Problem with Approximations Using Differentials

In summary, the component of a force in a direction to which it makes an angle is 500 nt cos(\frac{\pi}{3})
  • #1
rhdinah
17
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!
 
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  • #2
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.
 
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  • #3
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!
 
  • #4
What is the component of a force in a direction to which it makes an angle ##\theta##?
 
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  • #5
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##
 

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1. What is a problem with approximations using differentials?

The main problem with using differentials for approximations is that they are only accurate for small changes in the independent variable. As the change in the independent variable increases, the error in the approximation also increases.

2. How do differentials work for approximations?

Differentials use the derivative of a function to approximate the change in the value of the function for a small change in the independent variable. This is done by multiplying the derivative by the change in the independent variable, resulting in a linear approximation of the function.

3. Can differentials be used for all functions?

No, differentials can only be used for functions that are continuous and differentiable at the point of approximation. Functions with discontinuities or sharp turns cannot be accurately approximated using differentials.

4. How can I improve the accuracy of my approximations using differentials?

To improve the accuracy of an approximation using differentials, you can decrease the size of the change in the independent variable. This will reduce the error in the approximation, but it may also require more calculations.

5. Are there any other methods for approximating functions?

Yes, there are other methods for approximating functions such as using Taylor series or numerical integration. These methods may be more accurate for larger changes in the independent variable, but they also require more computational resources.

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