MHB Help solving Conditioning problem

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The discussion focuses on the numeric conditioning of the area of a triangle calculated using the formula S = 1/2 ab sin(γ). Participants explore how errors in the variables a, b, and γ affect the area S, specifically through the propagation of absolute and relative errors. The conversation highlights that the relative error in S can be expressed in terms of the contributions from each variable, with specific formulas provided for each. Additionally, it notes that for small angles γ, the relative error can be approximated. Overall, the discussion emphasizes understanding error amplification in the context of triangle area calculations.
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Hi,
I have the following problem : Area of a triangle is given by S = 1/2 ab sin(γ) (See figure).
Discuss numeric conditioning of S. Any tips appreciated :D
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natalia said:
Hi,
I have the following problem : Area of a triangle is given by S = 1/2 ab sin(γ) (See figure).
Discuss numeric conditioning of S. Any tips appreciated :D
https://www.physicsforums.com/attachments/2493

Hi natalia! ;)

I'm not quite sure what is intended with numeric conditioning...
Can you clarify?

What I can imagine is that we'd like to know how errors in $a, b, γ$ propagate.
The general formula for the absolute error is:
$$\Delta y \approx \frac{dy}{dx} \Delta x$$
For the relative error (as you might deduce) it is:
$$\frac{\Delta y}{y} \approx \frac x y \frac{dy}{dx} \frac{\Delta x} x$$

In your problem you would get:
For the contribution of $\Delta a$: $\frac{\Delta S}S = \frac{\Delta a}a$
For the contribution of $\Delta b$: $\frac{\Delta S}S = \frac{\Delta b}b$
For the contribution of $\Delta γ$: $\Delta S = \frac 1 2 a b \cos γ \Delta γ$
 
Yes, it refers to amplification of the relative error.
 
natalia said:
Yes, it refers to amplification of the relative error.

Aha! :D

Then, to complete it, we have for $Δγ$:

$$\frac{ΔS}{S}
= \frac{\frac 12 ab \cos γΔγ}{\frac 12 ab \sin γ}
= γ\cot γ \cdot \frac{Δγ}{γ}
$$

For small angles $γ$ this is approximately $\frac{Δγ}{γ}$.
 
Thank you again, I like Serena :)
 
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