Error estimation using differentials

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The discussion revolves around calculating the error in the component of a force measured at 500N with a potential error of 1N and an angle of 60° with an error of 0.5°. The participant initially applied differentials incorrectly in their calculations, leading to an estimated error of 3.75N instead of the correct value of 4.28N. The correct differential formula involves the angle's derivative being applied correctly to account for the sine and cosine relationships. The error was clarified through peer feedback, emphasizing the importance of accurate differentiation in error estimation.
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Homework Statement



A force of 500N is measured with a possible error of 1N. Its component in a direction 60° away from its line of action is required, where the angle is subject to an error of 0.5°. What (approximately) is the largest possible error in the component?

Homework Equations





The Attempt at a Solution



The component force is F_x = F cos \theta

so lnF_x~=~lnF~+~lncos\theta

applying differentials: \frac{dF_x}{F_x} = \frac{dF}{F} + \frac{d~cos\theta}{cos\theta} (-sin\theta)=\frac{dF}{F} + \frac{sin^{2} \theta}{cos\theta}d\theta

plugging in values \frac{dF_x}{F_x} = \frac{1}{500} + \frac{3}{4} \frac{2}{1} \frac{1}{2} \frac{\pi}{180} = 0.002 + 0.013 = 0.015
so the error is (0.015)(500)cos(60) = 3.75N

The solution says 4.28N, however, which I confirmed by checking each error combination. Where am I going wrong here?

Thanks in advance.
 
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You differentiated incorrectly. You should have
$$\frac{dF_x}{F_x} = \frac{dF}{F} + \frac{d\theta}{\cos\theta}(-\sin\theta).$$
 
vela said:
You differentiated incorrectly. You should have
$$\frac{dF_x}{F_x} = \frac{dF}{F} + \frac{d\theta}{\cos\theta}(-\sin\theta).$$


Gotcha, I figured it would be something simple like that! Thanks a lot!
 
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