October 10th, 2017, 02:43 AM  #1 
Member Joined: Dec 2015 From: England Posts: 30 Thanks: 0  Percentage errors
Please could someone help me with this question

October 10th, 2017, 04:03 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
What, exactly, is your difficulty? A "Taylor's series expansion" is just an expansion in powers of the variable. r and l are already powers so the only "Taylor's series expansion" needed is for $\displaystyle \cot(\alpha)$ about $\displaystyle \pi/4$. $\displaystyle \cot(\pi/4)= 1$. The derivative of cot is $\displaystyle \csc^2$ and $\displaystyle \csc^2(\pi/4)= \sqrt{2}$. "Neglecting second order terms" the Taylor's series for $\displaystyle \cot(\alpha)$ about $\displaystyle \pi/4$ is $\displaystyle 1+ \sqrt{2}(\alpha \pi/4)$. So the "Taylor's series expansion" for this, "neglecting second order terms", is $\displaystyle \pi r^2l+ \frac{1}{3}\pi r^3(1+ \sqrt{2}(\alpha \pi/4)$. At r= 1, l= 3, $\displaystyle \alpha= \pi/4$, that is $\displaystyle 3\pi$. If you want a very precise result, calculate that when r= 1.05, l= 3.03, and $\displaystyle \alpha= \pi/4+ \pi/40= 11\pi/40$. Then calculate the error, the difference between the two and divide by the actual value, $\displaystyle 3\pi$ to find the percentage error. You can get a good estimate using an "engineer's rule of thumb", that when you add quantities, you add the errors, and when you multiply quantities, you add the percentage error. Last edited by skipjack; October 10th, 2017 at 04:44 AM. 

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