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. The estimate of the maximum error in S is:

Solution

To estimate the maximum error in the calculated surface area SS, we need to apply the concept of differentials and relative errors.

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

  • ww is the weight in pounds
  • hh is the height in inches
  • SS is the surface area in square feet

To find the maximum relative error in SS, we'll differentiate the formula for SS with respect to both ww and hh. The general approach is to use the following formula for the differential:

dSSSwdwS+ShdhS\frac{dS}{S} \approx \left| \frac{\partial S}{\partial w} \cdot \frac{dw}{S} \right| + \left| \frac{\partial S}{\partial h} \cdot \frac{dh}{S} \right|

Step 1: Differentiate SS with respect to ww and hh

The partial derivatives of SS are:

Sw=0.1420.314w0.3141h0.79=0.044588w0.686h0.79\frac{\partial S}{\partial w} = 0.142 \cdot 0.314 \cdot w^{0.314 - 1} \cdot h^{0.79} = 0.044588 \cdot w^{-0.686} \cdot h^{0.79}

Sh=0.1420.79w0.314h0.791=0.11218w0.314h0.21\frac{\partial S}{\partial h} = 0.142 \cdot 0.79 \cdot w^{0.314} \cdot h^{0.79 - 1} = 0.11218 \cdot w^{0.314} \cdot h^{-0.21}

Step 2: Compute the relative error in SS

We know that the maximum relative errors in ww and hh are 2%, or 0.02 in decimal form. So, dww=dhh=0.02\frac{dw}{w} = \frac{dh}{h} = 0.02.

Now, the differential dSS\frac{dS}{S} becomes:

dSS0.314dww+0.79dhh\frac{dS}{S} \approx \left| 0.314 \cdot \frac{dw}{w} \right| + \left| 0.79 \cdot \frac{dh}{h} \right|

Substituting dww=0.02\frac{dw}{w} = 0.02 and dhh=0.02\frac{dh}{h} = 0.02:

dSS0.3140.02+0.790.02=0.00628+0.0158=0.02208\frac{dS}{S} \approx 0.314 \cdot 0.02 + 0.79 \cdot 0.02 = 0.00628 + 0.0158 = 0.02208

Step 3: Convert relative error into percentage

The maximum relative error in SS is 0.022080.02208, or approximately 2.21%. Thus, the maximum error in the calculated surface area is 2.21%.

Final Answer:

The estimate of the maximum error in SS is 2.21%.


Do you want more details or have any questions?

Here are 5 related questions:

  1. How is the formula for relative error derived in differentials?
  2. What happens if the relative errors in ww and hh are different?
  3. How would you calculate the error in volume for a similar problem?
  4. How does the exponent affect the sensitivity to error in this case?
  5. Can we calculate the absolute error instead of relative error?

Tip: When using differentials to estimate error, the relative error in a function depends on the exponents and coefficients of the variables in the equation.

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Math Problem Analysis

Mathematical Concepts

Differentials
Error Estimation
Surface Area Formula

Formulas

S = 0.142w^{0.314}h^{0.79}
Relative Error Formula: dS/S ≈ |∂S/∂w * dw/w| + |∂S/∂h * dh/h|
Partial Derivative with respect to w: ∂S/∂w = 0.044588 * w^{-0.686} * h^{0.79}
Partial Derivative with respect to h: ∂S/∂h = 0.11218 * w^{0.314} * h^{-0.21}

Theorems

Differentials and Relative Error Theorem

Suitable Grade Level

Grades 11-12 (High School Calculus)