Problem with Approximations Using Differentials

  • Thread starter Thread starter rhdinah
  • Start date Start date
  • Tags Tags
    Differentials
rhdinah
Messages
17
Reaction score
1

Homework Statement


I am working on this problem and having difficulty getting the required answer. It is the exact problem as here , but I’m still not getting it.

BTW this is problem 10, Section 4, Chapter 4 Partial Differentiation from M. Boas’s book Mathematical Methods in the Physical Sciences, 3rd edition.

A force 500 nt is measured with a possible error of 1 nt. 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?

Homework Equations

The Attempt at a Solution


We are going to use differentials here.

##Component = F_1*sin(\theta)##

##\frac{d Component}{Component} = \frac {d F_1}{F_1} + \frac {d sin(\theta)}{sin(\theta)} d \theta##

##\frac{d Component}{Component} = \frac {d F_1}{F_1} + \frac {cos(\theta)}{sin(\theta)} d \theta##

##Largest \left|\frac{d Component}{Component}\right| = \left|\frac {d F_1}{F_1}\right| + \left|\frac {cos(\theta)}{sin(\theta)} d \theta\right| = \frac{1}{500}+\frac{cos(\frac{\pi}{3})}{sin(\frac{\pi}{3})} * \frac{0.5}{60} = .002 + cot(\frac{\pi}{3}) * \frac{0.5}{60} = .002 + 0.00481 = 0.00681##

So the ##Component = 500 sin(\frac{\pi}{3}) = 433##

##Component Error = 433*0.00681 = 2.95## … which is not the correct answer of 4.28 nt

So if someone can direct me to my error I'd certainly appreciate it! Thank you!
 
Physics news on Phys.org
You have used ##d\theta/\theta## instead of ##d\theta## in your computation. Note that you also need to put it in radians based on how you treated the derivative.
 
  • Like
Likes rhdinah
Thank you Orodruin! Let me put your ideas in motion!

##Largest \left|\frac{d Component}{Component}\right| = \left|\frac {d F_1}{F_1}\right| + \left|\frac {cos(\theta)}{sin(\theta)} d \theta\right| = \frac{1}{500}+\frac{cos(\frac{\pi}{3})}{sin(\frac{\pi}{3})} * \frac{0.5\pi}{180} = .002 + cot(\frac{\pi}{3}) * \frac{0.5\pi}{180} = .002 + 0.00504 = 0.00704##

So the ##Component = 500 sin(\frac{\pi}{3}) = 433##

##Component Error = 433*0.00704 = 3.05## … which is still not the correct answer of 4.28 nt

So I suspect there's something seriously wrong with my approach ... my equation ... any further ideas that could lead me to a correct solution please? :-) Thank you!
 
What is the component of a force in a direction to which it makes an angle ##\theta##?
 
  • Like
Likes rhdinah
Yes, thank you so much! I looked at my original diagram and realized that I had the trig wrong! Shame on me! I was mislead a bit by the ref thread. I should have trusted my original diagram and not listened to the banter on that thread.
IMG_0248.jpg

To correct this:
##Largest \left|\frac{d Component}{Component}\right| = \left|\frac {d F_1}{F_1}\right| + \left|\frac {sin(\theta)}{cos(\theta)} d \theta\right| = \frac{1}{500}+\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})} * \frac{0.5\pi}{180} = .002 + tan(\frac{\pi}{3}) * \frac{0.5\pi}{180} = .002 + 0.0151 = 0.0171##

So the ##Component = 500 nt\hspace{.1cm} cos(\frac{\pi}{3}) = 250 nt##

##Component Error = 250 nt*0.0171 = 4.28 nt## … which is the correct answer.##\hspace{1cm}\square##
 

Attachments

  • IMG_0248.jpg
    IMG_0248.jpg
    28.8 KB · Views: 489
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top