Math Problem Statement

The two shorter sides (i.e. not the hypotenuse) of a right angle triangle are measured to be (5±0.1)

cm and (12±0.2)

cm, where ±0.1

and ±0.2

represent the size of the errors made in each measurement.

If we use values 5cm and 12cm as the side lengths, the area of the triangle and the length of the hypotenuse can be computed. Use a linear approximation to estimate the maximum error made in computing the area and hypotenuse length based on the errors in the side length measurements.

Error in hypotenuse length (enter a positive value rounded to two decimal places)

Answer = ?

Solution

To estimate the error in the hypotenuse length using a linear approximation, we need to apply differential calculus to the function for the hypotenuse of a right triangle.

Step 1: Hypotenuse Formula

The hypotenuse cc of a right triangle with side lengths aa and bb is given by the Pythagorean theorem: c=a2+b2c = \sqrt{a^2 + b^2} Here, a=5cma = 5 \, \text{cm} and b=12cmb = 12 \, \text{cm}.

Step 2: Calculate the Hypotenuse

Using the given side lengths a=5cma = 5 \, \text{cm} and b=12cmb = 12 \, \text{cm}: c=52+122=25+144=169=13cmc = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, \text{cm}

Step 3: Error in the Hypotenuse

The error in the hypotenuse Δc\Delta c is approximated by differentiating the hypotenuse formula with respect to aa and bb: dc=cada+cbdbdc = \frac{\partial c}{\partial a} da + \frac{\partial c}{\partial b} db The partial derivatives are: ca=aa2+b2,cb=ba2+b2\frac{\partial c}{\partial a} = \frac{a}{\sqrt{a^2 + b^2}}, \quad \frac{\partial c}{\partial b} = \frac{b}{\sqrt{a^2 + b^2}} Now, plugging in the values for a=5a = 5, b=12b = 12, and c=13c = 13: ca=513,cb=1213\frac{\partial c}{\partial a} = \frac{5}{13}, \quad \frac{\partial c}{\partial b} = \frac{12}{13}

The errors in aa and bb are given as Δa=0.1cm\Delta a = 0.1 \, \text{cm} and Δb=0.2cm\Delta b = 0.2 \, \text{cm}. So the error in the hypotenuse is: Δc513×0.1+1213×0.2\Delta c \approx \frac{5}{13} \times 0.1 + \frac{12}{13} \times 0.2 Now, let's calculate this.

Step 4: Calculation of Hypotenuse ErrorThe estimated maximum error in the hypotenuse length is approximately 0.22 cm (rounded to two decimal places).

Would you like more details or have any questions?

Here are some related questions you could explore:

  1. How do you estimate the error in the area of the triangle?
  2. What are the applications of error propagation in physics and engineering?
  3. How would the maximum error change if the measurement errors were smaller?
  4. How does this approximation compare with more advanced error analysis techniques?
  5. What is the role of partial derivatives in error propagation?

Tip: When calculating errors, always double-check that the units and rounding conventions are consistent to avoid misinterpretation.

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Math Problem Analysis

Mathematical Concepts

Error Propagation
Differential Calculus
Pythagorean Theorem

Formulas

Hypotenuse formula: c = √(a² + b²)
Error in hypotenuse: Δc ≈ (∂c/∂a)Δa + (∂c/∂b)Δb
Partial derivatives: ∂c/∂a = a / √(a² + b²), ∂c/∂b = b / √(a² + b²)

Theorems

Pythagorean Theorem
Error Propagation

Suitable Grade Level

College Level (Calculus)