# Thread: [SOLVED] error propagation: the approximation formula

1. ## [SOLVED] error propagation: the approximation formula

I can't seem to get the right answer to this one, and I hope someone can tell me what I am doing wrong!

"You measure the radius of a circle to be 12 cm and use the formula A = pi(r^2) to calculate the area. If your measurement of the radius is accurate to within 3%, approximately how accurate (to the nearest percent) is your calculation of the area?

$\displaystyle A(r) = \pi r^2$
$\displaystyle A'(r) = 2\pi r$
$\displaystyle \left|\frac{dA}{A}\right|= \left|\frac{2 \pi r dr}{ \pi r^2}\right|$
$\displaystyle \left|\frac{dA}{A}\right| = 2\left|\frac{dr}{r}\right|$
$\displaystyle \left|\frac{dA}{A}\right| = 2\left|\frac{\pm0.03(10)}{12}\right|$
$\displaystyle \left|\frac{dA}{A}\right| = 0.05$

The problem is that the answer given in the back of the book is 0.06. So, where did I mess up? Thanks for the help!

2. Hi
Originally Posted by sinewave85
I can't seem to get the right answer to this one, and I hope someone can tell me what I am doing wrong!

"You measure the radius of a circle to be 12 cm and use the formula A = pi(r^2) to calculate the area. If your measurement of the radius is accurate to within 3%, approximately how accurate (to the nearest percent) is your calculation of the area?

$\displaystyle A(r) = \pi r^2$
$\displaystyle A'(r) = 2\pi r$
$\displaystyle \left|\frac{dA}{A}\right|= \left|\frac{2 \pi r dr}{ \pi r^2}\right|$
$\displaystyle \left|\frac{dA}{A}\right| = 2\left|\frac{dr}{r}\right|$
Taking it from here, the relative error of $\displaystyle r$ is $\displaystyle \left|\frac{\mathrm{d}r}{r}\right|=0.03$ so $\displaystyle \left|\frac{\mathrm{d}A}{A}\right|=2\times 0.03=0.06$. Be careful to make a difference between absolute error ( $\displaystyle |\mathrm{d}r|$ ) and relative error. $\displaystyle \left( \left|\frac{\mathrm{d}r}{r}\right| \right)$

### if your measurement of the radius of a circle is accurate to within 3%

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