To solve this problem, we use the concept of linear approximation (also called the differential approximation) to estimate the maximum error in the volume of a cylinder due to potential errors in the radius and height.

Formula for the Volume of a Cylinder

The volume $V$ of a cylinder is given by: $V = \pi r^2 h$ where:

• $r$ is the radius,
• $h$ is the height,
• $\pi \approx 3.1416$.

Step 1: Find the Differential of the Volume

We need to find the differential $dV$ of the volume function $V = \pi r^2 h$. To do this, we use partial derivatives with respect to $r$ and $h$.

The total differential $dV$ is given by: $dV = \frac{\partial V}{\partial r} dr + \frac{\partial V}{\partial h} dh$

First, we compute the partial derivatives: $\frac{\partial V}{\partial r} = 2\pi r h \quad \text{and} \quad \frac{\partial V}{\partial h} = \pi r^2$

Thus, the total differential is: $dV = 2\pi r h \, dr + \pi r^2 \, dh$

Step 2: Estimate the Maximum Error

Given that the maximum errors in $r$ and $h$ are both 8%, we can express these errors as: $dr = 0.08r \quad \text{and} \quad dh = 0.08h$

Substitute these into the differential: $dV = 2\pi r h (0.08r) + \pi r^2 (0.08h)$ Simplifying: $dV = 0.16\pi r^2 h + 0.08\pi r^2 h$ $dV = 0.24\pi r^2 h$

Step 3: Calculate the Approximate Maximum Error in the Volume

Now, substitute the given values $r = 3.8 \, \text{m}$ and $h = 6.3 \, \text{m}$ into the equation for $dV$:

$dV = 0.24 \pi (3.8)^2 (6.3)$

First, compute $r^2$ and then multiply by $h$: $r^2 = (3.8)^2 = 14.44$ $r^2 h = 14.44 \times 6.3 = 90.972$

Now, calculate the error: $dV = 0.24 \times \pi \times 90.972 \approx 0.24 \times 3.1416 \times 90.972 \approx 65.044$

Thus, the approximate maximum error in the volume is: $dV \approx 65.04 \, \text{m}^3$

Final Answer:

The maximum error in the volume of the cylinder, based on the 8% possible errors in both radius and height, is approximately 65.04 cubic meters.

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Would you like a detailed explanation of any part of this process or have other questions? Here are some related questions you might find helpful:

1. How do you apply linear approximation to other geometric shapes?
2. How do you calculate the actual percentage error in the volume of a cylinder?
3. How does the size of the error in radius and height affect the overall volume error?
4. What are the advantages of using linear approximation in practical applications?
5. How can you extend this approach to multi-variable functions with more than two variables?

Tip: When dealing with measurements and errors, remember that the linear approximation works well when the errors are small. Larger errors may require higher-order approximations.