- #1
agnimusayoti
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- Homework Statement
- A force of 500 N is measured with a possible error of 1 N. 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 is (approximately) the largest possible error in the component?
- Relevant Equations
- For ##f(F,\theta) \rightarrow dF= \frac{\partial f}{\partial F} dF +\frac {\partial f}{\partial \theta} d\theta ##
From the question,
$$f(F,\theta)=F \cos \theta$$
1. If I use:
$$df=dF \cos{\theta} -F \sin {\theta} d\theta$$
and using radian,
$$df=dF \cos{\theta} -F \sin {\theta} d\theta \frac {\pi}{180^\circ}=5.28 N$$
2. But, if I take logarithm to both side:
$$ln f=ln F+ln \cos{\theta}$$
differentiate both sides:
$$\frac{df}{f}=\frac{dF}{F} + \frac{\sin\theta}{\cos \theta} d\theta$$
Using radian, it gives ##df=4.28 N##My question is, why the answers are different? Thanks.
$$f(F,\theta)=F \cos \theta$$
1. If I use:
$$df=dF \cos{\theta} -F \sin {\theta} d\theta$$
and using radian,
$$df=dF \cos{\theta} -F \sin {\theta} d\theta \frac {\pi}{180^\circ}=5.28 N$$
2. But, if I take logarithm to both side:
$$ln f=ln F+ln \cos{\theta}$$
differentiate both sides:
$$\frac{df}{f}=\frac{dF}{F} + \frac{\sin\theta}{\cos \theta} d\theta$$
Using radian, it gives ##df=4.28 N##My question is, why the answers are different? Thanks.
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