percentage Error Calculation

Percentage error is a measure of how inaccurate a measurement is, compared to the true value, and it is often expressed in terms of percentages. In our exercise with the cylinder, the percentage error in the measurement of volume is crucial due to variations in the radius and height.To find the percentage error:

• Determine the error amount (here \( dV \)).
• Calculate the ratio of the error to the original measurement (\( \frac{dV}{V} \)).
• Express this ratio as a percentage by multiplying by 100%.

In this context, by substituting the possible measurement errors into the formula, we observed that the maximum possible percentage error in the volume of the cylinder is 13%. While this value indicates potential deviation, it helps in assessing the reliability of the measurements.

Partial Derivatives in Error Analysis

Partial derivatives are a powerful tool in calculus used to find the rate of change of a function with respect to one of its variables, while keeping the others constant. This is particularly useful in error analysis where multiple variables contribute to an overall measurement.In the exercise's context, to approximate the error in the volume of the cylinder, we differentiated the volume formula with respect to its variables, \( r \) and \( h \):

• The partial derivative of \( V \) with respect to \( r \) is \( \frac{\partial V}{\partial r} = 2\pi rh \).
• With respect to \( h \), it is \( \frac{\partial V}{\partial h} = \pi r^2 \).

These derivatives provide insights into how small changes in the radius and height can affect the overall volume. By using these partial derivatives, we can accurately estimate the corresponding error in volume due to measurement inaccuracies, and thus calculate the maximum possible percentage error. This application highlights how calculus can be applied practically to solve real-world problems.