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 SS using the formula:

S=0.142w0.314h0.79S = 0.142 w^{0.314} h^{0.79}

Given that the errors in the measurements of weight (ww) and height (hh) 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 SS can be approximated using partial derivatives. Taking the natural logarithm of both sides of the equation S=0.142w0.314h0.79S = 0.142 w^{0.314} h^{0.79}, we get:

ln(S)=ln(0.142)+0.314ln(w)+0.79ln(h)\ln(S) = \ln(0.142) + 0.314 \ln(w) + 0.79 \ln(h)

Now, differentiating both sides with respect to ww and hh, we obtain:

dSS=0.314dww+0.79dhh\frac{dS}{S} = 0.314 \frac{dw}{w} + 0.79 \frac{dh}{h}

Step 2: Maximum error approximation

Since the error in both ww and hh is given as 2%, the maximum errors in dww\frac{dw}{w} and dhh\frac{dh}{h} are:

dww=0.02anddhh=0.02\frac{dw}{w} = 0.02 \quad \text{and} \quad \frac{dh}{h} = 0.02

Substitute these values into the equation for dSS\frac{dS}{S}:

dSS=0.314(0.02)+0.79(0.02)\frac{dS}{S} = 0.314(0.02) + 0.79(0.02)

Simplifying:

dSS=0.00628+0.0158=0.02208\frac{dS}{S} = 0.00628 + 0.0158 = 0.02208

Step 3: Express the result as a percentage

The relative error in SS is approximately 0.022080.02208, or 2.208%. Therefore, the maximum error in the calculated surface area is:

2.208%\boxed{2.208\%}

This means that the maximum error in the surface area SS is approximately 2.208%, given that the errors in both ww and hh are at most 2%.

Would you like further details or have any questions?

Here are 5 related questions to explore:

  1. How do you calculate the relative error in more complex models?
  2. What happens if the error in ww and hh differs from each other?
  3. How does the coefficient in the surface area formula affect the error calculation?
  4. What is the significance of the exponents on ww and hh in determining sensitivity to errors?
  5. 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