Math Problem Statement
A model for the surface area of some solid object is given by S=0.142w^(0.314)h^(0.79) , where w is the weight (in pounds), h is the height (in inches), and S is measured in square feet. If the errors in measurements of w and h are at most 2% , estimate the maximum error in the calculated surface area.
Solution
We are asked to estimate the maximum error in the calculated surface area using the formula:
Given that the errors in the measurements of weight () and height () are at most 2%, we need to estimate the maximum relative error in the surface area.
Step 1: Differentiate the equation
The relative error in can be approximated using partial derivatives. Taking the natural logarithm of both sides of the equation , we get:
Now, differentiating both sides with respect to and , we obtain:
Step 2: Maximum error approximation
Since the error in both and is given as 2%, the maximum errors in and are:
Substitute these values into the equation for :
Simplifying:
Step 3: Express the result as a percentage
The relative error in is approximately , or 2.208%. Therefore, the maximum error in the calculated surface area is:
This means that the maximum error in the surface area is approximately 2.208%, given that the errors in both and are at most 2%.
Would you like further details or have any questions?
Here are 5 related questions to explore:
- How do you calculate the relative error in more complex models?
- What happens if the error in and differs from each other?
- How does the coefficient in the surface area formula affect the error calculation?
- What is the significance of the exponents on and in determining sensitivity to errors?
- Can we find a general formula for maximum error estimation for any power-law relationship?
Tip: When estimating maximum error, small percentage errors in input values lead to approximations in the overall output using differential methods like partial derivatives.
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Math Problem Analysis
Mathematical Concepts
Differentiation
Error Analysis
Relative Error
Formulas
S = 0.142w^(0.314)h^(0.79)
dS/S = 0.314(dw/w) + 0.79(dh/h)
Theorems
Approximation of Relative Error using Partial Derivatives
Suitable Grade Level
Undergraduate Calculus
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